F(x)=20-x^2+8x
what are the minimum and maximum points and the axis of symmetry?

Vertex aka max or min point is found by -b/2a in form
f(x)=ax^2+bx+c
f(x)=-1x^2+8x+20
vertex x value is -8/(2)(-1)=-8/-2=4
input back to find y value
f(4)=-(4^2)+8*4+20
f(4)=-16+32+20
f(4)=36
max (since the graph opens down) is (4,36)
axis of symmetry is the x coordinate

max is (4,36)
axis of symmetrry is x=4

The length of what is left would be 1 1/2 feet long.

Really long apparently