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Lesson Menu 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

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Over Lesson 4–6 5-Minute Check 1 A.–81 B.–21 C.21 D.81 Find the value of the discriminant for the equation 5x 2 – x – 4 = 0.

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Over Lesson 4–6 5-Minute Check 2 A.2 irrational roots B.2 real, rational roots C.1 real, rational root D.2 complex roots Describe the number and type of roots for the equation 5x 2 – x – 4 = 0.

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Over Lesson 4–6 5-Minute Check 3 Find the exact solutions of 5x 2 – x – 4 = 0 by using the Quadratic Formula. A.1, – B. C.–1, D.

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Over Lesson 4–6 A. B. C. D. 5-Minute Check 4 Solve x 2 – 9x + 21 = 0 by using the Quadratic Formula.

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Over Lesson 4–6 5-Minute Check 5 Solve 2x 2 + 4x = 10 by using the Quadratic Formula. A. B. C. D.

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Over Lesson 4–6 5-Minute Check 6 A.about 3.2 s B.about 4.5 s C.about 5.1 s D.about 5.5 s 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?

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CCSS 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.

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Then/Now 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.

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Vocabulary vertex form

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

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

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

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

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

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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

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

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Example 2 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. Write an Equation Given a Graph

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Example 2 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. Write an Equation Given a Graph

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

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

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Concept

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

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Example 3 Graph Equations in Vertex Form Step 2The 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 3Plot additional points to help you complete the graph. Answer:

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

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End of the Lesson

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