Chapter 7.6, Problem 20WE

To determine

The maximum or minimum value of the function $f\left(x\right)=x^2+8x+6$ and then give the coordinates of the vertex of its graph.

Answer to Problem 20WE

The function $f\left(x\right)=x^2+8x+6$ will have minimum value of $-10$ and the vertex of graph of function is $\left(-4,-10\right)$ .

Explanation of Solution

Given information: The given function is,

  $f\left(x\right)=x^2+8x+6\mathrm{....}\left(1\right)$

Formula used: The function of the form $f\left(x\right)=a{x}^2+bx+c$ where $a\ne 0$ will have the maximum value if $a<0$ and minimum value when $a>0$ . Mathematically,

  $\left(\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\text{ is maximum for }a<0\\ \text{ is minimum for }a>0\end{array}\right)\mathrm{....}\left(2\right)$

Calculation: Comparing the formula (2) with (1) gives, $a=1$ which is greater than zero. Therefore the given function $f\left(x\right)=x^2+8x+6$ will give the minimum value.

Now the value for $\left(x\right)$ associated with minimum value can be evaluated from the formula $x=-\frac{b}{2a}$ .

  $\begin{array}{c}x=-\frac{\left(8\right)}{2\left(1\right)}\mathrm{....}\left(a=1,b=8\right)\\ =-4\end{array}$

Hence the value of function will give the minimum value at $x=-4$ . Now evaluate the minimum value at $x=-4$ from (1).

  $\begin{array}{c}{\left[f\left(-4\right)\right]}_{\min }={\left(-4\right)}^2+8\left(-4\right)+6\\ =-10\end{array}$

Again from (1) the given function can be written as,

  $\begin{array}{c}y=x^2+8x+\mathrm{6....}\left(\text{Replace }f\left(x\right)=y\right)\\ \text{or, }y-6=x^2+8x\\ y-6+\left(16\right)=\left(x^2+2\left(x\right)\left(4\right)+{\left(4\right)}^2\right)\mathrm{....}\left(\text{Complete square}\right)\\ y+10={\left(x+4\right)}^2\mathrm{....}\left(3\right)\end{array}$

Since (3) represents a parabola of the form $y-k=a{\left(x-h\right)}^2$ . Therefore the graph for the function (1) has vertex $\left(-4,-10\right)$ .