f(x)=(5x2+10x)+8
f(x)=5(x^2+2x)+8
f(x)=5(x^2+2x+1-1)+8
f(x)=5((x+1)^2-1)+8
f(x)+5(x+1)^2-5+8
f(x)=5(x+1)^2+3
for
f(x)=a(x-h)^2+k
vertex is (h,k)
we have
f(x)=5(x-(-1))^2+3
vertex is (-1,3)
minimum or max
hmm
let's evaluate another point that is not x=-1, if the y value is less than the y value of the vertex, then (-1,3) is a max
if the y value is greater than the y value of the vertex, then (-1,3) is a minimum
f(0)=0+0+8
8>3
minimum at (-1,3)