Question

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

Answer 1

Answer:

The vertex of the quadratic function is:

$(x_{v}, y_{v})=\left(-3,\:7\right)$

Step-by-step 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:

$x_v=-\frac{b}{2a}$

As the quadratic function of parabola params are

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

so

$x_v=-\frac{b}{2a}$

$x_v=-\frac{6}{2\cdot \:1}$

$x_v=-3$

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:

$(x_{v}, y_{v})=\left(-3,\:7\right)$