Question

the curve ax squared - 4x+1 has a stationary point where x equals -3
find the value of a and determine the nature of the stationary point

Answer 1

$f(x)=ax^2-4x+1\\ f'(x)=2ax-4\\ 0=2a\cdot(-3)-4\\ 6a=-4\\ a=-\dfrac{4}{6}=-\dfrac{2}{3}$

It's the quadratic equation where $a<0$ so the stationary point is a maximum.