[tex]\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
f(x)&=&{{ k}}&\cdot&x
&& \boxed{f(x)={{ k }}x}
\end{array}
\\\\
\textit{now, we know that }
\begin{cases}
f(x)=-70\\
x=-10
\end{cases}\implies \boxed{-70=k(-10)}[/tex]
solve for "k", to get the "constant of variation",
and plug it back in f(x) = kx
what's f(x) when x = 5? just make "x" to 5 to get f(x)
