We can find the y-intercept evaluating the function for x = 0, so:
[tex]\begin{gathered} y(x)=-5x^2+20x+60 \\ y(0)=-5(0)^2+20(0)+60=0+0+60 \\ y(0)=60 \end{gathered}[/tex]---------
We can find the zeros evaluating the function for y = 0. So using the factored form:
[tex]\begin{gathered} -5(x-6)(x+2)=0 \\ so\colon \\ x1=6 \\ and \\ x2=-2 \end{gathered}[/tex]-----------------------------------------------
The vertex V(h,k) is given by:
[tex]\begin{gathered} h=\frac{-b}{2a} \\ k=y(h) \end{gathered}[/tex]Or we can find it directly from the vertex form:
[tex]\begin{gathered} y=a(x-h)^2+k \\ so \\ for \\ y=-5(x-2)^2+80 \\ h=2 \\ k=80 \end{gathered}[/tex]So, the vertex is:
[tex](2,80)[/tex]---------
The symmetry axis is located at the same point of the x-coordinate of the vertex, so the axis of symmetry is:
[tex]x=2[/tex]-----------------------
The maximum value is located at the y-coordinate of the vertex (if it is positive) so, the maximum value is:
[tex]y=80[/tex]