1. Fireworks – Finding Intercepts
• The vertex is important, but it's not the only important
point on a parabola
y-intercept at (0, 10)
x-intercepts at (1,0) and (5, 0)
Vertex at (3, -8)

2. 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
Add 16 to both sides
16 = (x + 9)2
Take square root of both sides
Subtract 9 from both sides
= x + 9
= x + 9
-5= x
-13 = x
x-intercepts at x = -5 and x = –13

3. Fireworks – Finding Intercepts
• Another example, this time the parabola is concave down.
y = –(x – 7)2 + 3
0 = –(x – 7)2 + 3
Subtract 3 from both sides
–3 = –(x – 7)2
Divide both sides by -1
3 = (x – 7)2
Take square root of both sides
1.732 = x – 7
–1.732 = x – 7
Add 7 to both sides
8.732 = x
5.268 = x
x-intercepts at and 8.732

4. 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
Subtract 5 from both sides
–5 = 0.5(x + 3)2
Divide both sides by 0.5
–10 = (x + 3)2
Take square root of both sides
NO x-intercepts… can't take square root of a negative number.
Error = x
Error = x

5. Fireworks – Finding Intercepts
• Find the x-intercepts of the parabola for each of the quadratic equations.
1. y = (x – 7)2 – 9
x-intercepts at 10 and 4
2. y = 3(x + 4)2 + 6
NO x-intercepts
3. y = –0.5(x – 2)2 + 10
x-intercepts at and –2.472
• Is there a way to tell how many x-intercepts a parabola will have without doing any calculations?

6. 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-intercept at (0, 10)
y = 10
• The quadratic equation does not have to be vertex form to find the y-intercept.
y = x2 + 8x + 10
y-intercept at (0, 10)
y = (0)2 + 8(0) + 10
y = 10