Recent content by volc

When I plug in the x value (-9/2(1+x)^-1/2)/(1+x) I get: (-9/2)((10)^-3/2) and this is x or m(x) ? volc

(-9/2(1+x)^-1/2)/(1+x) = slope of the tgt line right ? And the equation for the tangental line is y= mx + b, where m = (-9/2(1+x)^-1/2)/(1+x). To find y I need to find b and I can't seen to be able to find b.

I don't understand what you mean. :/

Derivative of function = slope of the tgt line.

Find the y-intercept of the tangent line to the graph of y = (9)/sqrt(1 + x) at the point where x = 9. I used the quotient rule to find the derivative. I found that to be (-9/2(1+x)^-1/2)/(1+x) But then I am totally clued out... Thanks for helping me out. volc

Here is what I found, look at the end to see what my question is: At what point does the normal line to the graph of y= -4 - 3x + 4x^2 at (1, -3 ) intersect the parabola a second time? So far I have tried to find the derivative of the parabola which is: 8x-3 (this corresponds to the...