Chapter 3, Problem 1MC

In Exercises 1-4, graph the given quadratic function. Give each function’s domain and range.

1. f(x) = (x − 3)² − 4

To determine

To sketch: The graph of the function and then find the domain and range of the function $f\left(x\right)={\left(x-3\right)}^2-4$.

Answer to Problem 1MC

The domain of the function is $\left\{x\right|-\infty <x<\infty \}$.

The range of the function is $[-4,\infty )$.

Explanation of Solution

Definition used:

The domain of a function is a set of real values of the independent variable such that the corresponding dependent values also will be real.

The range of a function is defined as a set of all possible functional values, which is denoted by $f\left(x\right)=y$.

Calculation:

Consider the quadratic function $f\left(x\right)={\left(x-3\right)}^2-4$.

It is known that the standard form of the quadratic equation is $f\left(x\right)=a{\left(x-h\right)}^2+k$ and the point $\left(h,k\right)$ is the vertex of the parabola.

Compare the given quadratic equation $f\left(x\right)={\left(x-3\right)}^2-4$ with the standard form of the quadratic equation $f\left(x\right)=a{\left(x-h\right)}^2+k$ and obtain the values of a, h and k as 1, 3 and −4, respectively.

Thus, the vertex of the parabola is $\left(3,-4\right)$.

Find the x intercept by substituting 0 for $f\left(x\right)$ in $f\left(x\right)={\left(x-3\right)}^2-4$.

$\begin{array}{l}{\left(x-3\right)}^2-4=0\\ {\left(x-3\right)}^2=4\\ x-3=2\text{ or }x-3=-2\\ x=5\text{ or }x=1\end{array}$

The x intercepts are 1 and 5.

That is, the parabola passes through $\left(1,0\right)\text{ and }\left(5,0\right)$.

Find the y intercept by substituting 0 for x in $f\left(x\right)={\left(x-3\right)}^2-4$.

$\begin{array}{c}f\left(0\right)={\left(0-3\right)}^2-4\\ =9-4\\ =5\end{array}$

The y intercept is 5.

That is, the parabola passes through $\left(0,5\right)$.

Use the above information to draw the graph of the function $f\left(x\right)={\left(x-3\right)}^2-4$ as shown in below Figure 1.

From Figure 1, it is observed that the parabola is open upward and the vertex of the parabola is $\left(3,-4\right)$,

For all the real values of x, the function $f\left(x\right)={\left(x-3\right)}^2-4$ is well defined.

Therefore, there is no restriction for the domain of the function.

Hence, the domain of the function is $\left\{x\right|-\infty <x<\infty \}$.

Note that, the vertex of the parabola is $\left(3,-4\right)$, which is the lowest point on the graph.

Since y coordinate of the vertex is −4, the functional values will fall on or above −4.

Hence, the range of the function is $[-4,\infty )$.