Question

Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of
(– 2, – 16), and x intercepts at x = -6 and x = 2. (Do not include the negative sign in your answer.)

y = x2 + 4x − ___

Answer 1

i dont no da answer

Answer 2

then why did you comment? lol

Answer 3

The vertex form of a parabola which opens up or down is \[(x-h)^2=4p(y-k)\]with a focus at \((0,p)\) and vertex at \((h,k)\). To get the standard form (what you have), just multiply it out after plugging in the numbers. Then you'll have to solve for the value of \(p\) that makes y=0 when x = 2 or y = 0 when x = -6.

Answer 4

thanks so much!

Answer 5

Did you get an answer?

Answer 6

yes thank you my mom helped me but thanks for the help anyways

Answer 7

what did you get for the answer?