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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
you have to subtract first
well, first you have to expand, then you have to subtract
expand like 2a+2h.... and so on?
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
then you can factor out an \(h\) from the remaining terms in the numerator, and cancel with the \(h\) in the denominator
you should be left with \[4ah +14h+2h^2\] before you divide after you divide you get \[4a+14\] and next week you will be able to do this problem instantly in your head
ok, so I have 2a^2 +4ah + 2h^2 +14a +14h -13 and then I subtract 2a2+14a−13? how do I know to do that?
\(2a^2-2a^2=0\) \(14a-14a=0\) 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
which is one of my multiple choice answers...
well, I dunno if I'll be able to do that in my head instantly, but if I can work it out, then I'll be ok at least haha. Thanks @satellite73