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

2. 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.

3. Example 1: Graphing a Quadratic Function
Graph y = 3x2 – 6x + 1.
Step 1 Find the axis of symmetry.
Use x = Substitute 3 for a and –6 for b.
= 1
Simplify.
The axis of symmetry is x = 1.
Step 2 Find the vertex.
y = 3x2 – 6x + 1
The x-coordinate of the vertex is 1. Substitute 1 for x.
= 3(1)2 – 6(1) + 1
= 3 – 6 + 1
Simplify.
= –2
The y-coordinate is –2.
The vertex is (1, –2).

4. Example 1 Continued
Step 3 Find the y-intercept.
y = 3x2 – 6x + 1
y = 3x2 – 6x + 1
Identify c.
The y-intercept is 1; the graph passes through (0, 1).

5. Example 1 Continued
Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept.
Since the axis of symmetry is x = 1, choose x-values less than 1.
Substitute
x-coordinates.
Let x = –1.
Let x = –2.
y = 3(–1)2 – 6(–1) + 1
y = 3(–2)2 – 6(–2) + 1 = =
Simplify.
= 10
= 25
Two other points are (–1, 10) and (–2, 25).

6. Example 1 Continued Graph y = 3x2 – 6x + 1.
Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.
Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
x = 1
(–2, 25)
(–1, 10)
(0, 1)
(1, –2)
x = 1
(–1, 10)
(0, 1)
(1, –2)
(–2, 25)

7. Check It Out! Example 1b
Graph the quadratic function.
y + 6x = x2 + 9
y = x2 – 6x + 9
Rewrite in standard form.
Step 1 Find the axis of symmetry.
Use x = Substitute 1 for a and –6 for b.
= 3
Simplify.
The axis of symmetry is x = 3.

8. Check It Out! Example 1b Continued
Step 2 Find the vertex.
y = x2 – 6x + 9
The x-coordinate of the vertex is 3. Substitute 3 for x.
y = 32 – 6(3) + 9
= 9 –
Simplify.
= 0
The y-coordinate is
The vertex is (3, 0).

9. Check It Out! Example 1b Continued
Step 3 Find the y-intercept.
y = x2 – 6x + 9
y = x2 – 6x + 9
Identify c.
The y-intercept is 9; the graph passes through (0, 9).

10. Check It Out! Example 1b Continued
Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y- intercept.
Since the axis of symmetry is x = 3, choose x-values less than 3.
Let x = 2
Let x = 1
y = 1(2)2 – 6(2) + 9
Substitute
x-coordinates.
y = 1(1)2 – 6(1) + 9
= 4 –
= 1 – 6 + 9
Simplify.
= 1
= 4
Two other points are (2, 1) and (1, 4).

11. Check It Out! Example 1b Continued
y = x2 – 6x + 9
Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.
Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
x = 3
(3, 0)
(0, 9)
(2, 1)
(1, 4)
(0, 9)
(1, 4)
(2, 1)
x = 3
(3, 0)