1. Lesson 4.2B

2. Review Graph y = -4x2 + 8x + 2 vertex (1,6) y-intercept (0,2)

3. Let’s try to graph y = (x – 8)(x – 2).
Where does this graph cross the x-axis?
x = 8 and x = 2
What is the x-coordinate of the vertex?
x = 5
What are the coordinates of the vertex?
(5,-9)
What is the y-intercept?
(0,16)

4. Graphing Quadratics in Intercept Form
Intercept form y = a(x – p)(x – q) The x-intercepts are p and q. The vertex is (p+q, ?) 2 The y-intercept is (0,?)

5. Practice
1. y = (x – 3)(x – 7) x-int(s) (3,0) (7,0) vertex (5,-4) y-int (0,21) 2. y = 2(x + 3)(x – 1) x-int(s) (-3,0) (1,0) vertex (-1,-8) y-int (0,-6) 3. y = -(x + 1)(x – 5) x-int(s) (-1,0) (5,0) vertex (2,9) y-int (0,5)

6. One last problem…
Graph y = 2x(x + 4)
What form is this?

7. Looking ahead… 1. (x + 4)(x + 3) = 2. (x – 5)(x + 1) =

8. Review y = (x – 2)2 – 3 y = x2 + 4x + 5 y = 2(x – 1)(x + 5)

9. Does y = 2(x + 5)2 – 8 have a maximum or minimum value? What is it?

10. Applications
The height h of a volleyball t seconds after being hit is h = -16t2 + 48t + 4.
What is the maximum height of the volleyball? From what height was the volleyball hit?