Question

[2(-2+h)^{2} +3(-2+h)] -2= ?

Answer 1

\[[2(-2+h)^{2} +3(-2+h)] -2= ?\]

thanks

Answer 2

Do you want to factor:\[2(-2+h)^{2} +3(-2+h) -2\]

Assume -2+h=x

Answer 3

no don't factor, it's only a
part of a problem

Answer 4

just simplify

Answer 5

fyi, it's the whole bracket minus 2
[2(-2+h)^{2} +3(-2+h)] -2= ?

Answer 6

brackets don't matter ...it's still the same

Answer 7

oh hehe
:)

Answer 8

So just the expanded form is your answer

Answer 9

??

Answer 10

so what does it come to? my online hw has it coming to -5h+5h^2
how do they get that?

Answer 11

when i do it, it comes to 3h+2h^2

Answer 12

\[2\left( h^2-4h+4 \right)+(3h-6)-2\]

Answer 13

it can't be 5h^2

Answer 14

or something is wrong with the question you typed above

Answer 15

my mistake was distributing the square
do you want to see the whole problem?

Answer 16

sure

Answer 17

so
f'(a)=lim [f(a+h)-f(a)]/h
h-->0

Answer 18

\[f(x)=2x ^{2}+3x\]
a=-2

Answer 19

\[f'(-2)=\lim h-->0 [(2(-2+h)^{2}+3(-2+h))-2]/h\]

Answer 20

\[\Large f'(a)=\lim_{h \rightarrow 0}{f(a+h)-f(a) \over h}\]... that the definition of the derivative...ok

Answer 21

yes exactly!

Answer 22

\[f'(-2)=\lim_{h \rightarrow 0} (2(-2+h)^{?} +3(-2+h))-2\div h\]

Answer 23

then after that is the answer
\[\lim_{h \rightarrow 0} -5h+2h ^{2}\div h\]

Answer 24

that's right

Answer 25

but how do they get there?

Answer 26

back to your original question:
\[2(−2+h)^2+3(−2+h)−2\]\[=(2h^2-8h+8)+(3h-6)-2\]\[=2h^2+(-8h+3h)+(8-6-2)\]

Answer 27

oh my god
yeah i was doing the square wrong

Answer 28

you are a gentleman, thanks
so easy, i'm just not thinking thank you so much

Answer 29

\[so,\ \Large f'(-2)=-5\]