Fireworks – Finding Intercepts

• The vertex is important, but it's not the only important point on a parabola

Vertex at (3, -8)

x-intercepts at (1,0) and (5, 0)

y-intercept at (0, 10)

Fireworks – Finding Intercepts

• In addition to telling us where the vertex is located the vertex form can also help us find the x-intercepts of the parabola. Just set y = 0, and solve for x.

y = (x + 9)2 – 16

0 = (x + 9)2 – 16

16 = (x + 9)2

2)9(16 x

= x + 9 = x + 9

-5= x -13 = x

Add 16 to both sides

Take square root of both sides

Subtract 9 from both sides

x-intercepts at x = -5 and x = –13

4 4

Fireworks – Finding Intercepts

• Another example, this time the parabola is concave down.

y = –(x – 7)2 + 3

0 = –(x – 7)2 + 3–3 = –(x – 7)2

2)7(3 x

1.732 = x – 7 –1.732 = x – 7

8.732 = x 5.268 = x

Subtract 3 from both sides

Divide both sides by -1

Add 7 to both sides

x-intercepts at 5.268 and 8.732

3 = (x – 7)2

Take square root of both sides

Fireworks – Finding Intercepts

• Another example, this time the a value is 0.5.

y = 0.5(x + 3)2 + 5

0 = 0.5(x + 3)2 + 5–5 = 0.5(x + 3)2

2)3(10 x

Error = x Error = x

Subtract 5 from both sides

Divide both sides by 0.5

NO x-intercepts… can't take square root of a negative number.

–10 = (x + 3)2

Take square root of both sides

Fireworks – Finding Intercepts

• Find the x-intercepts of the parabola for each of the quadratic equations.

1. y = (x – 7)2 – 9

2. y = 3(x + 4)2 + 6

3. y = –0.5(x – 2)2 + 10

• Is there a way to tell how many x-intercepts a parabola will have without doing any calculations?

x-intercepts at 10 and 4

NO x-intercepts

x-intercepts at 6.472 and –2.472

Fireworks – Finding Intercepts

• Finding the y-intercept is a little more straightforward. Just set x = 0 and solve for y.

y = (x + 4)2 – 6 y = (0 + 4)2 – 6 y = 10

y-intercept at (0, 10)

• The quadratic equation does not have to be vertex form to find the y-intercept.

y = x2 + 8x + 10 y = (0)2 + 8(0) + 10

y = 10

y-intercept at (0, 10)