Question

For what numbers x is$6-x\ge x^2$?

Short answer

The solution set is$[-3,2]$

Step-by-step solution

Step 1. Given information

The given inequality is$6-x\ge x^2$

Step 2. Calculation

Rewrite the given inequality,

$x^2+x-6\le 0$

Find the y-intercept of the function $f\left(x\right)=x^2+x-6$by substituting $x=0$

$$\begin{gathered} f\left(0\right)={\left(0\right)}^2+0-6 \\ =-6 \end{gathered}$$

Factorize the equation as follows,

$$\begin{gathered} x^2+x-6=0 \\ (x-2)(x+3)=0 \\ x=2,-3 \end{gathered}$$

Thus, y-intercept is $-6$and x-intercepts are$2,-3$

Step 3. Calculation

Compare the given function with the standard form $a{x}^2+bx+c=0$,

$a=1,b=1,c=-6$

The x-coordinate of the vertex is as follows,

$-\frac{b}{2a}=-\frac{1}{2}$

Find the y-coordinate of the vertex,

$$\begin{gathered} f(-\frac{1}{2})={\left(-\frac{1}{2}\right)}^2-\left(\frac{1}{2}\right)+6 \\ =\frac{1}{4}-\frac{1}{2}+6 \\ =\frac{1-2-24}{4} \\ =-\frac{25}{4} \end{gathered}$$

Graph the function with vertex $\left(-\frac{1}{2},-\frac{25}{4}\right)$and the points $(-3,0),(0,-6)and(2,0)$

Thus, the solution set is$\left\{x|-3\le x\le 2\right\}$