# Math - Algebra II

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Hello, i'm currently doing functions and I was fine until I got to multiplication. I really don't know what i'm doing, and would really appreciate a step by step on a few questions.

f(x) = -3x + 9

g(x) = 8x + 7

Find g(f(x))

I got -24x + 16.

Another :

f(x) = -3x + 9

g(x) = 8x + 7

Find f(g(x))

• Math - Algebra II -

no, one term was right, but your constant is wrong.

do it this way :

f(x) = -3x + 9
g(x) = 8x + 7

then g(f(x))
= g(-3x+9)
= 8(-3x+9) + 7
= -24x + 72 + 7
= -24x + 79

and f(g(x))
= f(8x+7)
= -3(8x+7) + 9
= -24x - 21 + 9
= -24x - 12

• Math - Algebra II -

Thank you so much! I've been sitting here for hours. Only 4 to go. Thanks :)

• Math - Algebra II -

2 down.

I'm having trouble here

f(x) = 4x + 2

h(x) = -7x - 5

Find f(h(x))

I tried doing it how it was done above, but i'm not getting an answer listed.

= f(-7x - 5)
= 4x(-7x-5)+ 2
= 28x^2-20+2
= 28x^2-18

Where am I going wrong :| ?

ALSO -

f(x) = 4x-3

g(x) = x2-2

Find f(3a-4)

• Math - Algebra II -

from
= f(-7x - 5) to
= 4x(-7x-5)+ 2
you are replacing the x, so don't leave the x there.
should be
= 4(-7x-5)+ 2
= -28x - 20 + 2
= -28x - 18

for
f(x) = 4x-3

g(x) = x2-2

Find f(3a-4)
I will assume g(x) = x^2 - 2

then f(3a-4)
= 4(3a-4) - 3
= 12a - 16 - 3
= 12a - 19

the x in f(x) was merely replaced by 3a-4, so you have to replace it everywhere that you see an x.

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