Question

Find an equation for the line tangent to y=-2-4x^2
at the point (-3,-38)

i know how to set up the problem with f(x+h)-f(x)/h
but i think i keep messing up a step while simplifying because i've gotten the answer wrong twice now

Answer 1

Well is it a must f(x+h)=f(x)/h be used?

Answer 2

that's the only way that i know how to solve it and its f(x+h)-f(x)/h

Answer 3

if you know another way to solve it that would be fine

Answer 4

I know another way, but you'd be confused. Lets use this way. Show me what you have done.

Answer 5

i simplified it down to -2-4(h^2-6h+9)+38 and i don't know what to do next

Answer 6

all of that over h too

Answer 7

Well I continued from where you stopped, got a strange look! Let's start from the beginning.

Answer 8

\[\lim_{h \rightarrow 0}\frac{ (-2-4(x+h)^{2})-(-2-4x^{2}) }{ h }\]

Answer 9

ok i see where i messed up i didn't include the -2 in front of the f(x+h)

Answer 10

i just plugged x into 4x^2

Answer 11

Okay, try solving it now.

Answer 12

should i distribute the 4 before i take (x+)^2

Answer 13

(x+h) i mean

Answer 14

No. Expand the bracket first.

Answer 15

so -6(h^2-6h+9)-38 all over h?

Answer 16

is that what you got to so far

Answer 17

\[\lim_{h \rightarrow 0} \frac{ (-2-4(x^{2}+2xh+h^{2}))-(-2-4x ^{2}) }{ h }\]

Answer 18

shouldn't you plug the -3 in as x before squaring it

Answer 19

because it gives the point (-3,-38)

Answer 20

Not now. Plug in 3 when you have evaluated that limit.

Answer 21

so -4h-2x is what i got

Answer 22

\[\lim_{h \rightarrow 0} \frac{ -2-4x^{2}-8xh-4h^2+2+4x^{2} }{ h }\]

Answer 23

so everything but the -8xh and -4h^2 cancel out

Answer 24

over h

Answer 25

Simplifies to \[\lim_{h \rightarrow 0}\frac{ -8xh-4h^{2} }{ h }\]

Answer 26

then i took out -4h(h-2x) over h

Answer 27

Factoring out h from the top and cancelling gives..\[\lim_{h \rightarrow 0} (-8x-4h)=-8x\]

Answer 28

Now substitute -3 for x like you've always wanted to.

Answer 29

h=0?

Answer 30

At that point the function is continuous so direct substitution can be used.

Answer 31

what do i substitute?

Answer 32

-3 into the original function?

Answer 33

No. Into -8x. So you get -8(-3).

Answer 34

im confused

Answer 35

is 24 the slope then?

Answer 36

Yes.

Answer 37

would that make the final equation y=24x+34

Answer 38

No. Remember this? \[y-y_{1}=m(x-x_{1})\]

Answer 39

ya i put it into y=mx+b form from point slope form

Answer 40

y1=-38,x1=-3, m=24. Plug them in, re arrange and put it in slope intercept form (y=mx+b)

Answer 41

i still got the same final equation

Answer 42

Hopefully that's correct. That's the end of the problem.

Answer 43

yep i just put it in thanks for the help

Answer 44

Welcome.