13. find the points on the graph of $y^{3}-27y = x^{2}-90$ where the tangent line is vertical.

Explanation:

Step1: Differentiate implicitly

Differentiate $y^{3}-27y = x^{2}-90$ with respect to $x$. Using the chain - rule, we have $3y^{2}y'-27y'=2x$. Factor out $y'$: $y'(3y^{2}-27)=2x$. So, $y'=\frac{2x}{3y^{2}-27}$.

Step2: Find when the tangent is vertical

A tangent line is vertical when the slope is undefined, i.e., when the denominator of $y'$ is zero and the numerator is non - zero. Set $3y^{2}-27 = 0$. Solve for $y$: [ \begin{align*} 3y^{2}-27&=0\ 3y^{2}&=27\ y^{2}&=9\ y&=\pm3 \end{align*} ] When $y = 3$: Substitute $y = 3$ into the original equation $y^{3}-27y=x^{2}-90$. $3^{3}-27\times3=x^{2}-90$. $27 - 81=x^{2}-90$. $- 54=x^{2}-90$. $x^{2}=36$, so $x=\pm6$. When $y=-3$: Substitute $y = - 3$ into the original equation $y^{3}-27y=x^{2}-90$. $(-3)^{3}-27\times(-3)=x^{2}-90$. $-27 + 81=x^{2}-90$. $54=x^{2}-90$. $x^{2}=144$, so $x=\pm12$.

Answer:

The points are $(6,3),(-6,3),(12, - 3),(-12,-3)$