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this was already confusing enough without changing it to g(x)

replace \(g\) by \(f\) it makes no difference, it is all algabra

oh ok. so it's like: 2(a+h)^2+14(a)-13 over h?

which comes to...

no hold on

oh

you have to subtract first

well, first you have to expand, then you have to subtract

\(g(a+h)= 2(a+h)^2+14(a+h)-13\)

expand like 2a+2h.... and so on?

(2a+2h)^2?

for the first one

yeah it is
\[2(a^2+2ah+h^2)+14a+14h-13\]
\[=2a^2+4ah+2h^2+14a+14h-13\]

square first multiply by 2 second

oh ok, gotcha

from this you need to subtract \(2a^2+14a-13\) so everything without an \(h\) will be gone

subtract

\(2a^2-2a^2=0\) \(14a-14a=0\) and \(-13-(-13)=0\)

in other words, every term without an \(h\) is gone

oh, now I get it

and that leaves me with 4ah + 2h^2 +14h

yes, and now divide by \(h\)

as you said lol

ok

4a+2h+14

which is one of my multiple choice answers...