Where is the vertex of the parabola y = x^2 - 6x + 12?

Mar 13, 2016

color(blue)(" "x_("vertex")=+3)
I will let you solve for ${y}_{\text{vertex}}$ by substitution.

Explanation:

Consider the standard form equation of:
$\text{ } y = a {x}^{2} + b x + c$

Write this as
$\text{ } y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

In your case $a = 1$ so we have

$\frac{b}{a} = - 6$

Apply $\left(- \frac{1}{2} \times \frac{b}{a}\right) \to \left(- \frac{1}{2}\right) \times \left(- \frac{6}{1}\right) = + 3$

color(blue)(" "x_("vertex")=+3)

By substitution of $x = 3$ in the original equation you can determine
${y}_{\text{vertex}}$