Conditions from Stationary Points

Turn each condition into its own equation

A stationary point $(p,q)$ supplies two facts: the curve height is $f(p)=q$ , and its gradient is $f'(p)=0$ . Unknown coefficients remain constants when differentiating.

Worked example

y = x³ + ax² + b has a stationary point at (2, −3). Find a and b.

\begin{gathered}y'=3x^2+2ax\\y'(2)=0\\12+4a=0\\a=-3\\y(2)=-3\\8-12+b=-3\\b=1\end{gathered}

Check both conditions in the resulting function. Since $y''(2)=6>0$ , this point is a minimum.

Your turnPractice

$y=x^3+ax^2+b$ has a stationary point at $(4,-27)$ . Find the constants and the other stationary point, with its nature.

Show worked answer

$48+8a=0$ gives $a=-6$ ; $64-96+b=-27$ gives $b=5$ . Then $y'=3x(x-4)$ , so the other point is $(0,5)$ . Since $y''(0)=-12$ , it is a maximum.

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