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 ,