brandonloves
Compute and simplify the difference quotient
f(a+h)-f(a)
---------
h
g(x)=2x^2+14x -13
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brandonloves
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this was already confusing enough without changing it to g(x)
anonymous
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replace \(g\) by \(f\) it makes no difference, it is all algabra
brandonloves
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oh ok. so it's like: 2(a+h)^2+14(a)-13 over h?
brandonloves
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which comes to...
anonymous
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no hold on
brandonloves
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oh
anonymous
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you have to subtract first
anonymous
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well, first you have to expand, then you have to subtract
anonymous
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\(g(a+h)= 2(a+h)^2+14(a+h)-13\)
brandonloves
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expand like 2a+2h.... and so on?
brandonloves
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(2a+2h)^2?
brandonloves
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for the first one
anonymous
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yeah it is
\[2(a^2+2ah+h^2)+14a+14h-13\]
\[=2a^2+4ah+2h^2+14a+14h-13\]
anonymous
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square first multiply by 2 second
brandonloves
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oh ok, gotcha
anonymous
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from this you need to subtract \(2a^2+14a-13\) so everything without an \(h\) will be gone
anonymous
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then you can factor out an \(h\) from the remaining terms in the numerator, and cancel with the \(h\) in the denominator
anonymous
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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
brandonloves
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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?
anonymous
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subtract
anonymous
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\(2a^2-2a^2=0\) \(14a-14a=0\) and \(-13-(-13)=0\)
anonymous
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in other words, every term without an \(h\) is gone
brandonloves
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oh, now I get it
brandonloves
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and that leaves me with 4ah + 2h^2 +14h
anonymous
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yes, and now divide by \(h\)
brandonloves
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as you said lol
brandonloves
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ok
brandonloves
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4a+2h+14
brandonloves
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which is one of my multiple choice answers...
brandonloves
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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