Find values a,b given polynomial turning point

Question

So given the curve $\frac{ax-b}{x^2-1}$ has a turning point at (3,1), find values a and b

Just wondering if there are any elegant ways to solve this problem??

Answer 1

The fact that the function has a 'turning point' at $(3,1)$ implies that the derivative of your function at $x=3$ is $0$. That is,

$\dfrac{d}{dx}\bigg(\dfrac{ax-b}{x^2-1}\bigg)_{x=3}=0$, here you will need to use the quotient rule of differentiation.

Find the derivative, plug in $3$ for $x$, and set it equal to zero. Use this, and the fact that $\frac{a(3)-b}{(3)^2-1}=1$ to solve for both $a$ and $b$.

Answer 2

You have two unknown, namely $a$ and $b$ in $$ \frac{ax-b}{x^2-1}$$ and you also have two conditions which help you find the unknowns.

First of all $(3,1)$ is on your graph, and secondly your derivative at $x=3$ must be zero.

Two equations and two unknowns.

Can you take over from here?

I found $a=6$ and $b=10$