Title: Default for 2.2 Graphs of Functions and Parametric Form
Course: HELM Workbooks 2023-2024

The Graph of a Function

Consider the function \(f(x)=2x\). The output is obtained by multiplying the input by \(2\). We can choose several values for the input to this function and calculate the corresponding outputs. We have done this for integer values of \(x\) between \(-2\) and \(2\) and the results are shown in the table below.

input, \(x:\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\)
output, \(\:f(x):\) | \(-4\) | \(-2\) | \(0\) | \(2\) | \(4\)

To construct the graph of this function we first draw a pair of axes – a vertical axis and a horizontal axis. These are drawn at right-angles to each other and intersect as shown in the figure below.

[Image]
The two axes intersect at the origin

Each pair of input and output values can be represented on a graph by a single point. The input values are measured along the horizontal axis and the output values are measured along the vertical axis. The horizontal axis is often called the \(x\) axis. The vertical axis is commonly referred to as the \(y\) axis so that we often write the function as \[ y=f(x)=2x \] or simply \[y=2x.\]

Each pair of \(x\) and \(y\) values in the table is plotted as a single point, shown as \(\bullet\) in the figure above. A general point is often labelled as \((x,y)\). The values \(x\) and \(y\) are said to be the coordinates of the point. The points are then joined with a smooth curve to produce the required graph as shown in the figure above. Note that in this case the graph is a straight line. The graph can then be used to find function values other than those given in the table. For example, directly from the graph we can see that when \(x=1.5\), the value of \(y\) is \(3\).

Dependent and Independent Variables

Since \(x\) and \(y\) can have a number of different values they are variables. Here \(x\) is called the independent variable and \(y\) is called the dependent variable. Knowing or choosing a value of the independent variable \(x\), the function rule enables us to calculate the corresponding value of the dependent variable \(y\). To show this dependence we often write \(y(x)\). This is read as ‘\(y\) is a function of \(x\)’ or ‘\(y\) depends upon \(x\)’, or simply ‘\(y\) of \(x\)’. Note that it is the independent variable which is the input to the function and the dependent variable which is the output.

The Domain and Range of a Function

The set of values which we allow the independent variable to take is called the domain of the function. A domain is often an interval on the \(x\) axis. For example, the function \[y=g(x)=5x+2, \qquad -5\le x\le 20\] has any value of \(x\) between \(-5\) and \(20\) inclusive as its domain because it has been stated as this. If the domain of a function is not stated then it is taken to be the largest set possible. For example \[h(t)=t^2+1\] has domain \(-\infty<t<\infty\) since \(h\) is defined for every value of \(t\) and the domain has not been stated otherwise.

Later, you will meet some functions for which certain values of the independent variable must be excluded from the domain because at these values the function would be undefined. One such example is \( \displaystyle{f(x)={1\over x}}\) for which we must exclude the value \(x=0\), since \(\displaystyle{{1\over 0}}\) is meaningless.

Similarly, we cannot include the value \(x=2\) in the domain of \(f(x)=\dfrac{1}{x-2}\).

The set of values of the function for a given domain, that is, the set of \(y\) values, is called the range of the function. The range of \(g(x)\) above is \(-23\le g(x) \le 102\) and the range of \(h(t)\) is \(1\le h(t)<\infty\). Usually the range of a function can sometimes be identified by inspecting its graph. Note that the range depends on the domain, and taking a larger (or smaller) domain could result in a larger (or smaller) range.

There is a related concept, called the codomain of a function. The codomain of a function can be any set which contains the range of the function.

Example

Consider the function given by \(g(t)=2t^2+1\), \(-2\le t\le 2\).

1. State the domain of the function.
2. Plot a graph of the function.
3. Deduce the range of the function from the graph.

Solution

1. The domain is given as the interval \(-2\le t\le 2\), that is any value of \(t\) between \(-2\) and \(2\) inclusive.
2. To draw the graph a table of input and output values must be constructed first.
Table

\(t\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\)
\(y=g(t)\) | \(9\) | \(3\) | \(1\) | \(3\) | \(9\)

Each pair of \(t\) and \(y\) values in the table is plotted as a single point shown as \(\bullet\) in the figure below. The points are then joined with a smooth curve to produce the required graph.
[Image]
3. The range is the set of values which the function takes. By inspecting the graph we see that the range of \(g\) is the interval \(1\le g(t)\le 9\).

Parametric Representation of a Function

Suppose we write \(x\) and \(y\) in terms of \(t\) in the form \[x=4t \qquad y=2t^2, \qquad \hbox{ for } -1\le t\le 1\] For different values of \(t\) between \(-1\) and \(1\), we can calculate pairs of values of \(x\) and \(y\). For example when \(t=1\) we see that \(x=4(1)=4\) and \(y=2\times 1^2=2\). That is, \(t=1\) corresponds to the point with \((x,y)\) coordinates\((4,2)\).

A table of values is given below.

Table

\(t\) | \(-1\) | \(-0.5\) | \(0\) | \(0.5\) | \(1\)
\(x\) | \(-4\) | \(-2\) | \(0\) | \(2\) | \(4\)
\(y\) | \(2\) | \(0.5\) | \(0\) | \(0.5\) | \(2\)

If the resulting points are plotted on a graph then different values of \(t\) correspond to different points on the graph. The graph of is plotted in below.

[Image: Graph of the function defined parametrically by x=4t, y=2t^2]
Graph of the function defined parametrically by \(x=4t\), \(y=2t^2\), \(-1\le t\le1\)

It is often possible to convert a parametric representation of a function into the more usual form by combining the two expressions to eliminate the parameter. Thus if \(x=4t\) we can write \(\displaystyle{t={x\over 4}}\) and so

\[\begin{aligned} y=2t^2 &= 2\left({x\over 4}\right)^2 \\ &= {2x^2\over 16}\\ &= {x^2\over 8} \end{aligned}\]

Using \(y=\dfrac{x^2}{8}\) we can, by giving \(x\) values, find corresponding values of \(y\). Plotting these \((x,y)\) values gives, of course, exactly the same curve as in the figure above.

Exercise

Complete the table of values for the function \( f(x) = [value] \) for \(x\) between \(-3\) and \(3\).

input, \(x\) | output, \(f(x)\)
\(-3\) | [value]
\(-2\) | [value]
\(-1\) |
\(0\) |
\(1\) |
\(2\) |
\(3\) | [value]

From the table of values, what is the shape of the function \( f(x) = [value] \)?

Graph A | Graph B
[value] | [value]
Graph C | Graph D
[value] | [value]
Feedback: We fill in the values in the table by substituting the value of \(x\) into the function expression \(f(x) = [value]\). Hence: \(f(-1) = [value](-1)^3 = [value]\), \(f(0) = [value](0)^3 = [value]\), \(f(1) = [value](1)^3 = [value]\) and \(f(2) = [value](2)^3 = [value]\). To find the graph of \(f(x) = [value]\), you plot the \(x\) and \(f(x)\) points on a graph and pick the graph that is close to your sketch.

Exercise

First, a quick note on entering your answers. Remember 'and/or' here are logical 'and/or', so 'and' means the values satisfying both inequalities, 'or' means the values satisfying at least one of the inequalities.

Consider the function given by \(f(x) = [value],\quad [value] \leq x \leq [value]\)

1. State the domain of the function. Give your answer as an inequality in \(x\), e.g. "\( \alpha \leq x\) and/or \(x \leq \beta\)".
2. Complete the table of values for the function \( f(x) = [value] \) for \(x\) between \(-3\) and \(3\).

input, \(x\) | output, \(f(x)\)
\(-3\) | [value]
\(-2\) | [value]
\(-1\) |
\(0\) |
\(1\) |
\(2\) |
\(3\) | [value]

3. From the table of values, what is the shape of the function \( f(x) = [value] \)?

Choose from the graphs below:

Graph A | Graph B
[value] | [value]
Graph C | Graph D
[value] | [value]

4. Deduce the range of the function by inspecting the graph and table of values. Give your answer as an inequality in \(y\), e.g. "\( \alpha \leq y\) and \(y \leq \beta\)".
Feedback: 1. Since \( [value]\) can take any value of \(x\) in the given interval \([value] \leq x\leq [value]\) , the domain remains as \([value] \leq x\leq [value]\). 2. We fill in the values in the table by substituting the value of \(x\) into the function expression \(f(x) = [value]\). Hence: \(f(-1) = [value](-1)^3 = [value]\), \(f(0) = [value](0)^3 = [value]\), \(f(1) = [value](1)^3 = [value]\) and \(f(2) = [value](2)^3 = [value]\). 3. To find the graph of \(f(x) = [value]\), you plot the \(x\) and \(f(x)\) points on a graph and pick the graph that is close to your sketch. 4. From table and graph, the smallest value of \(f(x)\) occurs at \(x = 0\) which is \(f(0) = [value]\) and the biggest value at \(x=-3 \text{ or }3\), \(f(3) = [value]\), hence the range is \([value] \leq x\leq [value]\)

Exercise

Consider the function given by \(g(t) = [value],\quad [value] \leq t \leq [value]\),

1. State the domain of the function. Give your answer as an inequality in \(t\), e.g. "\( \alpha \leq t\) and \(t \leq \beta.\)". Remember 'and' here is a logical 'and', so your answer is only true when both inequalities are true.
2. Complete the table of values for the function \( g(t) = [value] \) for \(t\) between \(-3\) and \(3\).

input, \(t\) | output, \(g(t)\)
\(-3\) | [value]
\(-2\) | [value]
\(-1\) |
\(0\) |
\(1\) |
\(2\) |
\(3\) | [value]

3. From the table of values, what is the shape of the function \( g(t) = [value] \)?

Choose from the graphs below:

Graph A | Graph B
[value] | [value]
Graph C | Graph D
[value] | [value]

4. Deduce the range of the function by inspecting the graph and table of values. Give your answer as an inequality in \(y\), e.g. "\( \alpha \leq y\) and \(y \leq \beta\)". Remember 'and' here is a logical 'and', so your answer is only true when both inequalities are true.
Feedback: 1. Since \( g(t)=[value]\) can take any value of \(t\) in the given interval \([value] \leq t\leq [value]\) , the domain remains as \([value] \leq t\leq [value]\). 2. We fill in the values in the table by substituting the value of \(t\) into the function expression \(g(t) = [value]\). Hence: \(g(-1) = [value](-1)^3 = [value],\) \(g(0) = [value](0)^3 = [value],\) \(g(1) = [value](1)^3 = [value]\) and \(g(2) = [value](2)^3 = [value]\). 3. To find the graph of \(g(t) = [value]\), you plot the \(t\) and \(g(t)\) points on a graph and pick the graph that is close to your sketch. 4. From table and graph, the largest value of \(g(t)\) occurs at \(t=-3 \text{ or }3\) which is \(g(-3) =g(3) = [value].\) The biggest value occurs at \(t = 0\), \(\, g(0) = [value].\) Hence the range is \([value] \leq t\leq [value].\)

Exercise

Consider the function given by \(\displaystyle{h(t) = [value]},\)

1. State the domain of the function. Give your answer as an inequality in \(t\), e.g. "\( \alpha \leq t\) and \(t \leq \beta\)". Remember 'and' here is a logical 'and', so your answer is only true when both inequalities are true.
2. What is the shape of the function \( \displaystyle{h(t) = [value]} \)?

Choose from the graphs below:

Graph A | Graph B
[value] | [value]
Graph C | Graph D
[value] | [value]
Feedback: 1. The function \( h(t) = \displaystyle [value]\) cannot take the value of \(t\) for which the denominator is \(0\). When \(t = [value]\) , \(h([value])=\frac[value]{0}\) which is undefined, hence the domain is \(t < [value]\) and \(t > [value]\). 2. To find the graph of \(h(t) = \displaystyle [value]\), you plot the \(t\) and \(h(t)\) points on a graph and pick the graph that is close to your sketch.

Exercise

Consider the parametric function given by \[ x = [value], \quad y = [value].\]

1. Obtain an explicit equation for \(y\) in terms of \(x\)
\(y =\)
2. Find the graph of the function

Choose from the graphs below:

Graph A | Graph B
[value] | [value]
Graph C | Graph D
[value] | [value]
Feedback: 1. We have the curves \( \, x = \displaystyle [value]\) and \(\, y = \displaystyle [value]. \) Since \( x = \displaystyle [value]\) we may write \( t = \displaystyle [value]\). Substituting this into the expression for \(y\) gives \( y= \displaystyle [value]\). 2. Using the expression found, we can find some values of \(y\) at different values of \(x\). Using these values we can find the correct plot.

Exercise

Consider the parametric equations \[x=[value], \quad y=[value].\]

1. Obtain an explicit equation for \(y\) in terms of \(x\)
\(y =\)
2. Find the graph of the function

Choose from the graphs below:

Graph A | Graph B
[value] | [value]
Graph C | Graph D
[value] | [value]
Feedback: 1. We have the curves \( \, x = [value]\) and \(\, y = [value].\) Since \( x = [value]\) we may write \( t = \displaystyle [value] .\) Substituting this into the expression for \(y\) gives \(\, y = \displaystyle [value].\) 2. Using the expression found, we can find some values of \(y\) at different values of \(x\). Using these values we can find the correct plot.

Exercise

A particle with mass \(m\) falls under gravity so that at time \(t\) its distance from the \(x\)-axis is \(\displaystyle{[value]}\) and its distance from the \(y\)-axis is \(\displaystyle{[value]}\) where \(g\) is a constant (the acceleration due to gravity). Find the value of \(t\) when the particle crosses the \(x\)-axis and, at this time, find the distance from the \(y\)-axis.

1. The parametric equations of the path of the particle are
\( x = \)
\(y = \)
2. The value of \(t\) when the particle crosses the \(x\)-axis is
\(t=\)
3. The value of \(x\) at this value of \(t\) is
\(x=\)
Feedback: 1. The parametric equations of the path of the particle are \( x = [value]\) and \(y = \displaystyle [value]\). 2. When the particle crosses the \(x\)-axis, \(y=0\), hence you get the value of \(t\) by solving \( \displaystyle[value]=0\) to get \(t= \displaystyle[value]\). 3. Subtitute this value into the equation of \(x\) to get \( x = \displaystyle [value] \left([value]\right) = \displaystyle [value].\)