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.)

Answer 1

so
y=ax^2+bx+c

if a>0 the graph opens up .. so in this case we have a = 1 so the graph opens up and nothing we should do.

now, the vertex is given by -b/2a and in our case b=6 and a = 1
-b/2a = -3 nothing we should do here as well..

now we need to make sure y(-3) = -25 (because we know the vertex is (-3,-25))

our equation is y=x^2+6x-c
c is what we should find

y(-3) = (-3)^2 + 6*(-3) -c
(-3)^2 + 6*(-3) -c = -25

can you answer what is c now ?

Answer 2

I don't know how to solve this .... I really don't.

Answer 3

(-3)^2 + 6*(-3) -c = -25

9 -18 -c = -25

Answer 4

im sure you can tell what is c now

Answer 5

-25 ?

Answer 6

@Coolsector

Answer 7

9-18 -c = -25
-9 -c = -25
-9 + 25 = c
c = 16

Answer 8

so our equation is

y = x^2 +6x - 16

now they gave us another information.. we should make sure it's right
they said x -intercepts at x = -8 and x=2
so we have to make sure that
y(-8) = 0
(-8)^2 + 6 *(-8) - 16 = 64 - 48 - 16 = 0
so its good.
and
y(2) = 0
2^2 + 6*2 - 16 = 4 + 12 -16 = 0
so its good..

Answer 9

Thanks