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)

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

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 5.268 and 8.732

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

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 6.472 and –2.472 • Is there a way to tell how many x-intercepts a parabola will have without doing any calculations?

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