Let's take a look at the domain and range of some linear functions and some quadratic functions.
Domain and Range of \(f(x)=x\) and \(f(x)=x^2\)
From the graph, we note that for \(f(x)=x\):
- \(f(x)\) is defined for all \(x \in \mathbb{R}\)
- \(\)Therefore, the domain is \(\{x \mid x\in\mathbb{R}\}\). Alternatively, we can write: the domain is \(\mathbb{R}\).
- Every real number is an ouput of \(f(x)\)
- Therefore, the range is \(\mathbb{R}\).
From the graph, we note that for \(f(x)=x^2\):
- \(f(x)\) is defined for all \(x \in \mathbb{R}\)
- \(\)Therefore, the domain is \(\mathbb{R}\).
- Every real number greater than or equal to \(0\) is an output of \(f(x)\)
- Therefore, the range is \(\{y\in\mathbb{R} \mid y\ge0\}\).
Recall
The set \(\{y\in\mathbb{R} \mid y\ge0\}\) is read as: "The set of all \(y\) in \(\mathbb{R}\) such that \(y\) is greater than or equal to \(0\)".
Example 4
Determine the domain and range.
- \(f(x)=-3x(x-2)\)
- \(g(x)=-\dfrac{3}{4}(x-5)^2-1\)
- \(h(x)=3x^2-3x-18\)
Solution — Part A
Determine the domain and range of \(f(x)=-3x(x-2)\).
Since this quadratic function is written in factored form, we can note that:
- The zeros are \(0\) and \(2\).
- The axis of symmetry, which lies halfway between the zeros, is \(x=1\).
- The \(y\)-coordinate of the vertex is \(f(1)=-3(1)(1-2)=3\).
The graph of this parabola is:
We note from the graph that
- \(f(x)\) is defined for all \(x \in \mathbb{R}\), and
- all possible \(y\)-values are \(y\le 3\).
Therefore,
- \(D=\mathbb{R}\) (since the domain is all real numbers), and
- \(R=\{y\in\mathbb{R} \mid y\le3\}\).
Solution — Part B
Determine the domain and range of \(g(x)=-\dfrac{3}{4}(x-5)^2-1\).
Since this quadratic function is written in vertex form, we can note that:
- The vertex is \((5,-1)\).
- The parabola opens down since \(a=-\dfrac{3}{4}\) is negative.
- The parabola has a maximum \(y\)-value since it opens down.
- The maximum, which occurs at the vertex, is \(y=-1\).
The graph verifies these details.
We note that
- \(g(x)\) is defined for all \(x \in \mathbb{R}\), and
- all possible \(y\)-values are \(y\le -1\).
Therefore,
- \(D=\mathbb{R}\), and
- \(R=\{y\in\mathbb{R}|y\le-1\}\).
Solution — Part C
Determine the domain and range of \(h(x)=3x^2-3x-18\).
You may have begun to gather from the other parts in this example that
- the domain of a quadratic function is \(\mathbb{R}\) since quadratic functions are defined for all \(x\in\mathbb{R}\), and
- the range of a quadratic function is determined by the \(y\)-coordinate of the vertex and whether the parabola opens up or down.
We require the vertex of \(\)\(h(x)=3x^2-3x-18\).
Method 1: Find the vertex by first factoring
\(\begin{align*} h(x) &= 3x^2-3x-18 \\ &= 3(x^2-x-6) &\text{Remove a common factor of}~3\\ &= 3(x-3)(x+2) &\text{Determine a pair of integers with product}~{-6}~\text{and sum}~{-1}\\ \end{align*}\)
Since this quadratic relation is written in factored form, we can note that:
- The zeros are \(3\) and \(-2\).
- The axis of symmetry, which lies halfway between the zeros, is \(x=\dfrac{1}{2}\).
- The \(y\)-coordinate of the vertex is \(h\left(\dfrac{1}{2}\right)=3\left(\dfrac{1}{2}-3\right)\left(\dfrac{1}{2}+2\right)=\dfrac{-75}{4}\).
Therefore, the vertex is \(\left(\dfrac{1}{2}, \dfrac{-75}{4}\right)\).
Method 2: Find the vertex by first completing the square
\(\begin{align*} h(x) &= 3x^2-3x-18 \\[.5em] &= \class{hl1}{3}(x^2-x)-18 &\text{Common factor the}~a\text{-value from the first two terms.} \\ &= 3\left(x^2-x\class{hl2}{+\dfrac{1}{4}-\dfrac{1}{4}}\right)-18 &\text{Find the value that creates a perfect square} \\[.5em] &&\text{(take half of the}~x\text{-coefficient and square it).} \\ &&\text{Add and subtract that value.} \\[.5em] &= 3\left[\class{hl3}{\left(x-\dfrac{1}{2}\right)^2}-\dfrac{1}{4}\right]-18 &\text{Factor the perfect square.} \\[.5em] &= 3\left(x-\dfrac{1}{2}\right)^2\class{hl4}{-\dfrac{3}{4}}-18 &\text{Distribute the}~a\text{-value.} \\[.5em] &= 3\left(x-\dfrac{1}{2}\right)^2\class{hl1}{-\dfrac{75}{4}} &\text{Combine the remaining constants.} \end{align*}\)
Therefore, the vertex is \(\left(\dfrac{1}{2}, \dfrac{-75}{4}\right)\).
Method 3: Find the vertex by determining and applying a formula
Recall
The zeros of \(y=ax^2+bx+c\), where \(a \ne 0\), are given by
\(x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Recall how we can use this formula to remember the vertex formula, \(x_\text{vertex}=-\dfrac{b}{2a}\):
- The zeros are \(\dfrac{-b}{2a} + \dfrac{\sqrt{b^2-4ac}}{2a}\) and \(\dfrac{-b}{2a} - \dfrac{\sqrt{b^2-4ac}}{2a}\).
- The \(x\)-coordinate of the vertex is halfway between the two zeros.
- Since the zeros are of the form \(\dfrac{-b}{2a}+k\) and \(\dfrac{-b}{2a}-k\), halfway between the two zeros is \(\dfrac{-b}{2a}\).
Therefore, a formula for the \(x\)-coordinate of the vertex is \(x_\text{vertex}=\dfrac{-b}{2a}\).
Applying this formula to find the \(x\)-coordinate of the vertex of \(h(x)=3x^2-3x-18\):
\(\begin{align*} x_\text{vertex} &= \dfrac{-b}{2a}\\ &= \dfrac{3}{2(3)}\\ &= \dfrac{1}{2} \end{align*}\)
Therefore, the \(y\)-coordinate of the vertex is \(h\left(\dfrac{1}{2}\right)=3\left(\dfrac{1}{2}\right)^2 -3\left(\dfrac{1}{2}\right) -18=\dfrac{-75}{4}\).
Therefore, the vertex is \(\left(\dfrac{1}{2}, \dfrac{-75}{4}\right)\).
Method 4: Find the vertex by partial factoring
Step 1: Partially Factor
\(\begin{align*} h(x)&=3x^2-3x-18\\ &=3x(x-1)-18 \end{align*} \)
Step 2: Identify two points with the same \(y\)-coordinate
If we set each factored part equal to \(0\)\(\), we will find the input values that have output \(-18\):
\(\begin{align*} x = 0 && \text{ or } && x-1=0\\ &&&& x=1 \end{align*}\)
So the points \((0,-18)\) and \((1,-18)\) satisfy the relation.
Step 3: Determine the \(x\)-coordinate of the vertex
Since the points found in Step 2 share the same \(y\)-coordinate, the \(x\)-coordinate of the vertex is halfway between \(0\) and \(1\), at \(x=\dfrac{1}{2}\).
Step 4: Determine the \(y\)-coordinate of the vertex
Therefore, the \(y\)-coordinate of the vertex is \(h\left(\dfrac{1}{2}\right)=3\left(\dfrac{1}{2}\right)^2 -3\left(\dfrac{1}{2}\right) -18=\dfrac{-75}{4}\).
Determine the domain and range
We can observe that:
- Each method determined that the vertex is \(\left(\dfrac{1}{2}, \dfrac{-75}{4}\right)\).
- The parabola opens up since \(a=3\) is positive.
- The parabola has a minimum \(y\)-value since it opens up.
- The minimum, which occurs at the vertex, is \(y=\dfrac{-75}{4}\).
The graph verifies these details.
We note that
- \(h(x)\) is defined for all \(x \in \mathbb{R}\), and
- all possible \(y\)-values are \(y\ge \dfrac{-75}{4}\).
Therefore, for \(h(x)=3x^2-3x-18\),
- \(D=\mathbb{R}\), and
- \(R=\left\{y\in\mathbb{R} \mid y\ge \dfrac{-75}{4}\right\}\).
Remarks about Notation
In this question, when calculating the range of \(f(x)\), \(g(x)\), and \(h(x)\), we used \(y\) to represent each function without formally letting \(y\) equal the function. In this courseware, \(y\) always represents the output of a function unless \(y\) is explicitly defined in some other way.
Further, we will use \(D\) and \(R\) to denote the domain and range of a function, respectively.
Domain and Range of a Quadratic Function Defined for all \(x\in\mathbb{R}\)
Suppose a quadratic function \(f(x)=ax^2+bx+c\), where \(a \ne 0\), defined for all \(x \in\mathbb{R}\) has vertex \((h,k)\). Then,
- the domain is \(D=\mathbb{R}\), and
- the range is
- \(R=\{y\in\mathbb{R} \mid y \ge k\}\) if \(a \gt 0\), and
- \(R=\{y\in\mathbb{R} \mid y \le k\}\) if \(a \lt 0\).
To determine the range of a parabola, the vertex needs to be found. The method to find the vertex of a parabola differs depending on the given form:
- Vertex Form: \(f(x)=a(x-h)^2+k\) has vertex \((h,k)\).
- Factored Form: Determine the zeros, then the axis of symmetry (halfway between the zeros) and then the \(y\)-coordinate of the vertex.
- Standard Form: Determine the vertex in one of four ways:
- Change to factored form by factoring.
- Change to vertex form by completing the square.
- Use the formula \(x_\text{vertex}=\dfrac{-b}{2a}\) and then use the function to determine the \(y\)-coordinate of the vertex.
- Partially factor to determine a pair of points symmetric about the axis of symmetry.