Question

Rewrite the quadratic function f(x) = 2x^2+8x-3 in vertex form a(x-h)^2+k

Answer 1

I can see that I would factor out a 2 to start it and so I would have 2(x^2+4x) - 3 but I don't know what to do after that

Answer 2

f(x) = 2(x^2+4/2x)-3
b = 4/2 * 2 = 8/2 = 4
b^2 = 16

Is this on the right track?

Answer 3

@iambatman @Kainui

Answer 4

@agent0smith

Answer 5

Do you know how to complete the square?

f(x) = 2x^2+8x-3

f(x) = 2(x^2+4x) - 3

f(x) = 2(x^2+4x +4) - 3 - 2(4)

Then finish

Answer 6

I just watched a video on how to do it so I think this may be right

Answer 7

f(x) = 2(x^2 + 4/2x + 16 - 16) - 3
f(x) = 2(x+4)^2 - 32 - 3
f(x) = 2(x+4)^2 - 35

Answer 8

but when I put that in wolfram alpha, I get 2x^2+16x-3 instead of 2x^2+8x-3

Answer 9

hmm, I got 4/2x + 16 in one of my steps

Answer 10

Would it not be 4^2 = 16 and not +4?

Answer 11

@agent0smith

Answer 12

Divide the middle coefficient (here it's 4) by 2, and square it:

f(x) = 2x^2+8x-3

f(x) = 2(x^2+4x) - 3

f(x) = 2(x^2+4x +4) - 3 - 2(4)

f(x) = 2(x+2)^2 - 3 - 2(4)

Answer 13

f(x) = 2(x+2)^2 - 11

Answer 14

How did you get f(x) = 2(x+2)^2?

Answer 15

By factoring.

x^2 + bx + (b/2)^2
factors as
(x + b/2)^2