posted by Rudy .
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.
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.