start by writing the function in standard form
[tex]\begin{gathered} f(x)=5(x-35)(x+27) \\ f(x)=(5x-175)(x+27) \\ f(x)=5x^2+135x-175x-4725 \\ f(x)=5x^2-40x-4725 \end{gathered}[/tex]The axis of symmetry for a quadratic equation is found in the h of the vertex (h,k)
according to this the vertex can be found by:
[tex]\begin{gathered} (h,k)=(-\frac{b}{2a}\text{.f}(-\frac{b}{2a})) \\ h=-\frac{(-40)}{2\cdot5} \\ h=\frac{40}{10} \\ h=4 \end{gathered}[/tex]The axis of symmetry can be found at x=4
