Chapter 9 - Quadratic Functions and Equations Algebra I

Table of Contents 9.5 – Solving Quadratic Equations by Graphing 9.6 – Solving Quadratic Equations by Factoring 9.7 – Solving Quadratic Equations by Using Square Roots 9.8 - Completing the Square 9.9 – The Quadratic Formula and Discriminate

9.5 - Solving Quadratic Equations by Graphing Algebra I

9.5 Algebra 1 (bell work) One way to solve a quadratic equation in standard form is to graph the related function and find the x-values where y = 0. In other words, find the zeros of the related function. Recall that a quadratic function may have two, one, or no zeros.

9.5 Example 1 Solving Quadratic Equations by Graphing Solve the equation by graphing the related function. x = 0 2x2 – 18 = 0 ● ● (3, 0) Step 1 Write the related function. ● ● 2x2 – 18 = y, or y = 2x2 + 0x – 18 (2, –10) Step 2: Graph Axis of Symmetry Vertex Y - Intercept Two Other Points ● (0, –18) The axis of symmetry is x = 0. The vertex is (0, –18). Two other points (2, –10) and (3, 0) Graph the points and reflect them across the axis of symmetry.

Graph y = 2x2 + 0x – 18 9.5

9.5 Solve the equation by graphing the related function. 2x2 – 18 = 0 From Previous Slides Step 3 Find the zeros. The zeros appear to be 3 and –3. 0 0 Check 2x2 – 18 = 0 2(3)2 – 18 0 2(9) – 18 0 18 – 18 0  2x2 – 18 = 0 2(–3)2 – 18 0 2(9) – 18 0 18 – 18 0 

9.5 Solve the equation by graphing the related function. x = 3 –12x + 18 = –2x2 (3, 0) ● ● ● Step 1 Write the related function. y = –2x2 + 12x – 18 (4, –2) ● (5, –8) Step 2: Graph Axis of Symmetry Vertex Y - Intercept Two Other Points ● The axis of symmetry is x = 3. The vertex is (3, 0). Two other points (5, –8) and (4, –2). Graph the points and reflect them across the axis of symmetry.

Graph y = –2x2 + 12x – 18 9.5

9.5 Math Joke Q: What does a person have if they can easily find solutions to a quadratic equation? A: Zero Visibility

9.5 Solve the equation by graphing the related function. Optional x = –4 x2 – 8x – 16 = 2x2 Step 1 Write the related function. ● ● (–2 , 4) y = x2 + 8x + 16 Step 2: Graph Axis of Symmetry Vertex Y - Intercept Two Other Points ● ● (–3, 1) ● (–4, 0) The axis of symmetry is x = 0. The vertex is (0, –18). Two other points (2, –10) and (3, 0) Graph the points and reflect them across the axis of symmetry.

A frog jumps straight up from the ground. 9.5 Example 2 Application TI Emulator Software A frog jumps straight up from the ground. The quadratic function f(t) = –16t2 + 12t models the frog’s height above the ground after t seconds. About how long is the frog in the air? When the frog leaves the ground, its height is 0, and when the frog lands, its height is 0. So solve 0 = –16t2 + 12t to find the times when the frog leaves the ground and lands. Step 1 Write the related function 0 = –16t2 + 12t y = –16t2 + 12t The zeros appear to be 0 and 0.75. The frog leaves the ground at 0 seconds and lands at 0.75 seconds. The frog is off the ground for about 0.75 seconds.

What if…? A dolphin jumps out of the water. 9.5 What if…? A dolphin jumps out of the water. The quadratic function y = –16x2 + 32 x models the dolphin’s height above the water after x seconds. About how long is the dolphin out of the water? When the dolphin leaves the water, its height is 0, and when the dolphin reenters the water, its height is 0. So solve 0 = –16x2 + 32x to find the times when the dolphin leaves and reenters the water. Step 1 Write the related function 0 = –16x2 + 32x y = –16x2 + 32x The zeros appear to be 0 and 2. The dolphin leaves the water at 0 seconds and reenters at 2 seconds. The dolphin is out of the water for about 2 seconds.

HW pg. 645 9.5 3-7 (Odd, By hand), 14, 25-31, 49-57 (Odd)

9.6 - Solving Quadratic Equations by Factoring Algebra I

9.6 Algebra 1 (bell work)

(x – 7)(x + 2) = 0   Check (x – 7)(x + 2) = 0 9.6 Example 1 Using the Zero Product Property Use the Zero Product Property to solve the equation. Check your answer. (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 x = 7 or x = –2 The solutions are 7 and –2. Check (x – 7)(x + 2) = 0 (–2 – 7)(–2 + 2) 0 (–9)(0) 0 0 0  Check (x – 7)(x + 2) = 0 (7 – 7)(7 + 2) 0 (0)(9) 0 0 0 

9.6 Use the Zero Product Property to solve each equation. Check your answer. (x – 2)(x) = 0 (x)(x – 2) = 0 x = 0 or x – 2 = 0 x = 2 The solutions are 0 and 2. Check (x – 2)(x) = 0 (0 – 2)(0) 0 (–2)(0) 0 0 0  (x – 2)(x) = 0 (2 – 2)(2) 0 (0)(2) 0 0 0 

9.6 Use the Zero Product Property to solve each equation. Check your answer. Optional (x)(x + 4) = 0 x = 0 or x + 4 = 0 x = –4 The solutions are 0 and –4. Check (x)(x + 4) = 0 (0)(0 + 4) 0 (0)(4) 0 0 0  (x)(x +4) = 0 (–4)(–4 + 4) 0 (–4)(0) 0 0 0 

If a quadratic equation is written in standard form, ax2 + bx + c = 0 9.6 If a quadratic equation is written in standard form, ax2 + bx + c = 0 Then to solve the equation, you may need to factor before using the Zero Product Property.

Math Joke Citizen: Why is the quadratic equation lurking in our town? 9.6 Math Joke Citizen: Why is the quadratic equation lurking in our town? Sheriff: He’s searching for his roots

9.6 Example 2 Solving Quadratic Functions by Factoring Solve the quadratic equation by factoring. Check your answer. x2 – 6x + 8 = 0 (x – 4)(x – 2) = 0 x – 4 = 0 or x – 2 = 0 x = 4 or x = 2 The solutions are 4 and 2. x2 – 6x + 8 = 0 (4)2 – 6(4) + 8 0 16 – 24 + 8 0 0 0  Check x2 – 6x + 8 = 0 (2)2 – 6(2) + 8 0 4 – 12 + 8 0 0 0  Check

9.6 Solve the quadratic equation by factoring. Check your answer. x2 + 4x = 21 –2x2 = 20x + 50 +2x2 +2x2 0 = 2x2 + 20x + 50 –2x2 = 20x + 50 x2 + 4x = 21 –21 –21 x2 + 4x – 21 = 0 2x2 + 20x + 50 = 0 (x + 7)(x –3) = 0 2(x2 + 10x + 25) = 0 x + 7 = 0 or x – 3 = 0 2(x + 5)(x + 5) = 0 x = –7 or x = 3 The solutions are –7 and 3. 2 ≠ 0 or x + 5 = 0 x = –5

9.6 Solve the quadratic equation by factoring. Check your answer. 30x = –9x2 – 25 3x2 – 4x + 1 = 0 –9x2 – 30x – 25 = 0 (3x – 1)(x – 1) = 0 –1(9x2 + 30x + 25) = 0 3x – 1 = 0 or x – 1 = 0 –1(3x + 5)(3x + 5) = 0 or x = 1 –1 ≠ 0 or 3x + 5 = 0 The solutions are and x = 1.

h = –16t2 + 8t + 8 0 = –16t2 + 8t + 8 0 = –8(2t2 – t – 1) 9.6 Example 3 Application The height in feet of a diver above the water can be modeled by h(t) = –16t2 + 8t + 8, where t is time in seconds after the diver jumps off a platform. Find the time it takes for the diver to reach the water. h = –16t2 + 8t + 8 0 = –16t2 + 8t + 8 0 = –8(2t2 – t – 1) 0 = –8(2t + 1)(t – 1)

  9.6 –8 ≠ 0, 2t + 1 = 0 or t – 1= 0 Use the Zero Product Property. Solve each equation. 2t = –1 or t = 1  Since time cannot be negative, does not make sense in this situation. It takes the diver 1 second to reach the water. Check 0 = –16t2 + 8t + 8 0 –16(1)2 + 8(1) + 8 0 –16 + 8 + 8 0 0 Substitute 1 into the original equation. 

9.6 What if…? The equation for the height above the water for another diver can be modeled by h = –16t2 + 8t + 24. Find the time it takes this diver to reach the water. Optional The diver reaches the water when h = 0. h = –16t2 + 8t + 24 Factor out the GCF, –8. 0 = –16t2 + 8t + 24 Factor the trinomial. 0 = –8(2t2 – t – 3) 0 = –8(2t – 3)(t + 1)

  9.6 –8 ≠ 0, 2t – 3 = 0 or t + 1= 0 Use the Zero Product Property. Solve each equation. Since time cannot be negative, –1 does not make sense in this situation. t = 1.5 It takes the diver 1.5 seconds to reach the water. Check 0 = –16t2 + 8t + 24 Substitute 1 into the original equation. 0 –16(1.5)2 + 8(1.5) + 24 0 –36 + 12 + 24 0 0 

HW pg. 653 9.6- 1-19 (Odd), 46, 63-67 (Odd) Ch: 40-45, 50-61

9.7 - Solving Quadratic Equations by Using Square Roots Algebra I

9-7 Algebra 1 (bell work)

Positive and negative Square roots of 9 9.7 When you take the square root of a positive number and the sign of the square root is not indicated. You must find both the positive and negative square root. This is indicated by ±√ Positive Square root of 9 Negative Square root of 9 Positive and negative Square roots of 9

9.7 Example 1 Using Square Roots to solve x2 = a Solve using square roots. Check your answer. x2 = 169 x2 = 0 x = 0 x = ± 13 The solution is 0. The solutions are 13 and –13. x2 = –49 There is no real solution.

9.7 Example 2 Using Square Roots to Solve Quadratic Equations Solve using square roots. x2 + 7 = 7 16x2 – 49 = 0 –7 –7 x2 + 7 = 7 x2 = 0 16x2 – 49 = 0 +49 +49 The solution is 0.

9.7 Solve by using square roots. Check your answer. 100x2 + 49 = 0 36x2 = 1 100x2 + 49 = 0 –49 –49 100x2 =–49 . There is no real solution.

Math Joke Monster: Why did you rip up my garden? 9.7 Math Joke Monster: Why did you rip up my garden? Student: My math assignment is to find roots

–3x2 + 90 = 0 9.7 Solve. Round to the nearest hundredth. –3x2 + 90 = 0 –90 –90 x2 = 30 x  5.48 The approximate solutions are 5.48 and –5.48.

2x2 – 64 = 0 9.7 Optional Solve. Round to the nearest hundredth. + 64 + 64 x2 = 32 The approximate solutions are 5.66 and –5.66.

9.7 Example 4 Application Ms. Pirzada is building a retaining wall along one of the long sides of her rectangular garden. The garden is twice as long as it is wide. It also has an area of 578 square feet. What will be the length of the retaining wall? Let x represent the width of the garden. lw = A l = 2x , W=x 2x x = 578 ● 2x2 = 578 x = ± 17

HW pg. 659 9.7- 1-15 (Odd), 16, 36-38, 61-65 Ch: 41-53, 57-60

9.8 - Completing the Square Algebra I

9.8

9.8 In the previous lesson, you solved quadratic equations by isolating x2 and then using square roots. This method works if the quadratic equation, when written in standard form, is a perfect square. When a trinomial is a perfect square, there is a relationship between the coefficient of the x-term and the constant term. X2 + 6x + 9 x2 – 8x + 16

An expression in the form x2 + bx is not a perfect square. 9.8 An expression in the form x2 + bx is not a perfect square. However, you can use the relationship shown above to add a term to x2 + bx to form a trinomial that is a perfect square. This is called completing the square.

A. x2 + 2x + B. x2 – 6x + 9.8 Example 1 Completing the Square Complete the square to form a perfect square trinomial. A. x2 + 2x + B. x2 – 6x + x2 + 2x x2 + –6x x2 + 2x + 1 x2 – 6x + 9

9.8 Complete the square to form a perfect square trinomial. a. x2 + 12x + b. x2 – 5x + x2 + 12x x2 + –5x x2 – 5x + x2 + 12x + 36

Pg. 664 Step 1: Put in Standard Form 9.8 Solving a Quadratic Equation by Completing the Square Step 1: Put in Standard Form Step 2: Find and add it to both sides Step 3: Factor the trinomial and solve Pg. 664

9.8 Example 2 Solving x2 + bx = c by Completing the Square Solve by completing the square. x2 + 16x = –15 x2 – 4x – 6 = 0 Step 1 x2 + (–4x) = 6 Step 1 x2 + 16x = –15 Step 2 Step 2 Step 3 x2 – 4x + 4 = 6 + 4 Step 3 x2 + 16x + 64 = –15 + 64 Step 4 (x – 2)2 = 10 Step 4 (x + 8)2 = 49 Step 5 x – 2 = ± √10 Step 5 x + 8 = ± 7 Step 6 x + 8 = 7 or x + 8 = –7 x = –1 or x = –15 Step 6 x – 2 = √10 or x – 2 = –√10 x = 2 + √10 or x = 2 – √10

9.8 Example 3 Solving ax2 + bx = c by Completing the Square Day 2 Solve by completing the square. –3x2 + 12x – 15 = 0 a must = 1 Step 1 x2 – 4x + 5 = 0 x2 – 4x = –5 x2 + (–4x) = –5 Step 2 Step 3 x2 – 4x + 4 = –5 + 4 Step 4 (x – 2)2 = –1 There is no real number whose square is negative, so there are no real solutions.

5x2 + 19x = 4 9.8 Solve by completing the square. Step 4 Step 1 Step 5 The solutions are and –4.

3x2 – 5x – 2 = 0 9.8 Solve by completing the square. Step 1 Step 4 −

HW pg. 667 9.8- Day 1: 3-9 (Odd), 21-25 (Odd), 77-87 (Odd) Ch: 47-49, 59-61, 67-72

9.9 - The Quadratic Formula and Discriminant Algebra I

9.9 Algebra 1 (bell work)

6x2 + 5x – 4 = 0 9.9 Example 1 Using the Quadratic Formula Solve using the Quadratic Formula. 6x2 + 5x – 4 = 0 6x2 + 5x + (–4) = 0

x2 = x + 20 9.9 Solve using the Quadratic Formula. x = 5 or x = –4

–3x2 + 5x + 2 = 0 9.9 Solve using the Quadratic Formula. x = – or x = 2

2 – 5x2 = –9x 9.9 Solve using the Quadratic Formula. or x = 2

9.9 Example 2 Using the Quadratic Formula to Estimate Solutions Solve using the Quadratic Formula. x2 + 3x – 7 = 0 Use a calculator: x ≈ 1.54 or x ≈ –4.54.

2x2 – 8x + 1 = 0 9.9 Solve using the Quadratic Formula. Use a calculator: x ≈ 3.87 or x ≈ 0.13.

9.9 Algebra 1 (bell work) Day 2

3x2 – 2x + 2 = 0 2x2 + 11x + 12 = 0 x2 + 8x + 16 = 0 b2 – 4ac is zero. 9.9 Example 3 Using the Discriminant Find the number of solutions of each equation using the discriminant. A. B. C. 3x2 – 2x + 2 = 0 2x2 + 11x + 12 = 0 x2 + 8x + 16 = 0 a = 3, b = –2, c = 2 a = 2, b = 11, c = 12 a = 1, b = 8, c = 16 b2 – 4ac b2 – 4ac b2 – 4ac 112 – 4(2)(12) 82 – 4(1)(16) (–2)2 – 4(3)(2) 64 – 64 121 – 96 4 – 24 25 –20 b2 – 4ac is zero. There is one real solution. b2 – 4ac is negative. There are no real solutions. b2 – 4ac is positive. There are two real solutions.

9.9 Example 4 Application The height h in feet of an object shot straight up with initial velocity v in feet per second is given by h = –16t2 + vt + c, where c is the initial height of the object above the ground. The ringer on a carnival strength test is 2 feet off the ground and is shot upward with an initial velocity of 30 feet per second. Will it reach a height of 20 feet? Use the discriminant to explain your answer.

9.9 h = –16t2 + vt + c 20 = –16t2 + 30t + 2 Substitute 20 for h, 30 for v, and 2 for c. 0 = –16t2 + 30t + (–18) Subtract 20 from both sides. b2 – 4ac Evaluate the discriminant. 302 – 4(–16)(–18) = –252 Substitute –16 for a, 30 for b, and –18 for c. The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.

9.9 What if…? Suppose the weight is shot straight up with an initial velocity of 20 feet per second from 1 foot above the ground. Will it ring the bell? Use the discriminate to explain your answer. h = –16t2 + vt + c 20 = –16t2 + 20t + 1 Substitute 20 for h, 20 for v, and 1 for c. Subtract 20 from both sides. 0 = –16t2 + 20t + (–19) Evaluate the discriminant. b2 – 4ac Substitute –16 for a, 20 for b, and –19 for c. 202 – 4(–16)(–19) = –816 The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.

9.9

HW pg. 675 9.9- Day 1: 3-13 (Odd), 24-29 (Any Method), 55 Ch: 39, 52-54, 60, 61