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=-(4)/(6)=-(2)/(3)`

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