Chapter 2.1, Problem 20E

To determine

To write: write the given function $f\left(x\right)=x^2+12x+40$ in standard form and find the vertex, axis of symmetry and x-intercept.

Answer to Problem 20E

The vertex of the parabola is $\left(-6,4\right)$ , axis of symmetry is $x=-6$ and the graph is open upward.

Explanation of Solution

Given information: Given that the quadratic function is $f\left(x\right)=x^2+12x+40$

We know that the standard form of parabola is $f\left(x\right)=a{\left(x-h\right)}^2+k,a\ne 0$ whose axis of symmetry is $x=h$ and vertex $\left(h,k\right)$ .

For, the above form if $a>0$ then the graph is open upward and when $a<0$ the graph is open downward.

We have the given function is $f\left(x\right)=x^2+12x+40$

It can further written as

  $\begin{array}{c}f\left(x\right)=x^2+12x+40\\ ={\left(x+6\right)}^2+4\end{array}$

Comparing with above we get

  $\begin{array}{l}a=1>0\\ h=-6\\ k=4\end{array}$

Therefore, the vertex of the parabola is $\left(-6,4\right)$ axis of symmetry is $x=-6$ and the graph is open upward.

To get the x-intercept we put $f\left(x\right)=0$ and solve for x which gives

  $\begin{array}{l}{x}^2+12x+40=0\\ x=-6\pm 2i\end{array}$

The graph of the quadratic function is shown below.