Chapter 2.1, Problem 21E

To determine

The standard form of the quadratic function and the graph for it, also identify the graph, vertex axis of symmetry and x intercepts.

Answer to Problem 21E

The quadratic function in the standard form is $f\left(x\right)={\left(x+4\right)}^2-3$ , the vertex of the graph is $\left(-4,-3\right)$ and its axis of symmetry is $x=-4$ , the $x$ intercepts are at $\left(-4+\sqrt{3}\right),\left(-4-\sqrt{3}\right)$ and the intercepts of the graph is shown in Figure 1

Explanation of Solution

Given:

The given function is $f\left(x\right)=x^2+8x+13$ .

Calculation:

Consider the quadratic function is,

  $f\left(x\right)=a{\left(x-h\right)}^2+k$

The vertex of the above parabola for the standard form $f\left(x\right)=a{\left(x-h\right)}^2+k$ is,

  $\left(h,k\right)$

Consider the given equation is,

  $\begin{array}{c}f\left(x\right)=x^2+8x+13\\ =\left(x^2+8x+16\right)-16+13\\ ={\left(x+4\right)}^2-3\end{array}$

From the above equation the vertex of the function $f\left(x\right)=a{\left(x-h\right)}^2+k$ is,

  $\left(h,k\right)=\left(-4,-3\right)$

Thus, the axis of symmetry of the graph is $x=-4$ .

To find the $x$ intercept of the function $f\left(x\right)=x^2+8x+13$ equation the expression equal to zero as,

  $\begin{array}{l}{x}^2+8x+13=0\\ x^2+8x+16=16-13\\ {\left(x+4\right)}^2=3\\ x+4=\pm \sqrt{3}\end{array}$

Solve further as,

  $\begin{array}{l}x=-4\pm \sqrt{3}\\ x=\left(-4+\sqrt{3}\right),\left(-4-\sqrt{3}\right)\end{array}$

The sign of the coefficient $a$ is positive so the graph for the parabola will open in the upward direction with the x vertices at $\left(-4+\sqrt{3}\right),\left(-4-\sqrt{3}\right)$ ,

The graph is shown in Figure 1

  

Figure 1