Fireworks Finding Intercepts The vertex is important but

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 16 = (x + 9)2 Add 16 to both sides Take square root of both sides = x+ 9 -5= x = x+ 9 -13 = x Subtract 9 from both sides 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 8. 732 = x 5. 268 = x Add 7 to both sides 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 – 5 = 0. 5(x + 3)2 – 10 Subtract 5 from both sides Divide both sides by 0. 5 = (x + 3)2 Take square root of both sides Error = x NO x-intercepts… can't take square root of a negative number.

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