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

Question

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 $\frac{-3}{2}$?

Answer 1

Hint: Solve the equation $$f'(x)=-\frac{3}{2}$$ for $$x$$

Answer 2

I realize you didn't want a graph, but here: you can check your answer:

Answer 3

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 :)