1. ## Interpreting Limits

Evaluate the limit of the difference quotient. Interpret the limit as the slope of the tangent to a curve at a specific point.

lim [(2+h)^2-4]/h
h->0

my attempt:
I need help with the secondi part. I found the derivative which in this case is M_t=4 to the graph y=x^2 at x=_______? I am not sure how to find x =? at the specific point.

I tried to put M_t=4 into the original eqtn and solve for h but that didn't work. Please provide detailed steps to finishing this problem. Thank you!

2. Originally Posted by skeske1234
Evaluate the limit of the difference quotient. Interpret the limit as the slope of the tangent to a curve at a specific point.

lim [(2+h)^2-4]/h
h->0

my attempt:
I need help with the secondi part. I found the derivative which in this case is M_t=4 to the graph y=x^2 at x=_______? I am not sure how to find x =? at the specific point.

I tried to put M_t=4 into the original eqtn and solve for h but that didn't work. Please provide detailed steps to finishing this problem. Thank you!
$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

for your problem, note that $f(x) = x^2$

the limit is $f'(2)$

3. $\frac{{\left( {2 + h} \right)^2 - 4}}
{h} = \frac{{\left( {4 + 4h + h^2 } \right) - 4}}
{h} = 4 + h,\,h \ne 0$

4. Originally Posted by Plato
$\frac{{\left( {2 + h} \right)^2 - 4}}
{h} = \frac{{\left( {4 + 4h + h^2 } \right) - 4}}
{h} = 4 + h,\,h \ne 0$

the answer is: 4, slope of tangent to y=x^2 at x=2

My question is: how do I get x=2?

5. Originally Posted by skeske1234
the answer is: 4, slope of tangent to y=x^2 at x=2
My question is: how do I get x=2?
That question really do not make any sense. Does it?

If the question is, "find the x value where $y=x^2$ has slope 4."
You would look at the derivative: $2x=4$.
From which it i clear that $x=2$.

6. So you have already found the slope, m=4. I think you want the tangent line now. Based on your work, you used the difference quotient and pugged x=2 in to find the slope at this x value. So, at x=2 the slope is 4. So you have an x value, you now need a y-value in order to talk about the tangent line. Well, you know the tangent line has to touch our orinignal graph, $y=x^2$, so to find the y-value, just plug in our x-value. If x=2, y=4. So the point our line in tangent to is (2,4) with a slope of 4. Using point-slope form, our tangent line is:
$y-4=4(x-2)$
or
$y=4x-4$