Chapter 4.9, Problem 8E

Solution

To Find: The other possibilities of given graph.

The solution for $\frac{1}{4}{\left(x-2\right)}^2-1\ge 0$ is $x\le 0$ or $x\ge 0$

The solution for $\frac{1}{4}{\left(x-2\right)}^2-1\le 0$ is $0\le x\le 4$ .

Given:

Example 2:

The graph of $y=\frac{1}{4}{\left(x-2\right)}^2-1$ is shown below:

  

Explanation:

The other possibility of inequalities of $y=\frac{1}{4}{\left(x-2\right)}^2-1$ are as follows:

  $\begin{array}{l}\frac{1}{4}{\left(x-2\right)}^2-1=0\to \left(1\right)\\ \frac{1}{4}{\left(x-2\right)}^2-1\ge 0\to \left(2\right)\\ \frac{1}{4}{\left(x-2\right)}^2-1\le 0\to \left(3\right)\end{array}$

The equation $\left(1\right)$ solution set is set of values where the curve intersects the $x$ axis.

The curve intersects at two points $x=0$ and $x=4$ in the graph.

Thus, the solution to the equation $\left(1\right)$ is $x=0$ and $x=4$ .

The solution set for $\frac{1}{4}{\left(x-2\right)}^2-1\ge 0$ is all $x-$ values of points above the $x-$ axis on the parabola.

Thus, the solution for $\frac{1}{4}{\left(x-2\right)}^2-1\ge 0$ is $x\le 0$ or $x\ge 0$

The solution set for $\frac{1}{4}{\left(x-2\right)}^2-1\le 0$ is all $x-$ values of points above the $x-$ axis on the parabola.

Thus, the solution for $\frac{1}{4}{\left(x-2\right)}^2-1\le 0$ is $0\le x\le 4$ .