# Are there any points on the curve $y=\frac{x}{2}+\frac{1}{2x+4}$ where the slope is $\frac{-3}{2}$?

I am not sure how to find points of $$y=\frac{x}{2}+\frac{1}{2x+4}$$ where the slope is $$\frac{-3}{2}$$ without looking at a graph.

I can simplify the function by writing it as $$y=\frac{x(x+2)+1}{2(x+2)}$$, but I have no idea what to look for to discover the function.

I think that derivatives are equal to the slope of the line at a certain point, but I can't check every point in the function.

How can I find the points where the slope is $$\frac{-3}{2}$$?

Hint: Solve the equation $$f'(x)=-\frac{3}{2}$$ for $$x$$
• It is $$x_1=-\frac{5}{2}$$ or $$x_2=-\frac{3}{2}$$ – Dr. Sonnhard Graubner May 10 at 18:51
I realise you may know the answer but just for completeness sake, we have $$y=\frac{x}{2}+\frac{1}{2x+4}$$, as you said! Then differentiating this we get: $$y'(x)=\frac{1}{2}-\frac{1}{2(x+2)^2}$$ which is a formula for the gradient! We want the gradient at $$\frac{-3}{2}$$, so we set $$y'(x)=\frac{-3}{2}$$, which means: $$\frac{-3}{2}=\frac{1}{2}-\frac{1}{2(x+2)^2}$$
Then solving this we get the quadratic $$(x+2)^2=\frac{1}{4}$$, so solving this we get $$x=\frac{-5}{2}$$ and $$x=\frac{-3}{2}$$. If you don't understand the simplification which I skipped just ask :)