Answer
$$y=\frac{8}{(3x+5)^2}+1$$
Work Step by Step
$$\frac{dy}{dx}=\frac{-48}{(3x+5)^3}$$
$$dy=\frac{-48}{(3x+5)^3}dx$$
$u=3x+5$
$du=3dx$
$-16du=-48dx$
$$\int dy=\int -16u^{-3} du$$
$$y=8u^{-2}+C$$
$$y=\frac{8}{u^2} +C$$
$$y=\frac{8}{(3x+5)^2} +C$$
We are given that the graph of y passes through the point $(-1,3)$. Now we must plug in these values in order to find the value of $C$.
$3=\frac{8}{(-3+5)^2} +C$
$3=\frac{8}{4} +C$
$C=1$
$$y=\frac{8}{(3x+5)^2} +1$$
