We have the next function
[tex]f(x)=-10x^2-160x+9[/tex]
First we need to obtain the first derivate
[tex]f^{\prime}(x)=-20x-160[/tex]
Then we find the value of x when the derivate is 0
[tex]\begin{gathered} -20x-160=0 \\ -20x=160 \\ x=\frac{160}{-20} \\ x=-8 \end{gathered}[/tex]
The critical point of the function is x=-8
