Evaluate the limit of the difference quotient. Interpret the limit as the slope of the tangent to a curve at a specific point.
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!
Originally Posted by Plato
the answer is: 4, slope of tangent to y=x^2 at x=2
My question is: how do I get x=2?
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, , 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: