1. Warm Up x = 0 x = 1 (–2, 1) (0, 2) Find the axis of symmetry.
1. y = 4x2 – y = x2 – 3x + 1
3. y = –2x2 + 4x y = –2x2 + 3x – 1
Find the vertex.
5. y = x2 + 4x y = 3x2 + 2
7. y = 2x2 + 2x – 8
x = 0
x = 1
(–2, 1)
(0, 2)

2. Factor each trinomial. Check your answer.
x2 + 6x + 5

3. Objective
Graph a quadratic function in the form y = ax2 + bx + c.

4. Recall that a y-intercept is the y-coordinate of the point where a graph intersects the y-axis. The x-coordinate of this point is always 0. For a quadratic function written in the form
y = ax2 + bx + c, when x = 0, y = c. So the y-intercept of a quadratic function is c.

5. Graph y = 3x2 – 6x + 1.
Use x =
Find the axis of symmetry.
The axis of symmetry is x = 1.
= 1
The vertex is (1, –2).
Find the vertex.
y = 3(1)2 – 6(1) + 1
= 3 – 6 + 1
= –2
Find the y-intercept.
(0, 1).

6. Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry as its reflected point.
Helpful Hint

7. Use x =
Graph y = 2x2 + 6x + 2
The axis of symmetry is x

8. Graph y + 6x = x2 + 9
Rewrite in standard form.
Use x =
y = x2 – 6x + 9
= 3
The axis of symmetry is x = 3.
y = 32 – 6(3) + 9
= 9 –
= 0

9. = –16(1)2 + 32(1) = –16 + 32 = 16 f(x) = –16x2 + 32x + 0
The height in feet of a basketball that is shot can be modeled by f(x) = –16x2 + 32x, where x is the time in seconds after it is thrown. Find the basketball’s maximum height and the time it takes the basketball to reach this height. Then find how long the basketball is in the air.
= –16(1)2 + 32(1)
= –
= 16
f(x) = –16x2 + 32x + 0

10. The vertex is the highest or lowest point on a parabola
The vertex is the highest or lowest point on a parabola. Therefore, in the example, it gives the maximum height of the basketball.
Remember!