Chapter 9.6, Problem 1E

To determine

To find:

The maximum or minimum value of the function $f\left(x\right)$

Answer to Problem 1E

  $g\left(x\right)=x^2+x-12$ Is minimum at $a=1>0$

Explanation of Solution

Given:

  $\begin{array}{l}f\left(x\right)=-{\left(x-3\right)}^2+12,g\left(x\right)=x^2+x-12\\ -{\left(y+5\right)}^2=-12\left(x-2\right)\end{array}$

Concept used:

Theorem on the maximum or minimum value of a quadratic function $f\left(x\right)=a{x}^2+bx+c$ where $a\ne 0$ , then $f\left(-\frac{b}{2a}\right)$ is

 1.  The maximum value of $f$ if $a<0$

 2. The minimum value of $f$ if $a>0$

Calculation:

  $f\left(x\right)=-{\left(x-3\right)}^2+12,g\left(x\right)=x^2+x-12$

Comparing the value $f\left(x\right)=a{x}^2+bx+c$ where $a\ne 0$ , then $f\left(-\frac{b}{2a}\right)$ is

  $a=1,b=1,c=-12$

Since $f$ is a quadratic function and $a=1>0$

The minimum value of $f$ at

  $x=-\frac{b}{2a}$

Putting the value of $b,a$

  $x=-\frac{1}{2}$

Test one point in each of the region formed by the graph.

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function.

To draw the table

  $g\left(x\right)=x^2+x-12$

$x-axis$	$0$	$3$	$-3$	$-5$
$y-axis$	$12$	$24$	$18$	$32$

To draw a graph