Answer:
The vertex of the quadratic function is:
[tex](x_{v}, y_{v})=\left(-3,\:7\right)[/tex]
Step-by-step explanation:
Given the function
[tex]f\left(x\right)=x^2+6x+16[/tex]
As the vertex of the form [tex]y=ax^2+bx+c[/tex] is defined as:
[tex]x_v=-\frac{b}{2a}[/tex]
As the quadratic function of parabola params are
[tex]a=1,\:b=6,\:c=16[/tex]
so
[tex]x_v=-\frac{b}{2a}[/tex]
[tex]x_v=-\frac{6}{2\cdot \:1}[/tex]
[tex]x_v=-3[/tex]
Putting [tex]x_v=-3[/tex] to determine [tex]y_v[/tex]
[tex]y_v=\left(-3\right)^2+6\left(-3\right)+16[/tex]
[tex]y_v=3^2-18+16[/tex]
[tex]y_v=9-2[/tex]
[tex]y_v=7[/tex]
Therefore, the vertex of the quadratic function is:
[tex](x_{v}, y_{v})=\left(-3,\:7\right)[/tex]
