## Introductory Algebra for College Students (7th Edition)

$(3,-4)$
Completing the square of the right side of the given equation, $y=x^2-6x+5 ,$ results to \begin{array}{l}\require{cancel} y=(x^2-6x)+5 \\\\ y=\left( x^2-6x+\left( \dfrac{-6}{2} \right)^2 \right)+5-\left( \dfrac{-6}{2} \right)^2 \\\\ y=\left( x^2-6x+\left( -3 \right)^2 \right)+5-\left( -3 \right)^2 \\\\ y=\left( x^2-6x+9 \right)+5-9 \\\\ y=\left( x-3 \right)^2-4 .\end{array} The vertex of the parabola whose quadratic equation is in the form $y=a(x-h)^2+k$ is $(h,k)$. Hence, the vertex of the equation above is \begin{array}{l}\require{cancel} (3,-4) .\end{array}