# What is the vertex of y=x^2-6x+8?

May 23, 2016

color(blue)("Vertex"->(x,y)->(3,-1)

#### Explanation:

The given equation is in the format $y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

In your case $a = 1$

The following process is part way to completing the square

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a} \to \left(- \frac{1}{2}\right) \times \left(- 6\right) = + 3}$
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Substitute $x = + 3$ in the original equation to determine ${y}_{\text{vertex}}$

$\textcolor{b l u e}{{y}_{\text{vertex}} = {\left(3\right)}^{2} - 6 \left(3\right) + 8 = - 1}$
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color(blue)("Vertex"->(x,y)->(3,-1)