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Splash Screen

Five-Minute Check (over Lesson 4–6) CCSS Then/Now New Vocabulary Example 1: Write Functions in Vertex Form Example 2: Standardized Test Example: Write an Equation Given a Graph Concept Summary: Transformations of Quadratic Functions Example 3: Graph Equations in Vertex Form Lesson Menu

Find the value of the discriminant for the equation 5x 2 – x – 4 = 0. B. –21 C. 21 D. 81 5-Minute Check 1

Find the value of the discriminant for the equation 5x 2 – x – 4 = 0. B. –21 C. 21 D. 81 5-Minute Check 1

Describe the number and type of roots for the equation 5x 2 – x – 4 = 0. A. 2 irrational roots B. 2 real, rational roots C. 1 real, rational root D. 2 complex roots 5-Minute Check 2

Describe the number and type of roots for the equation 5x 2 – x – 4 = 0. A. 2 irrational roots B. 2 real, rational roots C. 1 real, rational root D. 2 complex roots 5-Minute Check 2

Find the exact solutions of 5x 2 – x – 4 = 0 by using the Quadratic Formula. 5-Minute Check 3

Find the exact solutions of 5x 2 – x – 4 = 0 by using the Quadratic Formula. 5-Minute Check 3

Solve x 2 – 9x + 21 = 0 by using the Quadratic Formula. 5-Minute Check 4

Solve x 2 – 9x + 21 = 0 by using the Quadratic Formula. 5-Minute Check 4

Solve 2x 2 + 4x = 10 by using the Quadratic Formula. 5-Minute Check 5

Solve 2x 2 + 4x = 10 by using the Quadratic Formula. 5-Minute Check 5

The function h(t) = –16t 2 + 70t + 6 models the height h in feet of an arrow shot into the air at time t in seconds. How long, to the nearest tenth, does it take for the arrow to hit the ground? A. about 3.2 s B. about 4.5 s C. about 5.1 s D. about 5.5 s 5-Minute Check 6

The function h(t) = –16t 2 + 70t + 6 models the height h in feet of an arrow shot into the air at time t in seconds. How long, to the nearest tenth, does it take for the arrow to hit the ground? A. about 3.2 s B. about 4.5 s C. about 5.1 s D. about 5.5 s 5-Minute Check 6

Mathematical Practices 7 Look for and make use of structure. Content Standards F.IF.8.a Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. F.BF.3 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f (kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Mathematical Practices 7 Look for and make use of structure. CCSS

You transformed graphs of functions. Write a quadratic function in the form y = a(x – h)2 + k. Transform graphs of quadratic functions of the form y = a(x – h)2 + k. Then/Now

vertex form Vocabulary

A. Write y = x 2 – 2x + 4 in vertex form. Write Functions in Vertex Form A. Write y = x 2 – 2x + 4 in vertex form. y = x2 – 2x + 4 Notice that x2 – 2x + 4 is not a perfect square. y = (x2 – 2x + 1) + 4 – 1 Balance this addition by subtracting 1. Example 1

y = (x – 1)2 + 3 Write x2 – 2x + 1 as a perfect square. Write Functions in Vertex Form y = (x – 1)2 + 3 Write x2 – 2x + 1 as a perfect square. Answer: Example 1

y = (x – 1)2 + 3 Write x2 – 2x + 1 as a perfect square. Write Functions in Vertex Form y = (x – 1)2 + 3 Write x2 – 2x + 1 as a perfect square. Answer: y = (x – 1)2 + 3 Example 1

y = –3x2 – 18x + 10 Original equation Write Functions in Vertex Form B. Write y = –3x 2 – 18x + 10 in vertex form. Then analyze the function. y = –3x2 – 18x + 10 Original equation y = –3(x2 + 6x) + 10 Group ax2 + bx and factor, dividing by a. y = –3(x2 + 6x + 9) + 10 – (–3)(9) Complete the square by adding 9 inside the parentheses. Balance this addition by subtracting –3(9). Example 1

y = –3(x + 3)2 + 37 Write x2 + 6x + 9 as a perfect square. Write Functions in Vertex Form y = –3(x + 3)2 + 37 Write x2 + 6x + 9 as a perfect square. Answer: Example 1

y = –3(x + 3)2 + 37 Write x2 + 6x + 9 as a perfect square. Write Functions in Vertex Form y = –3(x + 3)2 + 37 Write x2 + 6x + 9 as a perfect square. Answer: y = –3(x + 3)2 + 37 Example 1

A. Write y = x 2 + 6x + 5 in vertex form. A. y = (x + 6)2 – 31 B. y = (x – 3)2 + 14 C. y = (x + 3)2 + 14 D. y = (x + 3)2 – 4 Example 1

A. Write y = x 2 + 6x + 5 in vertex form. A. y = (x + 6)2 – 31 B. y = (x – 3)2 + 14 C. y = (x + 3)2 + 14 D. y = (x + 3)2 – 4 Example 1

B. Write y = –3x 2 – 18x + 4 in vertex form. A. y = –3(x + 3)2 – 23 B. y = –3(x + 3)2 + 31 C. y = –3(x – 3)2 – 23 D. y = –3(x – 3)2 + 31 Example 1

B. Write y = –3x 2 – 18x + 4 in vertex form. A. y = –3(x + 3)2 – 23 B. y = –3(x + 3)2 + 31 C. y = –3(x – 3)2 – 23 D. y = –3(x – 3)2 + 31 Example 1

Which is an equation of the function shown in the graph? Write an Equation Given a Graph Which is an equation of the function shown in the graph? Example 2

Write an Equation Given a Graph Read the Test Item You are given a graph of a parabola. You need to find an equation of the parabola. Solve the Test Item The vertex of the parabola is at (2, –3), so h = 2 and k = –3. Since the graph passes through (0, –1), let x = 0 and y = –1. Substitute these values into the vertex form of the equation and solve for a. Example 2

Substitute –1 for y, 0 for x, 2 for h, and –3 for k. Write an Equation Given a Graph Vertex form Substitute –1 for y, 0 for x, 2 for h, and –3 for k. Simplify. Add 3 to each side. Divide each side by 4. Example 2

The equation of the parabola in vertex form is Write an Equation Given a Graph The equation of the parabola in vertex form is Answer: Example 2

The equation of the parabola in vertex form is Write an Equation Given a Graph The equation of the parabola in vertex form is Answer: The answer is B. Example 2

Which is an equation of the function shown in the graph? B. C. D. Example 2

Which is an equation of the function shown in the graph? B. C. D. Example 2

Concept

Step 1 Rewrite the equation in vertex form. Graph Equations in Vertex Form Graph y = –2x 2 + 4x + 1. Step 1 Rewrite the equation in vertex form. y = –2x 2 + 4x + 1 Original equation y = –2(x 2 – 2x) + 1 Distributive Property y = –2(x 2 – 2x + 1) + 1 – (–2)(1) Complete the square. y = –2(x – 1)2 + 3 Simplify. Example 3

Step 3 Plot additional points to help you complete the graph. Graph Equations in Vertex Form Step 2 The vertex is at (1, 3). The axis of symmetry is x = 1. Because a = –2, the graph opens down and is narrower than the graph of y = –x 2. Step 3 Plot additional points to help you complete the graph. Answer: Example 3

Step 3 Plot additional points to help you complete the graph. Graph Equations in Vertex Form Step 2 The vertex is at (1, 3). The axis of symmetry is x = 1. Because a = –2, the graph opens down and is narrower than the graph of y = –x 2. Step 3 Plot additional points to help you complete the graph. Answer: Example 3

Graph y = 3x 2 + 12x + 8. A. B. C. D. Example 3

Graph y = 3x 2 + 12x + 8. A. B. C. D. Example 3

End of the Lesson