Question

Find the VERTEX of the parabola f(x) = x^2 - 16x + 63

Answer 1

Answer:

vertex = (8, - 1 )

Step-by-step explanation:

Given a parabola in standard form, f(x) = ax² + bx + c : a ≠ 0

Then the x- coordinate of the vertex is

$x_{vertex}$ = - $\frac{b}{2a}$

f(x) = x² - 16x + 63 ← is in standard form

with a = 1 and b = - 16, thus

$x_{vertex}$ = - $\frac{-16}{2}$ = 8

Substitute x = 8 into f(x)

f(8) = 8² - 16(8) + 63 = 64 - 128 + 63 = - 1

vertex = (8, - 1 )