Posted by **Rudy** on Wednesday, December 5, 2012 at 1:29am.

Given the function y=(x^2-5/x+3), answer the following questions.

a) Find dy/dx. Simplify your answer

b) Find the coordinates of any points on the graph of y where the tangent line is horizontal. Show your work.

c) Are there any points on the graph of y where the tangent line is parallel to the line y=x+7? Explain your answers.

- Calculus -
**Steve**, Wednesday, December 5, 2012 at 11:37am
y' = (x^2+6x+5)/(x+3)^2 = (x+1)(x+5)/(x+3)^2

or, you could say y = x-3+4/(x+3), so

y' = 1 - 4/(x+3)^2

so, where is y'=0? x = -1 or -5

where is y'=1? nowhere, since 1-4/(x+3)^2 < 1 for all x. Note that y has a slant asymptote with slope=1.

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