Let f(x)=x^2−18x+157 . what is the vertex form of f(x)? what is the minimum value of f(x)?

To find vertex form, complete the square
vertex form is
f(x)=a(x-h)²+k
vertex is (h,k)
and if a is positive, the vertex is a minimum

so

f(x)=x²-18x+157
f(x)=(x²-18x)+157
-18/2=-9, (-9)²=81
add negaitve and positie insides
f(x)=(x²-18x+81-81)+157
factor
f(x)=((x-9)²-81)+157
expand
f(x)=1(x-9)²-81+157
f(x)=1(x-9)²+76
(9,76) is vertex
1 is positive
vertex is a minimum
f(x) reaches its minimum value of 76 at x=9

minimum value of f(x) is 76