Question

What is the vertex of the quadratic function f(x)=x2+6x+16? Write your answer

as an ordered pair, (x,y).

Answer 1

The vertex of the quadratic function is:

`(x_(v), y_(v))=\left(-3,\:7\right)`

Explanation:

Given the function

`f\left(x\right)=x^2+6x+16`

As the vertex of the form
`y=ax^2+bx+c` is defined as:

`x_v=-(b)/(2a)`

As the quadratic function of parabola params are

`a=1,\:b=6,\:c=16`

so

`x_v=-(b)/(2a)`

`x_v=-(6)/(2\cdot \:1)`

`x_v=-3`

Putting
`x_v=-3` to determine
`y_v`

`y_v=\left(-3\right)^2+6\left(-3\right)+16`

`y_v=3^2-18+16`

`y_v=9-2`

`y_v=7`

Therefore, the vertex of the quadratic function is:

`(x_(v), y_(v))=\left(-3,\:7\right)`