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