Question

Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of
(– 3, – 25), and x intercepts at x = -8 and x = 2. (Do not include the negative sign in your answer.)

y = x2 + 6x − ___
HELP!

Answer 1

y = a(x-h)^2 + k is the basic vertex form where (h,k) is the vertex

The given vertex is (-3,-25), so h = -3 and k = -25

Answer 2

So

y = a(x-h)^2 + k

becomes

y = a(x-(-3))^2 + (-25)

y = a(x+3)^2 - 25

Answer 3

Finally, a = 1 since the leading coefficient of y = x2 + 6x − ___ is 1

Answer 4

so

y = 1(x+3)^2 - 25

y = (x+3)^2 - 25

Answer 5

Now expand and simplify

Answer 6

How would I expand that? and what not.

Answer 7

y = (x+3)^2 - 25

y = (x+3)(x+3) - 25

y = x(x+3)+3(x+3) - 25


Distribute that and combine like terms

Answer 8

Would part of the answer be x^2+6x-16?

Answer 9

yes it is, so the term that goes in the blank is 16

Answer 10

Makes sense, thank you!

Answer 11

yw