Question

Please help me, I can't seem to figure this out. :/
"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 − ___

Answer 1

opens up id defined by the sign of the first term; this one is already postive so thats taken care of for you

Answer 2

the simplest path might be to construct the vertex form and expand it tho

Answer 3

y = a(x-Vx)^2 + Vy

where your vertex is defined as the point (Vx,Vy)

Answer 4

Yes, I understand that part. :( Thank you for your help, I just need to know how to figure out the missing term. Should I use the vertex formula?

Answer 5

i would, just to keep the guess work at a minimum

Answer 6

And that's -b/2a, right?

Answer 7

y = a(x--3)^2 -25 ; lets assume a = 1 and adjust afterwards

y = (x^2+6x+9) -25

y = x^2+6x+9 -25

y = x^2+6x -16

then we have to adjust this if need be

Answer 8

So, do you plug in the x intercepts that were given? I'm sorry, I'm really confused.

Answer 9

to test it yes; we would plug in the x intercepts to see of we get zeros; in this case im sure we do tho

Answer 10

another method would have been to use the roots to construct a product of binomials with to get the same results:

x = -8 and x = 2

(x+8)(x-2) = x^2-2x+8x-16
= x^2 +6x -16

Answer 11

ah, so do you us that equation to find the missing variable?

Answer 12

*use

Answer 13

both strategies give the same equation to compare with

Answer 14

y = x^2 +6x − ___
y = x^2 +6x - 16

Answer 15

ooohhhhhhhh, I'm sorry, wow. I just got it. >.< Thank you so much.

Answer 16

yep :)