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 − ___
Help me on this one it's hard

Answer 1

You know that it will intersect the point (-2, -16), and you know that you're just missing the constant term. As it is, if you plug in -2 you get y=-4, which means you need to decrease that by 12 somehow to get (-2, -16). How would you do that?

Answer 2

4x-4=-16 and -2x1=-2?

Answer 3

Huh?

Answer 4

how would i decrease 12?

Answer 5

Okay, as is, if the equation is just y=x^2+4x, then it contains the point (-2,-4). You want it to contain the point (-2,-16) instead. That means y has to be 12 less than it is in the current equation. So, in what equation that starts with y=x^2+4x-___ do you get the point (-2,-16)?

Answer 6

not quite sure..

Answer 7

Okay, what if you put 2 in the blank, and had y=x^2+4x-2. Then if you plug in -2, you'll get -6, right? That will contain the point (-2,-6). So by subtracting 2 from the equation, you went from (-2,-4) to (-2,-6). With that in mind, how would you change it so that the equation contained (-2,-16)?

Answer 8

what's the answer