anonymous
  • anonymous
[2(-2+h)^{2} +3(-2+h)] -2= ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[[2(-2+h)^{2} +3(-2+h)] -2= ?\] thanks
PaxPolaris
  • PaxPolaris
Do you want to factor:\[2(-2+h)^{2} +3(-2+h) -2\] Assume -2+h=x
anonymous
  • anonymous
no don't factor, it's only a part of a problem

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anonymous
  • anonymous
just simplify
anonymous
  • anonymous
fyi, it's the whole bracket minus 2 [2(-2+h)^{2} +3(-2+h)] -2= ?
PaxPolaris
  • PaxPolaris
brackets don't matter ...it's still the same
anonymous
  • anonymous
oh hehe :)
PaxPolaris
  • PaxPolaris
So just the expanded form is your answer
PaxPolaris
  • PaxPolaris
??
anonymous
  • anonymous
so what does it come to? my online hw has it coming to -5h+5h^2 how do they get that?
anonymous
  • anonymous
when i do it, it comes to 3h+2h^2
PaxPolaris
  • PaxPolaris
\[2\left( h^2-4h+4 \right)+(3h-6)-2\]
PaxPolaris
  • PaxPolaris
it can't be 5h^2
PaxPolaris
  • PaxPolaris
or something is wrong with the question you typed above
anonymous
  • anonymous
my mistake was distributing the square do you want to see the whole problem?
PaxPolaris
  • PaxPolaris
sure
anonymous
  • anonymous
so f'(a)=lim [f(a+h)-f(a)]/h h-->0
anonymous
  • anonymous
\[f(x)=2x ^{2}+3x\] a=-2
anonymous
  • anonymous
\[f'(-2)=\lim h-->0 [(2(-2+h)^{2}+3(-2+h))-2]/h\]
PaxPolaris
  • PaxPolaris
\[\Large f'(a)=\lim_{h \rightarrow 0}{f(a+h)-f(a) \over h}\]... that the definition of the derivative...ok
anonymous
  • anonymous
yes exactly!
anonymous
  • anonymous
\[f'(-2)=\lim_{h \rightarrow 0} (2(-2+h)^{?} +3(-2+h))-2\div h\]
anonymous
  • anonymous
then after that is the answer \[\lim_{h \rightarrow 0} -5h+2h ^{2}\div h\]
PaxPolaris
  • PaxPolaris
that's right
anonymous
  • anonymous
but how do they get there?
PaxPolaris
  • PaxPolaris
back to your original question: \[2(−2+h)^2+3(−2+h)−2\]\[=(2h^2-8h+8)+(3h-6)-2\]\[=2h^2+(-8h+3h)+(8-6-2)\]
anonymous
  • anonymous
oh my god yeah i was doing the square wrong
anonymous
  • anonymous
you are a gentleman, thanks so easy, i'm just not thinking thank you so much
PaxPolaris
  • PaxPolaris
\[so,\ \Large f'(-2)=-5\]

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