Chapter 0.2, Problem 6E

(a)

To determine

To find: The domain and range of the given function.

Answer to Problem 6E

The domain is equal to $\left(-\infty ,\infty \right)$ and the range is equal to $\left[-9,\infty \right)$ .

Explanation of Solution

Given information:

The function is $y=x^2-9$ .

Calculation:

The given function is $y=x^2-9$ .

To find the domain: The domain for all quadratic functions is always all real values.

Since the given function is the quadratic function, so the domain of the given function is all real numbers.

  $\begin{array}{l}x\in R\\ x\in \left(-\infty ,\infty \right)\end{array}$

To find the range: Rewrite the equation of function and solve for $x$ .

  $\begin{array}{c}{x}^2=y+9\\ x=\pm \sqrt{y+9}\end{array}$

There does not exist a real square root of a negative number, so the inside of the square root, $y+9$ , must be greater than or equal to zero.

It can be simplified as below:

  $\begin{array}{c}y+9\ge 0\\ y\ge -9\end{array}$

So, the $x$ exists for all values of $y$ which are greater than or equal to $-9$ .

Therefore, the range is all real umber less than or equal to $-9$ .

i.e.

  $R=\left[-9,\infty \right)$

Hence, the domain is $\left(-\infty ,\infty \right)$ and the range is $\left[-9,\infty \right)$ .

(b)

To determine

To sketch: The graph of the function $y=x^2-9$ .

Explanation of Solution

Given information:

The function is $y=x^2-9$ .

Calculation:

The given function is $y=x^2-9$ .

The given function is the parabolic function with vertex $\left(0,-9\right)$ .

Now, sketch the graph of the function below: