We can find the y-intercept evaluating the function for x = 0, so:
![\begin{gathered} y(x)=-5x^2+20x+60 \\ y(0)=-5(0)^2+20(0)+60=0+0+60 \\ y(0)=60 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2ouj5wlu4oaxcxxbub24bvfqz0i57i324a.png)
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We can find the zeros evaluating the function for y = 0. So using the factored form:
![\begin{gathered} -5(x-6)(x+2)=0 \\ so\colon \\ x1=6 \\ and \\ x2=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/v6x1ndys0cy4xqj5890cuqzfhqihcadz6u.png)
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The vertex V(h,k) is given by:
![\begin{gathered} h=(-b)/(2a) \\ k=y(h) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/78hdbq1f6d44ovhm3lu0w9a90skd702148.png)
Or we can find it directly from the vertex form:
![\begin{gathered} y=a(x-h)^2+k \\ so \\ for \\ y=-5(x-2)^2+80 \\ h=2 \\ k=80 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8mpah92q0dbno7ogmhj3nv7e764wvhdobz.png)
So, the vertex is:
![(2,80)](https://img.qammunity.org/2023/formulas/mathematics/college/rggx2uoxlel5vxpz1yq47kzeq25un45i88.png)
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The symmetry axis is located at the same point of the x-coordinate of the vertex, so the axis of symmetry is:
![x=2](https://img.qammunity.org/2023/formulas/mathematics/college/6ij5lvx45qkbn22ki7umkb6rdcr9rugcgd.png)
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The maximum value is located at the y-coordinate of the vertex (if it is positive) so, the maximum value is:
![y=80](https://img.qammunity.org/2023/formulas/mathematics/high-school/7rpkh3od68zua4ailjrpcs8zs5makxfbqf.png)