Answer :
[tex]f(x)=ax^2-4x+1\\
f'(x)=2ax-4\\
0=2a\cdot(-3)-4\\
6a=-4\\
a=-\dfrac{4}{6}=-\dfrac{2}{3}[/tex]
It's the quadratic equation where [tex]a<0[/tex] so the stationary point is a maximum.
It's the quadratic equation where [tex]a<0[/tex] so the stationary point is a maximum.