Answer: \(\displaystyle y=(17-x^2)^3\).

Answer: \(\displaystyle y=(17-x^2)^3\).

Let \(Y=\sqrt[3]{y}\) and \(X=x^2\). The straight line in the \((X,Y)\)-plane passes through \((9,8)\) and \((16,1)\).

Its gradient is

\(\displaystyle \frac{1-8}{16-9}=-1\).

So the line has equation

\(Y=-X+c\).

Using \((X,Y)=(9,8)\),

\(8=-9+c\),

so \(c=17\).

Therefore

\(\sqrt[3]{y}=17-x^2\).

Cubing both sides gives

\(y=(17-x^2)^3\).