QUADRATIC FUNCTIONS: Problem Set 1 NOTES: The motion of projectiles can be represented by a quadratic equation. Here are two examples of this kind of motion: - A Ball being thrown upwards into the air. - An aircraft dropping a watermelon. *** You are going to plot a graph to show the motion of a cannonball being fired into the air at an angle of 45 degrees. The equation for the motion is: 2 y = ax + bx Where a = -1/500 and b = 1. 2 In the equation y = (-1/500)x + x, y represents the distance in meters and x is the time in seconds. The equation describing the motion of a projectile is a quadratic equation. The graph of a quadratic equation is a parabola. The value of y rises to a maximum and then falls back to zero. WORKSPACE: ANS: 45 , 80 , 105 , 120 , 125 , 120 , 105 , 80 , 45 , 0 SOLVING QUADRATIC EQUATIONS: Factoring - Solve quadratic equations by factoring and using the zero-product property NOTES: In past activities we factored and graphed equations of the form: 2 y = ax + bx + c This activity focuses on the particular case when y = 0. That is, we will learn to solve equations of the form: 2 0 = ax + bx + c To do this, we will use factoring. *** Let's see how to graph the following equations: 2 y = x - 5x - 14 2 x|y y = (-3) - 5(-3) - 14 --+-- 9 + 15 - 14 = 10 -3|10 -2|0 0|-14 *** Before pressing ENTER, describe in your own words the points where the graph crosses the x-axis. x|y --+-- -3|4 -2|0 0|-2 1|0 The points where the graph crosses the x-axis are the points where y = 0. *** Before pressing ENTER, describe in your own words the solutions to equations like 0 = x^2 - 7x + 12. The solutions to the equation are the x-values of the points where the graph crosses the x-axis. *** It is not always practical to graph an equation to determine the solutions when y = 0. In the next part of the lesson, algebraic procedures will be used to find solutions to equations like: 2 ax + bx + c = 0 Without graphing. First, let's review a few facts that will be needed. *** Complete the folling products: 3 * _ = 0 _ * 5 = 0 ? * ? = 0 Notice that the last product is correct for any pair of numbers as long as one of the numbers is zero. *** Before pressing ENTER, describe in your own words when: a * b = 0 Either a or b (or both) must be zero. This is known as the zero-product property. It also will be true that: (x - 3)(x + 5) = 0 If either: x - 3 = 0 OR x + 5 = 0 So x = 3 or x = -5. *** Before pressing ENTER, describe in your own words when: (ax + b)(cx + d) = 0 It will be true if either: ax + b = 0 or cx + d = 0 *** Notice what happens when we factor the following equation: 2 Start with... 3x - x = 0 Then factor... x(3x - 1) = 0 Then either... x = 0 OR 3x - 1 = 0 So either... x = 0 OR x = 1/3 Here are some more examples of solving equations: 2 4x - 9 = 0 (2x + 3)(2x - 3) = 0 x = -3/2 OR 3/2 And: 3 2 3x - 6x - 24x = 0 2 3x(x - 2x - 8) = 0 3x(x - 4)(x + 2) = 0 x = 0, 4 or -2 *** Before pressing ENTER, describe in your own words a way to solve equations of the form: 2 ax + bx + c = 0 Factor the equation, set each factor equal to zero and solve the resulting equations. WORKSPACE: 2 8x + 5x = 0 x(8x + 5) = 0 0 OR -5/8 *** 2 x + 8x + 16 = 0 -4 *** 2 x - 2x - 15 = 0 -3, 5 ANS: 0 , -14 , -18 , -18 , -14 , 0 , 4 , 0 , -2 , -2 , 0 , 4 , 10 , 0 , 0 , zero , 0 , 0 ,-9 , 5/2 , 0 , 0 , 0 , 4/3 , -3/2 , 0 -2 , -1/2 3 , 6/5 -4 , 7/3 -9/2 , 0 -2 5 , -3/2 5/4 , 0 -5/8 , -5 5 , -4 , 6 , -3 5 , 1 -1 , 3/2 -9 , ... , 7 3 , -9/5 -1 , 1/3 7 , -4 5 , 1/4 -1/4 , 8 2 , 10/3 -10/3 SOLVING QUADRATIC EQUATIONS: Completing the Square 2 - Solve equations of the form a(x + b) = c. - Solve quadratic equations by completing the square. NOTES: In the last activity we solved quadratic equations by factoring. 2 x + 7x + 12 = 0 (x + 3)(x + 4) = 0 However, not all equations are easily factored. We need some new methods to help us solve quadratics. The first method we will learn is called COMPLETING THE SQUARE. *** Sometimes our solutions will involve irrational numbers like: _ _ _ \/5 2\/3 3 + \/7 Let's see how we get these solutions! Look at this equation: 2 x = 9 If we take the square root of both sides, we find the two solutions. __ / 2 _ \/ x = \/9 |X| = 3 x = 3, -3 Here are some other examples: 2 x = 7 _ _ x = \/7, -\/7 And: 2 4x = 25 2 25 x = -- 4 5 -5 x = -, -- 2 2 *** Before pressing ENTER, describe in your own words how we solve the following quadratic equation: 2 ax = b Divide both sides by a and then take the square root of each side: 2 ax = b 2 b x = - a _ _ /b /b / - , - / - x = \/ a \/ a *** We can use this method to solve the following quadratic equations: 2 If, (x -4) = 36 Then, (x - 4) = 6 OR -6 Here are some more examples: 2 3(x - 6) = 75 2 (x - 6) = 25 (x - 6) = 5 OR -5 Here are some more examples: 2 (x - 2) = 5 _ _ (x - 2) = \/5 OR -\/5 _ _ x - 2 = \/5 OR x - 2 = -\/5 _ _ x = 2 + \/5 OR x = 2 - \/5 *** Before pressing ENTER, use your own words to describe how we solve the following quadratic equation: 2 a(x + b) = c First we divide by a. Next take the square root of both sides. Then we solve the resulting equations. 2 a(x + b) = c 2 c (x + b) = - a _ _ /c /c / - , - / - (x + b) = \/ a \/ a _ _ /c /c / - OR / - x = -b + \/ a x = -b - \/ a The solution steps we have been using don't always work well because some equations like: 2 ax + bx + c = 0 Are not easy to factor. To solve any equations of this form we will use two facts about PERFECT SQUARE TRINOMIALS. 2 2 2 (x + b) = x + 2bx + b 2 2 2 (x - b) = x - 2bx + b Notice how we solve the following equation: 2 x + 6x + 7 = 0 Since we can't factor this equation let's change it to a Perfect Square Trinomial which we can factor... 2 x + 6x + 7 = 0 [ 2 / x + 6x = -7 | \ 2 [ x + 6x + 9 = -7 + 9 2 (x + 3) = 2 _ _ (x + 3) = \/2 OR -\/2 _ _ x + 3 = \/2 OR x + 3 = -\/2 _ _ x = -3 + \/2 OR x = -3 - \/2 The procedure in brackets is a process called Completing the Square. Here are some more examples of completing the square: 2 x - 8x + 4 = 0 2 x - 8x = -4 2 x - 8x + 16 = -4 + 16 2 (x - 4) = 12 _ _ x - 4 = 2\/3 OR -2\/3 _ x = 4 +/- 2\/3 And: 2 x + 7x + 2 = 0 2 x + 7x = -2 2 49 -8 49 x + 7x + -- = -- + -- 4 4 4 2 41 (x + 7/2) = -- 4 WORKSPACE: 2 16x = 9 x = 3/4, -3/4 "This is correct because 16(3/4)(3/4) = 9 and 16(-3/4)(-3/4) = 9. *** ANS: 4 -4 , 10 -10 , 3\/2 -3\/2 , 2/3 -2/3 , 3/4 -3/4 , 5/7 -5/7 , \/3 -\/3 , 10 , -2 , 11 , 1 , 12 -2 , -6 + \/11 -6 - \/11 , -2 + \/3 -2 - \/3 , 5 -4 , 3 + \/7 3 - \/7 , 5 + \/7 5 - \/7 , 7/3 -5 , 1 + \/3 1 - \/3 The above is continued. Current Date: February 28, 2006 NOTES: Here are some more examples of completing the square: 2 x - 8x + 4 = 0 2 x - 8x = -4 2 x - 8x + 16 = -4 + 16 2 (x - 4) = 12 _ _ x - 4 = 2\/3 OR -2 \/3 _ x = 4 +/- 2\/3 And: 2 x + 7x + 2 = 0 2 x + 7x = -2 2 49 -8 49 x + 7x + -- = -- + -- 4 4 4 2 41 (x + 7/2) = -- 4 __ __ \/41 \/41 x + 7/2 = ---- OR - ---- 2 2 __ -7 +/- \/41 x = ----------- 2 *** Before pressing ENTER, describe in your own words one way to solve polynomials of this form: 2 x + bx + c = 0 Complete the square on the left side: 2 2 2 b b x + bx + -- = -c + -- 4 4 factor, 2 b 2 b (x + -) = -c + -- 2 4 Then take the square root of both sides and solve. *** Let's see how the transformation for solving an equation by completing the square affect the graph of the equation. 2 x - (3/2)x - 1 = 0 2 x - (3/2)x = 1 2 x - (3/2)x + 9/16 = 1 + 9/16 2 (x - 3/4) = 25/16 x - 3/4 = +/- 5/4 x - 3/4 = 5/4 OR x - 3/4 = - 5/4 x = 2 x = -1/2 Here is another graphic example of solving equations: 2 x + 2x - 1 = 0 2 x + 2x = 1 2 x + 2x + 1 = 1 + 1 2 (x + 1) = 2 _ x + 1 = +/- \/2 _ _ x + 1 = \/2 OR x + 1 = -\/2 _ _ x = -1 + \/2 x = -1 - \/2 ~ ~ x = .41 x = -2.41 *** Before pressing ENTER, describe in your own words how transformations of equations affect the solutions to the equations. As we have seen, the solutions are not affected, but always lie on the same lines. WORKSPACE: 2 x - 4x + 1 = 0 2 x - 4x = -1 2 x - 4x + 4 = -1 + 4 2 (x - 2) = 3 _ _ x - 2 = \/3 OR \/3 _ _ x = 2 + \/3, 2 - \/3 *Why do they pick that certain number for completing the square? I'm not sure... I'll have to figure that out* *** 2 x + 2x - 1 = 0 2 x + 2x = 1 2 x + 2x + 1 = 1 + 1 2 (x + 1) = 2 _ _ x + 1 = \/2 OR -\/2 _ x = -1 + \/2 *** 2 x + 3x - 8 = 0 2 x + 3x = -8 2 x + 3x + 9/4 = 41/4 2 (x + 3/2) = 41/4 __ __ x + 3/2 = \/41/2 OR -\/41/2 __ __ x = (-3 + \/41)/2, (-3 - \/41)/2 ANS: 2 + \/3 2 - \/3 , -1 + \/2 -1 - \/2 , (-3 + \/41)/2, (-3 - \/41)/2 , 9 -9 , 3/7 -3/7 , 2\/2 -2\/2 , 11 3 , -2 -8 , 9/2 -3/2 , -3 + \/2 -3 - \/2 , 4 + \/7 4 - \/7 , (-3 + \/33)/2 (-3 - \/33)/2 SOLVING QUADRATIC EQUATIONS: The Quadratic Formula 2 - Solve equations of the form ax + bx + c = 0 using the quadratic formula. NOTES: In the last activity we learned the method of completing the square which can use to solve all quadratic equations of the form: 2 ax + bx + c = 0 In this activity we will learn a "shortcut" for solving these same quadratic equations. This shortcut is the Quadratic Formula. *** Here is a quadratic equation: 2 3x + 2x + 1 = 0 Written in the form: 2 ax + bx + c = 0 In the above equation, a = 3 b = 2 c = 1 *** 2 ax + bx + c = 0 Determine the coefficients a, b, and c in the following quadratic equation: 2 4x + 6x + 2 = 0 a = 4, b = 6, c = 2 *** Using algebraic transformations, and completing the square, we can solve the following quadratic equation for x: 2 / ax + bx + c = 0 a = 0 The result is: _______ / 2 -b +/- \/ b - 4ac / x = ----------------- a = 0 2a This is a formula for finding the solution of all quadratic equations. It is called the quadratic formula. We will derive this formula in Algebra II. *** _______ / 2 -b +/- \/ b - 4ac x = ----------------- 2a The way we read the quadratic formula is: "x equals the opposite of b plus or minus the square root of b squared minus 4ac, all divided by 2a." *** _______ / 2 -b +/- \/ b - 4ac x = ----------------- 2a Let's see how the quadratic formula works. 2 x - 3x + 2 = 0 a = 1 b = -3 c = 2 _____________ / 2 -(-3) +/- \/(-3) - 4(1)(2) x = -------------------------- 2(1) _ 3 +/- \/1 = --------- 2 3 +/- 1 = ------- 2 So, 3 + 1 3 - 1 x = ----- OR ----- = 2 OR 1 2 2 *** Let's solve the same equation by completing the square. 2 x - 3x + 2 = 0 2 x - 3x = -2 2 x - 3x + (9/4) = -2 + (9/4) 2 [x - (3/2)] = 1/4 x - (3/2) = +/- 1/2 x - (3/2) = 1/2 OR x - (3/2) = -1/2 x = 4/2 = 2 x = 2/2 = 1 Notice that we get the same answers. *Huh? They look different to me...* *** Let's do an equation together... _______ / 2 -b +/- \/ b - 4ac x = ----------------- 2a 2 2x + 6x + 4 = 0 a = 2 b = 6 c = 4 ____________ / 2 -(6) +/- \/(6) - 4(2)(4) x = ------------------------ 2(2) _ -6 +/- \/4 x = ---------- 4 -6 + 2 -6 - 2 x = ------ OR ------ 4 4 x = -1 OR -2 *** Before pressing ENTER, use your own words to describe how we use the quadratic formula to solve equations like: 2 x - 5x + 9 = 0 Substitute 1, -5, and 9 for a, b, c in the formula, then simplify. _______ / 2 -b +/- \/ b - 4ac x = ----------------- 2a *** WORKSPACE: a = 1 b = 6 c = -16 _______ / 2 -6 +/- \/ 6 - 4(-16) x = -------------------- 2 ___ -6 +/- \/100 ------------ 2 -6 +/- 10 --------- 2 2 OR -8 *** a = 1 b = 7 c = -3 ______ / 2 -7 +/- \/ 7 + 12 x = ---------------- 2 __ -7 +/- \/61 ----------- 2 *** a = 3 b = 35 c = 50 ________ / 2 -35 +/- \/ 35 - 600 x = ------------------- 6 35 +/- 25 --------- 6 *** a = 5 b = 1 c = -18 _______ / 2 -1 +/- \/ 1 + 360 x = ----------------- 10 -1 +/- 19 --------- 10 9/5 OR -2 *** a = 1 b = 11 c = 24 _______ / 2 11 +/- \/ 11 - 96 x = ----------------- 2 11 +/- 5 -------- 2 3 OR 8 *** a = 6 b = 17 c = 7 ________ / 2 -17 +/- \/ 17 - 168 x = ------------------- 12 -17 +/- 11 ---------- 12 -1/2 OR -7/3 *** a = 5 b = 29 c = -6 _______ / 2 -b +/- \/ b - 4ac x = ----------------- 2a ANS: 4 6 2 , 3 -2 5 , 2 1/2 3 , 1 -4 -7 , 1 2 -4 , 2 6 4 , 6 , 6 , 2 , 4 , 2 , 4 , 2 , 2 , -1 , -2 , 1 -6 , 2 -8 , -1/2 1/3 , -3 + \/5 -3 - \/5 , 5/2 -7/3 , (-7 + \/61)/2 (-7 - \/61)/2 , 5/3 10 , 9/5 -2 , 3 8 , -1/2 -7/3 The above is continued. Current Date: March 6, 2006 NOTES: This is the format: 2 ax + bx + c = 0 _______ / 2 -b +/- \/ b - 4ac x = ----------------- 2a WORKSPACE: a = 5 b = 29 c = -6 ________ / 2 -29 +/- \/ 29 + 120 x = ------------------- 10 -29 +/- 31 ---------- 10 *** a = 2 b = 19 c = 35 ________ / 2 -19 +/- \/ 19 - 280 x = ------------------- 4 -19 +/- 9 --------- 4 *** a = 1 b = -7 c = 12 _______ / 2 7 +/- \/ -7 - 48 x = ---------------- 2 7 +/- 1 ------- 2 *** a = 28 b = -1 c = -15 _________ / 2 1 +/- \/ -1 + 1680 x = ------------------- 56 1 +/- 41 -------- 56 *** a = 9 b = -22 c = 8 _________ / 2 22 +/- \/ -22 - 288 x = ------------------- 18 22 +/- 14 --------- 18 *** a = 8 b = 25 c = 3 ________ / 2 -25 +/- \/ 25 - 96 x = ------------------ 16 -25 +/- 23 ---------- 16 *** ANS: 1/5 -6 , -5/2 -7 , 4 3 , -5/7 3/4 , 2 4/9 , -1/8 -3 QUADRATIC FUNCTIONS: Problem Set 2 NOTES: In Problem Set 1 we graphed points belonging to the quadratic equations that described the motion of a cannonball fired into the air. 2 y = (-1/500)x + x In this problem imagine that the cannonball is being fired on Jupiter where the gravity is about 2.5 times that of the Earth. As you might expect, gravity affects the flight of the cannonball. The equations for the motion of the cannonball on Jupiter is: 2 y = (-1/200)x + x Plot some points for this equation and later we'll compare the graphs for the effects of gravity. WORKSPACE: ANS: 0 , 22 , 38 , 47 , 50 , 47 , 38 , 22 , 0 QUADRATIC FUNCTIONS: The Distance Formula - Excursion NOTES: Suppose one day you find an old map that looks like this: --+----------+ R | | | a b | I| c | | | V | d | | e | E| | |f | R | g | -----R---A---- -------O---D-- Written on the map is the following description: "In a field 10 miles by 10 miles square, Are found 7 trees, A-B-C-D-E-F-G If you please. If you're smart, Find two trees 5 miles apart. Halfway between these two Is a treasure buried for you." Between which two trees will you find the buried treasure? At the end of this lesson we will return to this problem and you will be able to find the buried treasure! *** In a previous lesson you learned about the PYTHAGOREAN THEOREM and some of its applications. One of the most important applications of the Pythagorean Theorem is finding the DISTANCE between any two points in a plane. The distance between any two points is always a positive number. For example, to find the distance between these two points on the number line, you must compute the absolute value of the difference in their coordinates. The distance between two points will always be the same. This is true whether you go from point 1 to point 2 or from point 2 to point 1. *** If two points are on the same horizontal line, the distance between them is the absolute value of the difference between their x-coordinates. D = |10 - 2| = 8 If two points are on the same vertical line, the distance between them is the absolute value of the difference between their y-coordinates. D = |5 - 2| = 3 *** But what if two points are NOT on the same vertical or horizontal line? Now we have to find another way to determine the distance between the two points. Let's add a VERTICAL line to the graph. Now add a HORIZONTAL line to the graph. You have made a RIGHT triangle. The process of finding the lengths of the vertical and horizontal sides of a right triangle is similar to the process of finding the slope of a line, a skill that you learned earlier. To calculate the slope of a line, you found the change in y-coordinates and divided it by the change in x-coordinates. You can use the same information here to find the length of the third side of the triangle. Tx = x - x Ty = y - y 2 1 2 1 a = |Tx| b = |Tx| a - |x - x| b = |y - y| 2 1 2 1 Let side a equal 3 and side b equal 6, then solve for c: 2 2 2 c = a + b 2 2 2 c = 3 + 6 2 c = 9 + 36 2 c = 45 __ _ c = \/45 OR 3\/5 [This is the distance from (2,5) to (8,2)] *** Let's review the procedure for finding the distance between any two points. We will call these points point 1 (x^1,y^1) and point 2 (x^2,y^2) (Do not square, these are just labels!). But remember that it doesn't matter which point is chosen as point 1 or point 2. - First, make a right triangle. - Measure length of the vertical and horizontal sides. Now you can use the Pythagorean Theorem to find the length of the hypotenuse of this triangle. Let a = |x - x| and b = |y - y| 2 1 2 1 Then solve for c: 2 2 2 c = |x - x| + |y - y| 2 1 2 1 Since any number squared is positive, the absolute value signs are not needed. 2 2 2 c = (x - x) + (y - y) 2 1 2 1 Finally, this is the result when we solve for c. _________________ / 2 2 c = \/(x - x) + (y - y) 2 1 2 1 This is the length of the third side of the triangle (the hypotenuse). It is also the distance from (x^1,y^1) to (x^2,y^2). _________________ / 2 2 c = \/(x - x) + (y - y) 2 1 2 1 If we let D represent the distance between any two points (x^1,y^1) and (x^2,y^2) on a coordinate plane then we have the following equation: _________________ / 2 2 D = \/(x - x) + (y - y) 2 1 2 1 This equation is known as the Distance Formula. The distance is the same no matter which point you start with. *** To find the distance from the point (4,-2) to (-5,3), let: x = 4 x = -5 1 2 y = -2 y = 3 1 2 Use these values in the Distance Formula. _________________ / 2 2 D = \/(x - x) + (y - y) 2 1 2 1 ____________________ / 2 2 D = \/(-5 - 4) + (3- (-2)) __________ / 2 2 D = \/(-9) + (5) _______ D = \/81 + 25 ___ D = \/106 *** At the beginning of the lesson you were examining an old map. To locate the buried treasure, you must find two trees that are 5 miles apart. Tree A. (4,8) Tree B. (8,8) Tree C. (2,7) Tree D. (5,5) Tree E. (9,4) Tree F. (1,2) Tree G. (7,1) Now you can use the Distance Formula to find the distance between any two trees. Which two trees are 5 miles apart? Tree D and F. X marks the spot halfway between the two trees where treasure is buried. HAPPY DIGGING!!! *Editor's Note: Hehe, :-) ... O_o ... ;-)* WORKSPACE: ANS: 11 , 8 , 12 , 7 , 10 , 7 , 3 , 6 , \/40 , #3 , #1 , 5 , 13 , 10 , 4,8 , 8,8 , D F QUADRATIC EQUATIONS AND THE DISTANCE FORMULA II: Test This lesson will evaluate your skills in the solutions to quadratic equations and using the distance formula. NOTES: WORKSPACE: ANS: a , a , NOT:a,d , b , d , c , d , a , c , b DIVINDING POLYNOMIALS: Long Division - Divide polynomials using long division. NOTES: In our unit on polynomials we learned how to factor some polynomails by "guess and test." In this topic we are going to investigate another method for determining factors of polynomials. That method is the method of long division. We say that 13 is a factor of 32513 if 13 divides into 32,513 evenly. 32513/13 = 2501 Since the remainder is zero, 13 is a factor of 32,513. We can use the method of long division to determine if 3K + 1 is a factor of 3k^2 + 4k + 1. Notice how it is similar to dividing 341 by 31. 1k +1 ______________ 3k + 1 )3k^2 + 4k + 1 -(3k^2 + k) -------- 3k + 1 -(3k + 1) ------ 0 11 ____ 31 )341 -(31) -- 31 -(31) -- 0 *** Before pressing ENTER, describe in your own words one way to tell if "polynomial a" is a factor of "polynomial b." Divide "polynomial b" by "polynomial a" using long division. If the remainder is zero, then "a" is a factor of "b." *** WORKSPACE: ANS: k , 2k^2 + 5k , 4 , 8k + 20 , 3k^2 , 3k^3 + 12k^2 , -2k , -2k^2 - 8k , -5 , -5k - 20 , 2x , 10x^2 + 8x , -3 , -15x - 12 , x^2 , x^3 - 4x^2 , 2x^2 - 7x , 2x , 2x^2 - 8x , x - 4 , +1 , x - 4 , 0 , 3x^2 , 3x^3 + 6x^2 , -5x^2 - 3x , -5x , -5x^2 - 10x , 7x + 8 , +7 , 7x + 14 , -6 , 3x^2 , 6x^3 - 12x^2 , 12x^2 - 38x , +6x , 12x^2 - 24x , -14x + 28 , -7 , -14x + 28 , 0 , 3x^2 , 15x^3 - 6x^2 , -35x^2 - x , -7x , -35x^2 + 14x , -15x + 6 , -3 , -15x + 6 , 0 , 2x^2 , 6x^3 + 12x^2 , x^2 - 10x , +(1/3)x , x^2 + 2x , -12x - 4 , -4 , -12x - 24 , 20 , x^2 + 3x + 4 , 0 , 3x^3 - 2x - 6 , 7 , 5x^2 + 2x - 9 , 26 , 2x^2 + (1/2)x - 8 , 0 , 7x^3 + 5x^2 + 6x + 1 , 5 The above is continued (some parts are from this date, though). Current Date: March 14, 2006. DIVIDING POLYNOMIALS: Exponents - Simplify expressions of the form a^m / a^n. - Review the meaning of zero and negative exponents. NOTES: In this topic we will learn how to simplify expressions like these: 7 3 3 a a a -- -- -- 3 7 3 a a a Let's see how we simplify the quotient of expressions with the same base. *** Write a general rule for the quotient of two expressions with the same base, where the exponent in the numerator is greater than the expeonent in the denominator. m - n *** Write a general rule for the quotient of two expressions with the same base, where the exponent in the denominator is greater than the exponent in the numerator (REVERSED). n - m *** Here is another rule for simplifying expressions like a^m / a^n where n > m. We subtract the exponent in the denominator from the exponent in the numerator. 5 x 5-10 -5 --- = x = x 10 x Negative exponents can be used to simplify expressions like x^5 / x^10. *** 8 x 8-14 -6 --- = x = x 14 x Negative exponents can be used to simplify expressions like x^8 / x^14. We know that: 5 x 1 1 --- = ----- = -- 10 10-5 5 x x x And: 5 x 5-10 -5 --- = x = x 10 x So: -5 1 x = -- 5 x *** Notice how we simplify the following expressions. 5 6 x y 2 5 ---- = x y 3 x y 3 5 3 a b -1 3 b ---- = a b = -- 4 2 a a b *** Before pressing ENTER, describe in your own words how we simplify expressions like: 7 a b ---- 4 2 a b Subtract the exponents of corresponding variables and then simplify. WORKSPACE: ANS: a*a*a , a^2 , k*k*k , k^5 , a^4 , b^6 , x , y^7 , m-n , a*a*a*a*a*a*a , 4 , k*k*k*k*k*k , 2 , 1 a^3 , 1 b^5 , 1 x , 1 y^8 , n-m , x^-1 , x^-4 , x^-5 , x^-3 , m-n , x^-6 , 3^-4 , d^-2 , r^-7 , x^-8 , -n , n t^2 , z^5 y , 1 n^2k^2 , g^4h , n m^2 , 1 , 2 , 4 , 3 , 1 , 3 , 9 , 1 , 1 , r^3y , x^-5 , 1 , n^3m^-6 , x^-1 , y^-3 , p^2k^-4 , x^9 , 1 RATIONAL EXPRESSIONS: Problem Set 1 NOTES: Problem 1: A 1979 television commericial for Timex watches contained the statement that the watches were 99.97% accurate. In a day, how much time might be lost or gained by a watch with that accuracy? Round to the nearest second. ANS: 26 seconds *** In 1983 an economist stated that all of the world's population could easily fit in the state of Texas, with plenty of room for homes and yards. The world's population is 4.7 billion. The area of texas is 267,300 square miles. How many square feet would there be for each person? 3 ANS: 1.585 * 10 sq ft/person (OR: 1,585 sq ft/person) *** A 30-foot by 60-foot ballroom is to be turned into a discotheque by placing an 18-foot by 42-foot dance floor in the middle and carpeting the remainder. What percent of the floor is dance area? Round the nearest percent. ANS: 42% WORKSPACE: ANS: 26 , 1.585 * 10^3 , 42 POLYNOMIALS AND EXPONENTIAL OPERATIONS II: Test This lesson will evaluate your skills in dividing polynomials and simplifying exponential expressions. NOTES: WORKSPACE: ANS: *Editor's Note: Made some mistakes, taken out* OPERATIONS ON RATIONAL EXPRESSIONS: Simplifying Rational Expressions - Simplify rational expressions by canceling common factors - Analyze graphs of rational functions NOTES: In this activity we will simplify rational expressions like: 2 x + 8x + 7 ---------- 2 x - 2x - 3 By finding common factors and canceling. We also will investigate how this affects the graph of the expression. *** Let's talk about equivalence. Sometimes things that are equivalent under one condition are not necessarily equivalent under another condition. Let's look at an example... Two halves are not always equivalent to one whole. Suppose we are given these two halves of a coat. *Picture of left and right side of coat torn in half* We can say that the sum of the halves is equivalent to one whole coat. However, suppose we are given these two halves. *Picture of two left sides ^_^ * We cannot say that the sum of these halves is equivalent to one whole coat. So, under some conditions we can say that two halves are equivalent to one whole, and other conditions they are not. *** A similar case arises in simplifying rational expressions. 2x 2(x) 2 ------ = -------- = ----- 2 x(x - 2) x - 2 x - 2x According to the example above, 2x ------ should be equivalent to, 2 x - 2x 2 ----- x - 2 We will see that for some values of x the expressions are equivalent, but this is not true for ALL values of x. *Shows picture of a graph impossible to reproduce* These graphs are equivalent EXCEPT at x = 0. For the graph on the left there is a hole at x = 0, and for the one on the right there is not. If we want the graph on the right to be equivalent to the one on the left we RESTRICT the expression. 2 / ----- x = 0 x - 2 By excluding the x-value where the hole occurs, our graph on the left and our graph on the right become equivalent. *** Let's look at some more examples: x + 1 x + 1 1 ----- = -------------- = ----- 2 (x + 1)(x - 1) x - 1 x - 1 Here are the graphs of these expressions. *Sigh, can't show pictures...* The hole in the left graph indicates that in order for these two expressions to be equivalent: / x = -1 *** Let's simplify this expression: 2 x + 5x + 6 (x + 2)(x + 3) x + 3 ----------- = -------------- = ----- 2 (x + 2)(x - 5) x - 5 x - 3x - 10 How must we restrict x + 3 ----- x - 5 so that it is equivalent too 2 x + 5x + 6 / ----------- ? x = -2 2 x - 3x - 10 *** Before pressing ENTER, descirbe in your own words one way to make the simplified form of an expression equivalent to the original expression. Look at the graphs of both expressions. If there are "holes" in the original graph, restrict the simplified expression at the x-value of the holes. *** We can find the "holes" of a graph WITHOUT graphing! *Yay!* x - 3 x - 3 ----------- = -------------- 2 (x - 3)(x + 7) x + 4x - 21 This expression has a hole in its graph at x = 3. 2 x - 2x - 8 (x + 2)(x - 4) ---------- = -------------- 2 (x + 1)(x - 4) x - 3x - 4 This expression has a hole in its graph at x = 4. 2 x - 4 (x - 2)(x + 2) ------ = -------------- 3 x(x - 2)(x + 2) x - 4x This expression has holes in its graph at x = 2 and x = -2. *** At what x-values are there "holes" in the graphs of the following expressions? x + 2 --------- 2 x + x - 2 There is a hole at x = ... -2. *Next time factor it, sigh.* *** Before pressing ENTER, describe in your own words one way to determine the values for the holes in the graph of an expression. The holes occur at the x-values which make both the numerator and the denominator zero. This occurs because there are common factors in the numerator and the denominator. *** In addition to restricting a simplified expression at the x-values where there are holes, we must also restrict it at those x-values that make it undefined. x + 4 x + 4 1 ---------- = -------------- = ----- 2 (x + 4)(x - 3) x - 3 x + x - 12 / x = -4 Notice: If x = 3, the expression is undefined since the denominator would be zero. 2x - 8 2(x - 4) 2 ------ = -------------- = ----- 2 (x + 4)(x - 4) x + 4 x - 16 / x = 4, -4 *** Before pressing ENTER, describe in your own words the restricted values of a simplified rational expression. The restricted values of the simplified expression are the values of x which make the denominator of the original expression zero. WORKSPACE: ANS: -1 , x + 3 x - 5 , -2 , -2 , 1 , -3 , -2 , 1 , -3 , 4 , -5 , 3 x - 2 , -2 2 , x + 3 , 5 , x - 1 x + 2 , -6 -2 , 2x - 2 x + 8 , 3 -8 , x + 2 x - 1 , 4 1 , x + 5 x + 2 , -2 5 , 1 2x^3 , -3 -1 , x + 3 x + 1 , -3 -1 , x , 0 3 , x - 1 x + 9 , -1 -9 , x + 2 x - 2 , 2 4 , x + 4 x - 4 , 4 -5 , x + 1 x - 5 , 5 6 , x + 7 x^3 , 0 -2 , x - 8 x - 7 , 7 -7 The above is continued (some parts are from this date, though). Current Date: March 15, 2006. OPERATIONS ON RATIONAL EXPRESSIONS: Multiplication and Division - Multiply and divide rational expressions NOTES: In this activity we will use what we've learned about simplifying rational expressions to help us multiply and divide rational expressions. Remember how we multiply rational numbers. (We multiply numerator by numerator and denominator by denominator.) 3 8 24 2 - * - = -- = - 4 9 36 3 We could simplfy this process by canceling common factors first. 3 8 2 * 3 * 4 2 - * - = --------- = - 4 9 3 * 3 * 4 3 We follow the same process for rational expressions. x + 4 x - 3 (x + 4)(x - 3) x + 4 ----- * ----- = -------------- = ----- x - 3 x + 6 (x - 3)(x + 6) x + 6 Here we multiply and cancel common factors. - We factor first on this one. 2 2 x + 4x - 5 x + 5x - 14 ---------- * ----------- 2 x + 5 x + 6x - 7 Equals... (x + 5)(x - 1) (x + 7)(x - 2) -------------- * -------------- (x - 1)(x + 7) (x + 5) - Then cancel common factors. - Finally we simplify. = x - 2 *** Before pressing ENTER, describe in your own words one way to multiply rational expressions. Factor the numerator and denominator of each expression and "cancel" like factors before multiplying. *** Recall how we divide rational numbers. 3 9 3 4 12 1 - / - = - * - = -- = - 8 4 8 9 72 6 /\ \\ || \\ We first Then we simplify. invert and multiply. We could simplify this process by canceling common factors first. / / 3 9 3 4 3 * 4 1 - / - = - * - = ------------- = - 8 4 8 9 2 * 4 * 3 * 3 6 / / / We follow the same process for the rational expressions. x + 3 2x + 6 x + 3 3x + 12 ----- / ------- = ----- * ------- x + 4 3x + 12 x + 4 2x + 6 \\ \\ We invert and multiply. /////// /////// (x + 3) * 3(x + 4) 3 = ------------------ = - We cancel (x + 4) * 2(x + 3) 2 to simplify. /////// /////// We follow the same process for the rational expressions. 2 x - 2 x + 3x - 10 ---------- / ----------- 2 2 x + 2x - 3 x + 8x + 15 We invert and multiply. \\ 2 x - 2 x + 8x + 15 = ---------- * ----------- 2 2 x + 2x - 3 x + 3x - 10 ///////////////////// (x - 2)(x + 5)(x + 3) 1 = ---------------------------- = ----- (x + 3)(x - 1)(x + 5)(x - 2) x - 1 /////// ////////////// - We cancel to simplify. *** Before pressing ENTER, describe in your own words one way to divide rational expressions. Change the quotient to a product by inverting the divisor, factor each expression completely, cancel common factors and simplify. *** WORKSPACE: 3(x + 1) -2 -------- * ------- (x + 4) (x + 5) *Note to self: Simplify ALL THE WAY* *** ANS: 6 , -6x - 6 x^2 + 9x + 20 , 5 x - 4 , x + 4 x + 1 , 3x^2 - 9x - 30 2x - 2 , 1 3 , x + 2 x + 3 , x - 2 x + 4 , 3x - 15 2x - 8 , x + 3 x - 1 , x - 5 x + 4 , 9x - 18 2x - 6 , 1 1 , 1 1 , x^2 - 6x + 9 x^2 + 7x + 10 , x^2 - 7x + 6 x^2 - 4 , -x + 2 x - 1 , x^2 - 9x + 20 -2x^2 - 16x - 30 , x^2 - 2x + 1 x^2 + 4x + 3 , x + 1 x + 3 , The above is continued (some parts are from this date, though). Current Date: March 16, 2006. OPERATIONS ON RATIONAL EXPRESSIONS: Addition and Subtraction - Add and subtract rational expressions NOTES: In this activity we will extend what we learned about multiplying and dividing rational expressions to help us add and subtract rational expressions. *** Remember how we add and subtract common denominators. 3 5 8 - + - = - = 4 5 2 2 We simply add the numerators and simplify. 14 2 12 4 -- - - = -- = - 9 9 9 3 Here we subtract the numerators. *** We follow the same process for rational expressions with common denominators. Addition: 2x + 9 4x - 7 (2x + 9) + (4x - 7) ------ + ------ = ------------------- 2x - 4 2x - 4 2x - 4 6x + 2 3x + 1 = ------ = ------ 2x - 4 x - 2 Subtraction: 5x + 3 4x - 5 (5x + 3) - (4x - 5) ------ - ------ = ------------------- 3x + 1 3x + 1 3x + 1 x + 8 = ------ 3x + 1 *** Before pressing ENTER, describe in your own words how we add and subtract rational expressions with common denominators. We add or subtract the numerators and leave the denominator the same. We then simplify if possible. *** Remember how we add and subtract rational numbers with different denominators. 5 2 5 7 2 4 35 8 43 - + - = - * - + - * - = -- + -- = -- 4 7 4 7 7 4 28 28 28 - We find a common denominator. - We add the numerators. Another: 4 5 4 5 - - - = ----- - ----- 9 6 3 * 3 2 * 3 - First we find a common denominator. 4 2 5 3 = - * - - - * - 9 2 6 3 - Next we subtract the numerators. 8 15 -7 -- - -- = -- 18 18 18 We follow the same process for rational expressions with different denominators. x - 6 x + 5 ----- + ----- x + 4 x - 4 - Find the common denominator. x - 6 (x - 4) x + 5 (x + 4) ----- * ------- + ----- * ------- x + 4 (x - 4) x - 4 (x + 4) 2 2 x - 10x + 24 x + 9x + 20 = -------------- + -------------- (x + 4)(x - 4) (x - 4)(x + 4) - Add the numerators and simplify. 2 2x - x + 44 = ----------- 2 x - 16 Another: 2 x + 3x + 2 1 ---------- - ----- 2 x + 3 x + 4x + 3 - We find a common denominator. 2 x + 3x + 2 1 (x + 1) -------------- - ----- * ------- (x + 3)(x + 3) x + 3 (x + 1) 2 x + 3x + 2 - (x + 1) -------------------- (x + 3)(x + 1) - Then we subtract and simplify. (x + 1)(x + 1) x + 1 -------------- = ----- (x + 3)(x + 3) x + 3 *** Before pressing ENTER, describe in your own words how we add and subtract rational expressions with different denominators. We factor each denominator to determine the factors of the least common denominator. Then we change each expression so it has the same denominator and then simplify. *** Sometime's it is best to reduce each expression BEFORE finding a common denominator. 2 x - 9 8x ----- + ------ x + 3 2 x + 2x (x + 3)(x - 3) 8x = -------------- + -------- (x + 3) x(x + 2) x - 3 8 ----- + ----- 1 x + 2 (x - 3)(x + 2) 8 = -------------- + ----- (x + 2) x + 2 2 2 (x - x - 6) + 8 x - x + 2 = --------------- = --------- x + 2 x + 2 Another: 2 x - 3x - 10 x + 6 ----------- - ----------- x - 5 2 x + 3x - 18 (x + 2)(x - 5) (x + 6) = -------------- - -------------- (x - 5) (x + 6)(x - 3) 1 (x + 2)(x - 3) 1 = x + 2 - ----- = -------------- - ----- x - 3 (x - 3) x - 3 2 2 (x - x - 6) - 1 x - x - 7 = --------------- = --------- x - 3 x - 3 *** Before pressing ENTER, describe in your own words one way to simplify a sum or difference of rational expressions. Reduce each expression to its simplest form before finding a common denominator. WORKSPACE: x x - 1 ----- - ----- x + 2 x + 3 x(x + 3) (x - 1)(x + 2) -------------- - -------------- (x + 2)(x + 3) (x + 3)(x + 2) 2 2 x + 3x x + x - 2 ---------- - ---------- 2 2 x + 5x + 6 x + 5x + 6 *** 4x 2 ----- + ----------- x + 2 2 x + 7x + 10 4x 2 ----- + -------------- x + 2 (x + 2)(x + 5) 4x 2 ----- + -------------- x + 2 (x + 2)(x + 5) 2 4x + 20x + 2 ------------ 2 x + 7x + 10 *** 2 3x - 4 x + 3x + 2 ------ - ---------- x - 1 2 x - 1 2 3x - 4 x + 3x + 2 ------ - ---------- x - 1 (x + 1)(x - 1) 2x - 6 ------ x - 1 *** 2 2x - 7x - 4 x + 3 ----------- + ----- x - 4 x - 7 2 2x - 7x - 4 x + 3 ----------- + ----- x - 4 (x - 7)(x - 4) (2x + 1)(x - 4) (x + 3) --------------- + ------- (x - 4) (x - 7) (2x + 1)(x - 4) (x + 3) --------------- + ------- (x - 4) (x - 7) 2 2x - 12x - 4 ------------ x - 7 *** 2 2 3x + x - 2 x + 5x - 24 ---------- - ----------- 2 x - 3 x + 9x + 8 (3x - 2)(x + 1) (x + 8)(x - 3) --------------- - -------------- (x + 1)(x + 8) (x - 3) 2 -x - 13x - 66 ------------- x + 8 *** 2 2 x + 6x + 13 x + x - 12 ----------- - ---------- x + 5 x + 5 5x + 25 5(x + 5) ------- = -------- = 5 x + 5 (x + 5) *** 2 x - 4x + 3 3x + 12 ---------- + ------- 2 2 x - 3x + 2 x - 16 2 x - 4x + 6 ---------- 2 x - 6x + 8 *** ANS: -x + 1 2x + 1 , x + 4 , 12x - 11 x + 4 , 6x + 2 2x - 3 , 10x - 5 3x - 1 , 2x + 2 x^2 + 5x + 6 , 2x^2 - 6x + 3 x^2 + 2x - 15 , -2x - 1 x^2 - 5x - 6 , 4x^2 + 20x + 2 x^2 + 7x + 10 , 2x - 6 x - 1 , 2x^2 - 12x - 4 x - 7 , -x^2 - 13x - 66 x + 8 , x^2 + 16x + 48 x^2 + 3x - 54 , 9x + 54 x^2 - 8x - 20 , x^2 - 3x - 27 x - 7 , 5 , 2 x^2 + 8x + 12 , x^2 - 4x + 6 x^2 - 6x + 8 , 3x^2 - 10x - 23 3x + 6 , 2x - 3 x + 8 , -x - 5 12x + 36 , 2x^2 - 4 x - 2 , x^2 - 2x - 12 x + 2 , -4x + 10 x^2 - 2x - 8 , -6x - 22 x^2 + 12x + 35 The above is continued (some parts are from this date, though). Current Date: March 20-21, 2006. RATIONAL EXPRESSIONS: Problem Set 2 NOTES: Problem 3: The 1970 census from the western states showed a population size of 34,838,243. The 1980 census for the same region showed 43,172,490. What is the percent increase? Round your answer to the nearest percent. ANS: 24% 43,172,490 - 34,838,243 = 8,334,247 The increase is a fraction of the original population. We'll convert the fraction to a decimal and the decimal to a percent. 8,334,247 ---------- = .2392 = 24% 34,838,243 WORKSPACE: ANS: Biology , 1/3 , 24 RATIONAL EXPRESSIONS: Number Bases - Excursion NOTES: Different Bases: Our numbering system is known as the base, ten, or decimal system. It's called that because all numbers can be represented by the combination of ten different symbols: 0 1 2 3 4 5 6 7 8 9 The group of symbols 362 represents the same thing to everyone using the base ten system. The symbols 362 stand for: 3 - 100's 6 - 10's 2 - 1's Why don't the columns stand for fives, or twenties instead of ones, tens, and hundreds? The reason is that the ones column represents 10^0, the tens column represents 10^1, the hundreds column 10^2, etc. Base 5 uses only these symbols: 0 1 2 3 4 What would the group of symbols 341 (base 5) represent in our normal base ten system? Let's first identify what each column stands for. 0 5 = 1 1 5 = 1 2 5 = 25 3 5 = 125 GRAPH: 3 (5) 125's 0 = 0 2 (5) 25's 3 = 75 1 (5) 5's 4 = 20 0 (5) 1's 1 = 1 ---------------------- TOTAL: 96 (in base 10) Now try it for 1432 (base 5): 125's 1 = 125 25's 4 = 100 5's 3 = 15 1's 2 = 2 ----------------- TOTAL: 242 (in base 10) On to base 8... *** We have seen that the symbols 1432 can mean different things in different bases. 1432 (base 10) = 1432 1432 (base 5) = 96 1432 (base 8) = 794 Now we'll look at a very practical numbering scheme using base two. Base Two: Pick a number from 1 to 15. Now answer the questions below using this table: A | B | C | D -----+-----+-----+----- 8 | 4 | 2 | 1 9 | 5 | 3 | 3 10 | 6 | 6 | 5 11 | 7 | 7 | 7 12 | 12 | 10 | 9 13 | 13 | 11 | 11 14 | 14 | 14 | 13 15 | 15 | 15 | 15 The trick is that the table is cleverly constructed so that your yes/no responses are changed into 1's or 0's (base 2 symbols). If you want to learn a magic trick to impress your friends copy the table down on a card or a piece of paper. Give the table to a friend and ask them to think of a number from 1 to 15. Then tell them that you will ask them a series of questions about their number. To guess their number just ask them if their number is in each of the columns. If their answer is yes, add the power of two at the top of the column. If the answer is no, add nothing. You will quickly come up with their answer, almost magically! It takes only four guesses to identify any number from 1 to 15. If you had a computer to help add the powers of 2 you could guess any number up to a million with just twenty guesses! Now, if you can specify (guess) such large numbers so easily you could do the same thing with letters (if you gave each letter a number). You could do the whole alphabet with 6 guesses, or a whole keyboard of letters and punctuation makrs with 7 guesses. You could even identify every person in the United States with just 30 guesses the same way! Another nice thing about base 2 (or binary) numbers is that it's easy to build electronic devices that use them. If you say that when a switch is off that means 0 and when a switch is on that means 1, you can represent huge amounts of numbers or people or words with just 20 or 30 switches. These electronic switches don't have to be as large as ordinary mechanical switches, like a light switch. In fact, in modern computers switches have been made so small that this computer's microchip can hold hundreds of thousands of them. Each switch is either on (1) or off (0). That's why computers can hold so much information, because they use base 2. WORKSPACE: ANS: 1 , 125 , 100 , 15 , 2 , 0 1 2 3 4 5 6 7 , 1 , 8 , 64 , 512 , 512 , 256 , 24 , 2 , NO , YES , YES , NO The above is continued (some parts are from this date, though). Current Date: March 22, 2006. RATIONAL OPERATIONS II: Test This lesson will evaluate your skills in performing operations with polynomials and simplifying expressions in the form of fractions. NOTES: WORKSPACE: 125 + 75 + 15 + 4 ANS: a , a , c , b , d , c , c , d , 9 , 219 Now Entering: PRE-CALC. ALGEBRA II LINEAR SYSTEMS AND MATRICES: Linear Functions and Graphs Pre-Test This test has 10 questions, each worth 10 points. You have one chance to answer each question, so be sure the answer you choose is the answer you want graded. NOTES: WORKSPACE: ANS: (2,1) , NOT:one , (1,-2) , NOT:1 , NOT:#4 , NOT:#4 , #3 , #2 , NOT:#3 , NOT:#1 LINEAR FUNCTIONS AND GRAPHS: Forms of Equations - Determine the slope and the intercepts of a line from its equations or graph - Transform a linear equation in one form to an equivalent equation in standard form NOTES: In Algebra I you studies equations and graphs of linear functions. In this activity you will review graphs of linear functions and write their equations in several different forms. *** Use your own words to describe the graph and the equation of a direct variation. The graph is a LINE through the origin. The slope m is equal to y/x for any point P(x,y) on the line, so: y = mx. *** The graph of a direct variation will be shifted if a constant is added to the equation. *** The slope-intercept form of a linear equation is y = mx + b where m is the slope and b is the y-intercept. The graph of the equation crosses the y-axis at (0,b) and the slope is: y - b m = ----- x Where P(x,y) is another point on the line. *** You used the y-intercept to determine the slope-intercept form of a linear equation. You use both intercepts to determine the two-intercept form of the equation. You can find the two-intercept form of the equation for this line by locating the intercepts. Since the x-intercept is -3 and the y-intercept is 2, the equation is: x y -- + - = 1 -3 2 *** Sometimes it is difficult to determine the intercepts from the graph of a line. To find the equation of the line, you can use any two points on the line and the two-point form of the equation. y - y y - y 1 1 2 ------ = ------ x - x x - x 1 1 2 To find the equation for this line, use the points (-1,8) and (3,-3) and the two-point form. y - y y - y 1 1 2 ------ = ------ x - x x - x 1 1 2 y - 8 8 + 3 ----- = ------ x + 1 -1 - 3 or y - 8 -11 ----- = --- x + 1 4 *** There are many forms for writing linear equations. y = mx ----------------- direct variation y = mx + b -------------- slope-intercept x y - + - = 1 ----------------- two-intercept a b y - y y - y 1 1 2 ------ = ------ --------------- two-point x - x x - x 1 1 2 Each of these forms can also be written as an equivalent equation in standard form, ax + by = c. *** You can transform the equation y = (4/3)x - 4 from slope-intercept form to standard form. y = (4/3)x - 4 3y = 4x - 12 (multiply by 3) -4x + 3y = -12 (subtract 4x) This equation is in standard form. *** Subtracting x from both sides of the equation completes the transformation from two-point form to standard form. y - 3 1 ----- = - x + 2 3 x + 2 y - 3 = ----- 3 3y - 9 = x + 2 3y = x + 11 -x + 3y = 11 *** By performing the same operations on both sides of the equation, you can transform any form of a linear equation into an equivalent equation in standard form. WORKSPACE: ANS: 15 , origin , 4 , -2/3 , line , mx , 0 , -3 , b , -1 , 2 , a , 5 , -3 , multiplication , multiplication , addition , 6 , 7x , 0 , -10 , -2 , -3 , 4 and 3 , 7y , -6 LINEAR FUNCTIONS AND GRAPHS: Algebraic and Geometric Transformations - Build the graph of a linear equation by adding the graphs of its terms - Transform a linear equation algebraically and geometrically NOTES: In the last activity you studies linear equations and their graphs. Algebraically you can build a linear equation by adding terms. y = x + 3 Geometrically, you can build the graph of this linear function by adding the graphs of its terms. Watch as the graphs y and y = 3 are added 1 2 together. At x = 1, 3 + 1 = 4. At x = 2, 3 + 2 = 5. At x = 3, 3 + 3 = 6. At x = -1, 3+(-1) = 2. At x = -2, 3+(-2) = 1. At x = -3, 3+(-3) = 0. The equation of the sum of y = x and y 1 2 = 3 is y = x + 3. Here are the graphs of y = x and y = -4. 1 2 At x = -1, the value for y is -1 and the 1 value for y is -4. 2 y + y = -1+(-4) 1 2 = -5 *** You have added the graphs of a constant term and a linear term to determine the graph of a linear equation of the form y = mx + b. Conversely, you can determine the two terms that were added to form the graph of a linear function. *** Algebraically, a linear equation like y = 7x + 3 is the sum of a linear term and a constant term. Geometrically, you can graph this function by adding the graphs of: y = 7x and y = 3 *** Now you will transform and graph linear equations. As you perform the algebraic transformations, the graph shows the geometric transformations. *** Here are the graphs of y = x - 4 and y = -x + 6. Substituting for y, you get x - 4 = -x + 6. Notice that the lines intersect at x = 5, the solution of x - 4 = -x + 6. *** Performing algebraic operations on both sides of an equation transforms it into equivlanet equations. Although the equations change in form, the graphs show that the solutions are preserved. WORKSPACE: ANS: -4 , -2 , 4 , 2 , 1 , #1 , #4 , 3 , 6 , 10 , 5 , -7 , 5/4 , 28/5 , ... , #2 , #3 , #1 , 3 , -1 LINEAR FUNCTIONS AND GRAPHS: Solving Systems of Linear Equations - Solve systems of linear equations by using the substitution method - Solve systems of linear equations by using the linear combination method NOTES: In the last activity, you performed algebraic transformations on equations and followed the corresponding geometric transformations on the graphs. This mapping back and forth between algebra and geometry can often be used to study and solve mathematical problems. *** Sometimes you can solve systems of equations by reading the solution from the graph. y = x - 4 y = -x + 6 The point (5,1) lies on both lines and satisfies each of the equations. 1 = 5 - 4 1 = -5 + 6 *** When the solution is difficult to determine from the graph, you can use algebraic methods to find the ordered pair that satisfies both equations. The substitution method can be used to solve this system. 5y = 4x + 20 y = 2x + 3 You can substitute 2x + 3 for y in the first equation. 5(2x + 3) = 4x + 20 10x + 15 = 4x + 20 6x = 5 x = 5/6 y = 2(5/6) + 3 = 10/6 + 18/6 = 28/6 = 14/3 The solution is the ordered pair (5/6,14/3). *** Use your own words to describe how the substitution method solves a system of equations like these. x - 2y = 10 y = 3x Substitute 3x for y in the first equation and transform it into a single equation for x. Then solve for y. x - 6x = 10 -5x = 10 x = -2 y = 3x y = 3(-2) y = -6 Point (-2,-6) satisfies both equations. *** Sometimes neither equations in a system has x or y alone on one side of the equation. 3x + 6y = 4 y - 3x = -4 When this happens, one of the equations must be transformed before the substitution method can be used. 3x + 6y = 4 y - 3x = -4 You can isolate y in the second equation. y = 3x - 4 *** 5x - 2y = 4 3x + 2y = 1 Although you solved this system of equations using the substitution method, there is another way to eliminate one of the variables. To form a single equation in x, you can add the equations. 5x - 2y = 4 3x + 2y = 1 ----------- 8x + 0y = 5 8x = 5 x = 5/8 3(5/8) + 2y = 1 15/8 + 2y = 1 2y = 8/8 - 15/8 2y = -7/8 y = -7/16 The ordered pair (5/8,-7/16) is the same solution obtained by using the substitution method. *** This linear combination method of solving systems is convenient to use when both equations are in standard form. Sometimes one or both equations must be multiplied by a constant before they can be added or subtracted to eliminated one of the varaiables. *** 2x + y = -2 x - 3y = 2 You can multiply the first equation by 3. 6x + 3y = -2 x - 3y = 2 ------------ 7x = -4 x = -4/7 -4/7 - 3y = 2 -3y = 14/7 + 4/7 -3y = 18/7 y = -6/7 *** Systems of linear equations can be solved geometrically by graphing the lines and determining the point of intersection. Systems of equations can also be solved algebraically by using the substitution method or the linear combination method. *** Use your own words to describe when you might use each algebraic method. If one of the equations can be easily expressed in terms of a single varaiable, the substitution method works well. If both equations are in standard form, with coefficients other than 1, the linear combination method may be easier to use. WORKSPACE: ANS: #4 , #3 , 4/3 , 8/3 , 4/5 , -12/5 , 4/3 , 0 , -6/7 , 16/7 , 5/8 , -7/16 , 0 , -5/2 , 10/3 , 1/12 , 7/2 , -8 , standard , 9/4 , -15/8 , -1/6 , 19/6 , 9/4 , 1/2 , -1/3 , -4 , 9/2 The above is continued (some parts are from this date, though). Current Date: March 23, 2006. LINEAR FUNCTIONS AND GRAPHS: Problem Set 1 NOTES: WORKSPACE: ANS: 135 , 8 , 38 LINEAR SYSTEMS AND MATRICES: Linear Functions and Graphs Post-Test NOTES: WORKSPACE: The point (2,-2) could NEVER be the solution to a system of equations which includes the line... #4 y = (1/2)x - 1 *** A system of two linear equations, whose graphs intersect in a single point, has HOW MANY solution? #2 *** Choose the equations that is NOT equivalent to the other equations. #2 5y + 30 = 3x *** The ordered pair which satisfies this system of linear equations is... #2 (3,2) *** Use the substition method to solve this system of linear equations. #4 (-3,-5) *** The standard form for the linear equation -x y -- + - = 1 3 2 is... #4 *** The standard form for the linear equation y + 2 3 ----- = - x - 1 4 is... #2 *** Use the linear combination method to solve this system of equations. #4 (3,0) *** Use the linear combination method to solve this system. #3 *** To eliminate the y-term, multiply the first equation by... #3 2 ANS: See above. SOLVING SYSTEMS WITH MATRICES: Row Operations - Represent systems of linear equations using matrices - Solve a linear system by transforming the matrix of the system In the last lesson you saw that linear equations can always be transformed to the standard form of the equation. You also solved systems of two linear equations. In this lesson you will use matrices to solve systems of two linear equations in standard form. *** The pair of linear equations: 3x + 2y = 5 2x + 7y = 1 Is represented by this matrix: [3 2 5] [2 7 1] NOTES: Linear equations such as these can be represented by a matrix because the equations are in standard form. ax + by = c dx + ey = f The matrix which corresponds to this system is written: [a b c] [d e f] Watch what happens to the corresponding matrix as linear combinations are used to solve this pair of equations. 2x + 3y = 5 [2 3 5] 4x + 3y = 9 [4 3 9] -2x - 3y = -5 [-2 -3 -5] 4x + 3y = 9 [ 4 3 9] 2x + 0 = 4 [2 0 4] 4x + 3y = 9 [4 3 9] -4x + 0 = -8 [-4 0 -8] 4x + 3y = 9 [ 4 3 9] The second row is replaced by adding the two equations. -4x + 0 = -8 [-4 0 -8] 0 + 3y = 1 [0 3 1] The first equation is multiplied by -1/4 and the second equation by 1/3. x + 0 = 2 [1 0 2 ] 0 + y = 1/3 [0 1 1/3] x = 2, y = 1/3 x = 2, y = 1/3 The point (2,1/3) satisfies the system. *** When transforming a system of equations or its representative matrix to solve the system, you can:' 1. Multiply either equation (or row) by a constant. 2. Add the equations (or rows) to eliminate varaiables (produce zeros). To solve a system of equations such as: x + 3y = 7 2x + 9y = 20 You can perform ROW OPERATIONS on the matrix. [1 3 7] [2 9 20] The first row is multiplied by -3. [1 3 7] x + 3y = 7 [2 9 20] 2x + 9y = 20 [-3 -9 -21] [ 2 9 20] | | | | | | Indicate the operation performed. ANS: Addition [-1 0 -1] [ 2 9 20] | | | | | | Indicate the operation performed. ANS: Multiplied [-2 0 -2] [ 2 9 20] | | | | | | Indicate the operation performed. ANS: Addition [-2 0 -2] [ 0 9 18] *** Sometimes two or more operations can be applied in one step to speed up the matrix method of solving a system. To solve the system: 3x - 2y = -6 -2x + y = 5 Determine the matrix: [ 3 -2 -6] [-2 1 5] Multiply 2nd row by 2 and add rows. [ 3 -2 -6] [-1 0 4] Multiply 2nd row by 3 and add rows. [ 0 -2 6] [-1 0 4] Multiply 1st row by -1/2, 2nd row by -1. [0 1 -3] [1 0 -4] The solution point is (-4,-3). *** To solve a system of linear equations using a matrix of the form: [a b c] [d e f] You can perform row operations on the matrix until it is transformed to a solution matrix of the form: [0 1 p] OR [1 0 q] [1 0 q] [0 1 p] Where x = q and y = p. If a matrix is transformed to an equivalent solution matrix of the form: [0 1 p] OR [1 0 q] [1 0 q] [0 1 p] The values for x and y can be checked in the original equations to see if they are correct. The matrix... [0 1 4/3] [1 0 -1/3] ...is the solution matrix for the system: 3x + 6x = 7 -2x + 4y = 6 because 3(-1/3) + 6(4/3) = -3/3 + 24/3 = 7 and -2(-1/3) + 4(4/3) = 2/3 + 16/3 = 6 *** To solve a system of linear equations using matrix methods, first determine the matrix for the system, then find a solution matrix of the form: [0 1 q] OR [1 0 p] [1 0 p] [0 1 q] By performing ROW operations on the matrix. WORKSPACE: ANS: #4 , #2 , #4 , #3 , #2 , #2 , standard , addition , multiplication , addition , multiplication , 1,2 , -6 , 2 , -2 , #3 , #3 , #2 , row , #3 , #2 , #4 , #1 , #4 SOLVING SYSTEMS WITH MATRICES: Cramer's Rules and Determinants - Solve a system of linear equations using Cramer's Rule - Find the determinant of a 2 by 2 matrix NOTES: In the last activity you saw that a matrix can be used to represent a system of linear equations. ax + by = c --\ [a b c] dx + ey = f --/ [d e f] Furthermore, row operations can be performed to transform that matrix. [a b c] --\ [1 0 p] [d e f] --/ [0 1 q] The solution x = p and y = q can be read from the matrix. Now, you can use matrix transformations to find a general rule that gives the solution to any pair of linear equations in standard form. ax + by = c dx + ey = f Start with the matrix for this system and perform row operations to find the solution for the system. [a b c] [d e f] Multiply the top row by e. [a b c] ax + by = c [d e f] dx + ey = f [ae be ce] [ d e f] [ ae be ce] [-bd -be -bf] [ae-bd 0 ce-bf] [ -bd -be -bf ] [ 1 0 ce-bf] -----] ---\ ce - bf ae-bd] ---/ x = ------- [-bd -be -bf ] ae - bd A similar process can be used to obtain a solution matrix for y. [a b c] --\ ax + by = c [d e f] --/ dx + ey = f [-ad -bd -cd ] [ 0 1 af-cd] [ -----] [ ae-bd] The solution for y read from this matrix is: af - cd y = ------- ae - bd The general formulas for x and y can be used to solve this system. 2x + 3y = 2 4x - 9y = -1 ce - bf af - cd x = ------- y = ------- ae - bd ae - bd 2(-9) - 3(-1) -18 + 3 x = ------------- = -------- = 1/2 2(-9) - 3(4) -18 - 12 2(-1) - 2(4) -2 - 8 y = ------------ = -------- = 1/3 2(-9) - 3(4) -18 - 12 The solution to this system is the ordered pair (1/2,1/3). *** Use the general formulas for x and y to solve this system of equations. 4x - 3y = 11 5x - 6y = 9 ax + by = c dx + ey = f ce - bf af - cd x = ------- y = ------- ae - bd ae - bd 11(-6) + 3(9) x = ------------- 4(-6) + 3(5) ... -39 13 x = --- = -- -9 3 4(9) - 11(5) y = ------------ 4(-6) + 3(5) ... -19 19 y = --- = -- -9 9 *** The formulas for finding the solution of any system of two linear equations are known as CRAMER'S RULE. ce - bf af - cd x = ------- y = ------- ae - bd ae - bd Because Cramer's Rule is difficult to remember, the concept of "determinant" may be of help. The determinant of the matrix: [5 3] is denoted by |5 3| [7 2] |7 2| And the value of the determinant is 5(2) - 7(3) = 10 - 21 = -11. *** Now you can apply derminants to Cramer's Rule and to the solution of a system of linear equations. Once again, consider the linear equations: ax + by = c dx + ey = f And the corresponding matrix: [a b c] [d e f] *** IMPORTANT, KEEP THIS: Cramer's Rule: ce - bf x = ------- ae - bd af - cd y = ------- ae - bd [a b c] [d e f] Find the determinant of the x and y coefficient matrix. [a b] D = ae - bd [d e] Notice that this determinant is the denominator for both values of Cramer's Rule. NEXT: Find the determinant of the matrix formed by replacing the x coefficients with the constant terms. [c b] D = ce - bf [f e] Notice that this determinant is the "x-numerator" of Cramer's Rule. NEXT: Find the determinant of the matrix formed by replacing the y coefficients with the constant terms. [a c] D = af - cd [d f] Notice that this determinant is the "y-numerator" of Cramer's Rule. THINK: This is the determinant form of Cramer's Rule. |c b| |a c| |f e| |d f| x = ----- y = ----- |a b| |a b| |d e| |d e| [a b c] [d e f] ...[animation of taking the c and f away]... [a b] [d e] The denominator of both equations is the determinant of the coefficient matrix. ...[another animation, switched some stuff around]... [c b] [f e] The numerator of the x equation is the determinant of the matrix formed by replacing the x coefficients. ...[another animation, switched more around]... [a c] [d f] The numerator of the y equation is the determinant of the matrix formed by replacing the y coefficients. *** To solve the system: 3x + 4y = 8 2x - 2y = 7 Using the determinant form of Cramer's Rule, we follow these steps: Evaluate the DENOMINATOR DETERMINANT. |3 4| = 2(-2) - 4(2) = -14 |2 -2| Then the X-NUMERATOR DETERMINANT. |8 4| = 8(-2) - 4(7) = -44 |7 -2| Then the Y-NUMERATOR DETERMINANT. |3 8| = 3(7) - 8(2) = 5 |2 7| We simplify to find the solution. -44 5 x = --- = 22/7 y = --- = -5/14 -14 -14 *** The general formulas used to solve a system of two linear equations are known as Cramer's Rule. ax + by = c dx + ey = f ce - bf af - cd x = ------- y = ------- ae - bd ae - bd Since the rule can difficult to remember, the determinant form of the rule is more easily applied. |c b| |a c| |f e| |d f| x = ----- y = ----- |a b| |a b| |d e| |d e| WORKSPACE: 2x - 3y = 3 4x + 6y = -2 x = 1/2 | 3 -3| |-2 6| x = ------- | 2 -3| | 4 6| ... x = 12/24 = 1/2 |2 3| |4 -2| y = ------ |2 -3| |4 6| ... y = -16/24 = -2/3 ANS: 13/3 , 19/9 , -2 , 11 , ae - bd , ce - bf , af - cd , 5 , 20 , 10 , 1/2 , -2/3 , #3 , #4 , #1 , #2 , #1 The above is continued (some parts are from this date, though). Current Date: March 28, 2006. QUADRATIC FUNCTIONS: Plotting Points - Evaluate quadratic functions - Plot points using x and f(x) values NOTES: In Unit 1 you graphed linear functions of the form y = mx + b. All the points which satisfy a linear function lie on a straight line. If only two points of the function have been plotted, you can predict the location of other points. These points lie on the graph of a linear function. For x = 3, you can guess the value of y. y = 3 All three points satisfy the function y = 2x - 3. *** Next, you will be given points on the graphs of other functions which may not be linear. Try to guess the location of other points on the graphs of these "mystery" functions. *** Use your own words to describe the results of predicting the location of other points on the graph of a function when only a few of the points have been plotted. Unless the function is known to be linear, errors may result when predicting the location of other points. To avoid errors, the rule or function must be known. *** When a function is given, it can be evaluated to plot points on its graph. You can plot some points of f(x) = x^2 - 6x + 11 by completing the table. x | f(x) --+----- 1 | 6 2 | 3 3 | 2 4 | 3 5 | 6 Complete this table to plot points of: 2 f(x) = x - 2x - 2 x | f(x) --+----- -1| 1 0 | -2 1 | -3 2 | -2 3 | 1 Complete this table to plot points of: 2 f(x) = -2x + 4x - 3 x | f(x) --+----- -1| -9 0 | -3 1 | -1 2 | -3 3 | -9 *** Describe how to plot points of the graph when the rule or function is given. To plot points of a given function, make a table of x and f(x) values that satisfy that function. *** You have been evaluating functions of the form: 2 y = ax + bx + c These equations are called QUADRATIC EQUATIONS. These are examples of QUADRATIC FUNCTIONS: 2 2 f(x) = 2x - 5x - 12 f(x) = x + 4 2 2 f(x) = -7x f(x) = 3x - x 2 2 f(x) = 16 - x f(x) = 6x - 2 + 3x Notice the x-squared terms. These are NOT quadratic functions: f(x) = 2x + 5 f(x) = 5x 3 f(x) = -3 f(x) = x - x + 2 *** Before pressing ENTER, use your own words to describe a quadratic function. It is a second degree function that can be written in the form: 2 / f(x) = ax + bx + c, a = 0 It must have an x-squared term but no term raised to a higher power. WORKSPACE: ANS: 9 , 1 , 3 , 2 , -1 , 2 , 3 , 6 , 1 , -2 , -3 , -2 , 1 , -9 , -3 , -1 , -3 , -9 , NO , YES , NO , NO , YES , 1 , 6 , .25 , YES , NO , #2 , #2 , #3 , #4 , #4 The above is continued (some parts are from this date, though). Current Date: March 29, 2006. GRAPHING QUADRATIC FUNCTIONS: Patterns in Graphs - Find the maximum or minimum point of a quadratic function - Identify the line of symmetry - Determine where the function is increasing or decreasing NOTES: In the last activity, you plotted points of a quadratic function by evaluating the function for several values of x. To sketch the entire graph, you could plot many points using x and f(x). However, to save time, you can look for critical points and patterns when evaluating the function. *** You may have noticed that the x-values in the tables clearly indicated a maximum or minimum point for the functions. Unless certain x-values are chosen, the maximum or minimum point may not be evident. *** Given the funtion: 2 f(x) = x + 4x + 6 One value of x whose f(x)-value is in the shaded area is x = -1. f(-1) = 3 The point (-1,3) is the minimum point of this function. *** Use your own words to describe how you can identify the maximum or minimum point of a quadratic function. If the appropiate values for x have been chosen, you can look at the f(x)-values of the table or graph. *** Once the maximum or minimum point of a quadratic function has been located, you can look for patterns to find other points that satisfy the function. *** Describe how the axis of symmetry can be used to predict other points that lie on the graph of a quadratic function. Any point on one side of the axis of symmetry has a mirror image on the other side of the axis. *** You have seen how maximum or minimum points and the axis of symmetry produce patterns in the graphs of quadratic functions. Now, you will learn about another pattern that allows you to complete the sketch of a graph. These points satisfy: 2 f(x) = x x | f(x) --+----- 0 | 0 1 | 1 2 | 4 3 | 9 5 | 25 For x > 0, as x increases f(x) also increases. When x > 0, f(x) = x^2 is an INCREASING FUNCTION. These points also satisfy: 2 f(x) = x x | f(x) --+----- -5| 25 -3| 9 -2| 4 -1| 1 0 | 0 For x < 0, as x increases f(x) decreases. When x < 0, f(x) = x^2 is a DECREASING FUNCTION. *** Describe how you can determine where a quadratic function is increasing or decreasing. If f(x) increases as x increases, then the function is increasing. If f(x) decreases as x increases, then the function is decreasing. *** Using the maximum or minimum point, the line of symmetry, and the interval where the function increases or decreases, you are now ready to sketch the graph of a quadratic function. The graph of any quadratic function is a PARABOLA. *** Given a quadratic function that increases for all values of x > 4 and has a minimum point, only parabola B satisfied these conditions. [B is a graph of a line opening towards the north] *** Before pressing ENTER, describe how the graph of a quadratic function is sketched. To sketch the graph, plot a few points to determine the maximum or minimum point and the line of symmetry. Decide where the function is increasing and decreasing to connect the points of the parabola. WORKSPACE: ANS: #1 Min. , #1 Min. , #2 Max. , #1 Min. , #2 Max. , #4 , #1 , 5 , 4 , 6 , 0 , 3 , 8 , 3 , -4 , 2 , #1 Inc. , #2 Dec. , #2 Dec. , #1 Inc. , #1 Inc. , #2 Dec. , 4,2 , 4 , 4 , #2 , #3 , #3 , #3 (4,1) , #4 , 4 , #2 , #4 THE CRITICAL POINT FORM: Exploring Graphs of: 2 y = a(x - h) + k - Learn how values for a, h, and k in the equations... 2 y = a(x - h) + k ...determine the shape and location of the graph NOTES: In the last activity you plotted points to graph quadratic function in STANDARD FORM, f(x) = ax^2 + bx + c. In this activity you will graph quadratic functions in the CRITICAL POINT FORM: 2 f(x) = a(x - h) + k Noting the effects of a, h, and k on the graph. *** In the next 2 explorations, you will choose values for h and k to explore their graphs. These are examples of some of the values you might try. VALUE EXAMPLE ----------------------------------------- positive integer 4 negative integer -6 fraction 2/3 decimal numeral 5.7 WORKSPACE: ANS: THE CRITICAL POINT FORM: Analyzing y = a(x - h)^2 + k - Graph quadratic equations in critical point form - Determine the critical point form of a quadratic equation from its graph NOTES: In the last activity you learned values for "a", "h", and "k" effect the graph of: 2 y = a(x - h) + k Now you will use these values to graph parabolas. Given the equation: 2 y = 2(x + 1) + 3 / | \ a = 2 h = -1 k = 3 The point (-1,3) is the vertex of the graph. *** You can determine the value of "a" and identify the vertex (h,k) from the equation y = a(x - h)^2 + k. After plotting the vertex, the value of "a" can be used to plot other points and sketch the graph. *** The vertex of: 2 y = 2(x - 3) + 4 Is (3,4) and a = 2. Since "a" is POSITIVE, the graph opens UPWARD. With a = 2, points one over and 2 up on either side of the vertex also like on the graph. 2 y = 2(x - 3) + 4 These three points will show the direction and location of the parabola. *** Use the value for a to find the points whose x-values are -1 and -3. 2 y = -(x + 2) + 5 Vertex: (-2,5) a = -1 (-1,4),(-3,4) AND: What is the vertex of? 2 y = 3(x - 3) - 4 Vertex: (3,-4) a = 3 (2,-1),(4,-1) *** Given a quadratic equation, you used the vertex and the value of "a" to plot some points and sketch the graph. Even if the quadratic equation is not known, you can sketch the graph if you know the vertex and the value of "a". *** To sketch a graph of an equation of the form: 2 y = a(x - h) + k Plot the vertex (h,k) and find the value of "a". Then use the value of "a" to determine the direction of the parabola and to plot other points. Given a quadratic equation in critical point form, you have learned to graph the parabola. Now, given the graph of a quadratic function, you can determine the values of "a", "h", and "k" to write the equation for the parabola. *** The vertex (h,k) of this graph is (3,-2). Points one over and 1 up also like on the graph so a = 1. Substituting these values, the critical point form is: 2 y = (x - 3) - 2 *** Describe how you can determine the equation of a quadratic function from its graph, then press ENTER. Locate the vertex (h,k), determine the value of a, and substitute these values in the equation: 2 y = a(x - h) + k WORKSPACE: ANS: -1 , 2,4 , 1 , -3,-2 , 3 , 0,1 , -2 , 1,0 , 1/2 , -5,-9 , -2,5 , -1 , 4,4 , 3,-4 , 3 , -1,-1 , 3 , 3 , 1 , 1 , -1 , -1 , 0 , 0 , h,k , -4,6 , -2 , -2,4,6 , #1 , #4 , #4 , #2 , 3,1 , -4,3 , 0,-2 , 1,0 , 2 , -1 , 1 , #1 , #3 , #2 THE CRITICAL POINT FORM: Transforming Quadratic Equations - Convert quadratic equations from standard form to critical point form - Build quadratic functions graphically - Illustrate algebraic transformations with graphs NOTES: In previous activities, you graphed quadratic functions using the standard form f(x) = ax^2 + bx + c and the critical point form f(x) = a(x - h)^2 + k. In this activity, you will study transformations of these functions algebraically and graphically. *** To convert equations like... 2 y = 2x + 20x + 47 ...from standard form to critical point form, you can use the method of COMPLETING THE SQUARE. First, factor out the coefficient of the x^2 term from the first two terms. 2 y = 2x + 20x + 47 2 y = 2(x + 10x) + 47 Next divide 10 by 2 and square the result. 10/2 = 5 5^2 = 25 2 y = 2(x + 10x + 25) + 47 - 50 | / / 2 * 25 = 50 2 y = 2(x + 5) - 3 The equation has been transformed from standard form to critical point form by completing the square. *** Convert this equations from standard form to critical point form. 2 y = 3x - 6x - 1 2 y = 3(x - 2x) - 1 2 y = 3(x - 2x + 1) - 1 - 3 2 y = 3(x - 1) - 4 Another: 2 y = -4x + 24x - 35 2 y = -4(x - 6x) - 35 2 y = -4(x - 6x + 9) - 35 + 36 2 y = -4(x - 3) + 1 *** Before pressing ENTER, use your own words to describe how to transform a quadratic equation from standard form to critical point form. To transform: 2 y = ax + bx + c Into: 2 y = a(x - h) + k Complete the square and simplify. *** Notice, in standard form a quadratic equation is the sum of three terms: 2 ax + bx + c || || | ||2 || | ax bx c ...[animation]... You can build quadratic expressions algebraically by adding the terms. 2 3x - 4x + 8 You can also build the graph of a quadratic function by adding the graphs of its terms. This is the graph of a quadratic term. 2 y = x 1 This is a graph of a constant term. y = 3 2 And... 2 y = x 1 y = 3 2 For x = -2, the sum of the two y-values is 7. For x = 0, the sum of the two y-values is 3. For x = 1, the sum of the two y-values is 4. If the two y-values are added for each x, the resulting graph is the graph of: 2 y = x + 3 *** Desribe how the graph of a quadratic function can be built by adding the graphs of its terms. Sketch the graph of each term. Add the y-values for any given x. These sums determine the graph of the sum of the terms. *** In Unit 1, you transformed linear equations and graphed each step of the transformation. Now you will transform and graph quadratic equations. As you perform the algebraic transformations, the graph shows the geometric transformations. CONTINUE HERE. WORKSPACE: ANS: 2 , 1,3 , 1,4 , 6 , 9,36 , 3,1 , 2 , 5 , 0 , -3 , -4 , -3 , 0 , -2 , -2 , 0 , 4 ,