I know it's kind of late but... can anyone help?

Question

If given
f(x) = 4x to the 2nd power -x -4
(4^2-x-4)

Find f(x +2)?

If anyone knows it'll be really heplful

Answer 1

You can just replace x with x+2.

So, with f(x) = (4x^2-x-4), then
f(x+2) = [4(x+2)^2-(x+2)-4]
=4(x^2+4x+4) - x - 2 - 4
=4x^2+16x+16 - x - 2 - 4
=4x^2+15x+10


That's assuming you mean 4x^2 and not (4x)^2.

Answer 2

F(x+2) = 4(x+2)^2-(x+2)-4
=4(x^2+4x+4)-x-2-4
=4x^2+16x+16-x-6
=4x^2+15x+10

Wal C .. check your distribution....Rev, i would say the same to you, but i see you noticed your mistake and fixed it.

Answer 3

f(x) = 4x² - x - 4
So f(x + 2) = 4(x + 2)² - (x + 2) - 4
= 4(x² + 4x + 4) - x - 2 - 4
= 4x² + 16x + 4 - x - 6
= 4x² + 15x - 2

Answer 4

f(x) = 4x^2 - x - 4
f(x+2) = 4(x+2)^2 - (x+2) - 4
f(x+2) = 4(x^2 + 4x + 4) - x - 2 - 4
f(x+2) = 4x^2 + 16x + 16 - x - 6
f(x+2) = 4x^2 + 15x + 10

Answer 5

in your original equation substitute y for x.
f(y) = 4*y^2 - y - 4

now let y = x + 2

f(y) = 4* (x+2) ^2 - (x + 2) -4

now simplify.

I find that making a substitution of variables as the first step helps me to not get confused.

Answer 6

I agree with JEELTZ