Question

Seriously guys come on help me with this please, I'd do the same for you.

Find the equation of the tangent to the graph at the indicated point

f(x)= x^2-9; a=2

Answer 1

Aren't you sweet? >:)

Tangent to the graph? This screams *derivative*.
Can you find the derivative of f(x) ?

Answer 2

I can solve this problem up to a point then I just get stuck, I am somewhat familiar with derivatives but we are just barely learning them now in class. So I'm doing the best I can with how very little my teacher has gone over.

Answer 3

Okay, so show me what you've got so far, then.

Answer 4

So far I have gotten to this point 5). 4h+h^2-10/h

1). And I got that far by using this form f(a+h)-f(a)/h

2). (2+h)^2-9-(2^2-9)/h

3). 4+4h+h^2-9-5/h

4). 4+4h+h^2-14/h

After step 5 I just get completely lost

Answer 5

I suggest using this form instead:
\[\Large f'(a) = \lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}\]
It might actually be a little easier

Answer 6

Unfortunately our teacher never taught us that formula only f(a+h)-f(a)/h and then even still I'm generally getting it I just get stuck at the second to last step.

Answer 7

Okay, no problem.
\[\Large \lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}\]

Answer 8

You get
\[\Large \lim_{h\rightarrow0}\frac{\color{blue}{(a+h)^2-9}-\color{red}{(a^2-9)}}{h
}\]
Please simplify :)

Answer 9

1). (2+h)^2-9-(2^2-9)/h

2). 4+4h+h^2-9-5/h

3). 4+4h+h^2-14/h

4). 4h+h^2-10/h <------this is where I get stuck not sure what to do after this point

Answer 10

algebra mistake

Answer 11

how do you mean?

Answer 12

if you do it right, using your method, everything without an \(h\) in it should cancel (add up to zero) in the numerator

Answer 13

Oh I see I have reworked this problem so many times not sure where I went wrong in my math.

Answer 14

\[f(2)=2^2-9=-5\]
\[f(2+h)=(2+h)^2-9=4+4h+t^2-9=-5+4h+4h^2\]
\[f(2+h)-f(2)=-5+4h+4h^2-(-5)=4h+h^2\]

Answer 15

Great thanks for that, I am kind of confused as to how you got what you got. When I do the problem I get something totally different.