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

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Answer 1 · 2016-03-13T20:49:11

$color(blue)(" "x_("vertex")=+3)$
I will let you solve for $y_("vertex")$ by substitution.

Consider the standard form equation of:
$" "y=ax^2+bx+c$

Write this as
$" "y=a(x^2+b/ax)+c$

In your case $a=1$ so we have

$b/a =-6$

Apply $(-1/2xxb/a) -> (-1/2)xx(-6/1) = +3$

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

By substitution of $x=3$ in the original equation you can determine
$y_("vertex")$

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