given
[tex]y=x^2-12x+14[/tex]The vertex form is a special form of a quadratic function. From the vertex form, it is easily visible where the maximum or minimum point (the vertex) of the parabola is, it has the form
[tex]\begin{gathered} y=(x-h)^2+k \\ where \\ the\text{ vertex is \lparen h,k\rparen} \end{gathered}[/tex]Step 1
a) complete the square
[tex]\begin{gathered} y=x^{2}-12x+14 \\ y=x^2-12x+14+(-6)^2+6^2 \\ us\text{ ethe binomial formula} \\ y=\left(x-6\right)^2-(-6^2)+6^14 \\ y=(x-6)^2-22 \end{gathered}[/tex]so , the equation in vertex form is
[tex]y=(x-6)^2-22[/tex]I hope this helps you
