Given the function:
![y=x^2+16x+71](https://img.qammunity.org/2023/formulas/mathematics/college/9rnhmrf9xz5tdasbdrmh6chuk4l5prx42i.png)
We need to find the vertex of the parabola
So, we will complete the square of the function as follows:
![\begin{gathered} y=x^2+16x+71+((16)/(2))^2-((16)/(2))^2 \\ \\ y=x^2+16x+71+64-64 \\ y=(x^2+16x+64)+(71-64) \\ \\ y=(x+8)^2+7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b721x28xsi8uikl0mpbqo0pjhywovqwpud.png)
The general equation of the parabola has the form:
![y=(x-h)^2+k](https://img.qammunity.org/2023/formulas/mathematics/high-school/h4kqjwwshdhi39of5qh206bvx52ti58zvg.png)
Where ( h, k) is the coordinates of the vertex
By comparing the equations:
![\begin{gathered} x-h=x+8 \\ -h=8 \\ h=-8 \\ k=7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tgsum19brfwc5uakn6dyuoe9x2z0vlz7w4.png)
So, the answer will be:
The coordinates of the vertex = ( -8, 7)