Question

(a) Solve $x^2-8x+12\ge 0$ graphically and (b) write the solution in interval notation.

Short answer

The solution in interval notation is$\hspace{0.17em}(-\infty \hspace{0.17em},\hspace{0.17em}2]\cup \hspace{0.17em}[6,\hspace{0.17em}\infty \hspace{0.17em})$

Step-by-step solution

Step 1. Given information

$x^2-8x+12\ge 0$

Step 2. Graph the function

$$\begin{gathered} f\left(x\right)=x^2-8x+12 \\ a=1,b=\left(-8\right),c=12 \end{gathered}$$

The parabola opens upward as $a$is positive.

Find the Axis of symmetry.

$$\begin{gathered} f\left(x\right)=-\frac{b}{2a} \\ f\left(x\right)=-\frac{\left(-8\right)}{2} \\ f\left(x\right)=4 \end{gathered}$$

Find the Vertex

$$\begin{gathered} f\left(x\right)=x^2-8x+12 \\ f\left(4\right)={\left(4\right)}^2-8\left(4\right)+12 \\ f\left(4\right)=-4 \end{gathered}$$

Vertex $\left(4,-4\right)$

Step 3. y-intercept

$f\left(0\right)=0^2-8\left(0\right)+12$

y-intercept is $\left(0,12\right)$

Point symmetric to y-intercept is $\left(8,12\right)$

Find x-intercepts

$$\begin{gathered} f\left(x\right)=x^2-8x+12 \\ 0=x^2-8x+12 \\ 0=\left(x-2\right)\left(x-6\right) \\ x=2orx=6 \end{gathered}$$

The x-intercepts are $\left(2,0\right)and\left(6,0\right)$

The graphical solution is

Step 4. Solution

The solution in interval notation is$\hspace{0.17em}(-\infty \hspace{0.17em},\hspace{0.17em}2]\cup \hspace{0.17em}[6,\hspace{0.17em}\infty \hspace{0.17em})$