Question

What is the vertex of x^2 -10x + 21

Answer 1

Answer:

(5, - 4)

Step-by-step explanation:

Given a quadratic in standard form : ax² + bx + c : a ≠ 0

Then the x- coordinate of the vertex is

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

x² - 10x + 21 ← is in standard form

with a = 1, b = - 10, so

$x_{vertex}$ = - $\frac{-10}{2}$ = 5

Substitute x = 5 into the quadratic for the corresponding value of y

y = 5² - 10(5) + 21 = 25 - 50 + 21 = - 4

vertex = (5, - 4)