Chapter 6.8, Problem 57PPSE

Solution

To draw the graph for the given expression.

  $\text{Domain}=\left[-\frac{3}{2},\infty \right)$$\text{Range}=\left[-3,\infty \right)$

Given:

The given expression is:

  $y=-3-\sqrt{12x+18}$

Calculation:Simplifying the given expression as:

  $\begin{array}{l}y=-3-\sqrt{12x+18}\\ \Rightarrow y=-3-\sqrt{12\left(x+\frac{18}{12}\right)}\\ \Rightarrow y=-3-\sqrt{4\cdot 3\left(x+\frac{3}{2}\right)}\\ \Rightarrow y=-3-2\sqrt{3}\sqrt{\left(x+\frac{3}{2}\right)}\end{array}$

Now, to draw the graph using the transformations, the following transformations are involved:

Let us consider the parent function as:

  $f(x)=\sqrt{x}$

The first transformation is to shift the parent function in the horizontal direction towards left by $\frac{3}{2}$ units:

  $\begin{array}{l}g(x)=f\left(x+\frac{3}{2}\right)\\ \Rightarrow g(x)=\sqrt{\left(x+\frac{3}{2}\right)}\end{array}$

The second transformation is to vertically stretch the graph by $2\sqrt{3}$ units:

  $\begin{array}{l}h(x)=2\sqrt{3}g(x)\\ \Rightarrow h(x)=2\sqrt{3}\sqrt{\left(x+\frac{3}{2}\right)}\end{array}$

The third transformation is to reflect the graph about $x-$ axis:

  $\begin{array}{l}k(x)=-h(x)\\ \Rightarrow k(x)=-2\sqrt{3}\sqrt{\left(x+\frac{3}{2}\right)}\end{array}$

The fourth transformation is to shift the graph vertically down by $3$ units:

  $\begin{array}{l}l(x)=k(x)-3\\ \Rightarrow l(x)=-3-2\sqrt{3}\sqrt{\left(x+\frac{3}{2}\right)}\end{array}$

Graph:The graph for the function is shown as:

  

Interpretation:

The graph is defined for all real numbers equal or greater than $-\frac{3}{2}$ . Hence, the domain is:

  $\text{Domain}=\left[-\frac{3}{2},\infty \right)$

Also, the graph covers all values of real numbers equal or less than $-3$ on $y-$ axis. Hence, the range is:

  $\text{Range}=\left(-\infty ,-3\right]$