Question

Convert into vertex form: y=x^2+6x-13

Answer 1

The general vertex form looks like this: \( a\left(x-h\right)^2 + k\)
Where the vertex is given by \( \left(h,k\right)\).
To reach this, you have to complete the square. See this link for more information: http://www.purplemath.com/modules/sqrvertx.htm

Completing the square of your given polynomial:
\[y = x^2 + 6x - 13\]
Take the 6 (the coefficient in front of the \(x\) term) and divide it by 2, then square it: \(\left(\frac{6}{2}\right)^2 = 9\)
Now add this number immediately after the \(6x\) term and subtract the number immediately after the \(-13\) term:
\[y=\left(x^2 + 6x + \color{red}{9}\right) - 13 - \color{red}{9}\]
The above equation can be simplified into the general vertex form listed above, which you should be able to do now.