Alternative Format — Lesson 1: Graphing Three Common Functions

Let's Start Thinking

Graphing a Function

Graphing is an important skill in dealing with functions.

We can use a graph to:

  • Identify whether or not a relation is a function (using the vertical line test).
  • Visualize the domain and range.
    • Which values of \(x\) and \(y\) are associated with the function?
    • What is the minimum/maximum value of \(y\)?
    • What values of \(x\) cannot be used as an input?

In this lesson, we will sketch the graphs of three common functions. Being able to draw these three graphs will allow us to sketch more complicated functions later on.


Lesson Goals

  • Sketch the graphs of \(f(x)=x^2\), \(f(x)=\sqrt x\), and \(f(x)= \dfrac1x\).
  • Introduce the idea of an asymptote on a graph.
  • Identify the domain and range of the functions \(f(x)=x^2\), \(f(x)=\sqrt x\), and \(f(x)= \dfrac1x\) using their graphs. 

The Quadratic Function


Review of the Quadratic Function

\(f(x)=x^2\) is the simplest possible case of a quadratic function. It is sometimes called a base function, as it can be used to generate the graphs of all quadratic functions. 

Math in Action

The function \(f(x)=x^2\) can be used to represent the area of a square with side lengths \(x\).

In this section of the lesson, we will sketch the graph of \(f(x)=x^2\), and identify its domain and range.

Graphing the Quadratic Function

To graph, first create a table of values. Select particular values of \(x\) and then evaluate \(f(x)=x^2\).

\(x\) \(f(x)=x^2\)
\(-3\) \(9\)
\(-2\) \(4\)
\(-1\) \(1\)
\(0\) \(0\)
\(1\) \(1\)
\(2\) \(4\)
\(3\) \(9\)

Sample Calculation:

\(\)When \(x=-3\), evaluate \(f(-3)\).

\(\begin{align*} f(-3)&= (-3)^2 \\ &= 9\end{align*}\)

We can use this table to create the graph of \(y=x^2\), or \(y=f(x)\).  

We often talk about sketching the graph of \(y=f(x)\), where \(f(x)\) is a defined function.

This means that the \(y\)-coordinate of a point on the graph is the output value \(f(x)\) associated with a particular input value (i.e., the \(x\)-coordinate of the point).

Create the graph of \(y=f(x)\) by plotting the points and joining them with a smooth curve. As usual, show that the graph continues infinitely in both directions by adding arrows to the graph or by extending the graph to the edges of the grid (as has been done here).  

The graph of y equals x squared.

Finding the Domain and the Range for the Quadratic Function

The domain of \(f(x)=x^2\) is \(\{x \: | \: x \in \mathbb{R}\}\).  (Alternatively, we can write: the domain is \(\mathbb{R}\).)

Reasons:

  • It is possible to square any real number. So the input can be any real number.
  • The graph continues infinitely into the positive and negative \(x\)-values.

The range of \(f(x)=x^2\) is \(\{y \in \mathbb{R} ~ | ~ y\geq 0\}\). 

Reasons:

  • When any real number is squared, the result is greater than or equal to \(0\). So the output of the function will be greater than or equal to \(0\).
  • The graph has a minimum \(y\)-value of \(0\), which occurs at the vertex \((0, 0)\). The rest of the points have \(y \gt 0\).  

The Square Root Function


Graphing the Square Root Function

Recall

\(f(x)=\sqrt x\) is called the square root function.

Math in Action

If a square room in an apartment is listed as having an area of \(x\), \(f(x)=\sqrt x\) can be used to represent the dimensions of the room.

If we know how to sketch a graph \(f(x) = \sqrt{x}\), it can be used as a base graph to help us sketch other functions involving a square root.

We can sketch the graph of this function by creating a table of values. 

When creating tables of values for linear or quadratic functions, there are no limitations on the \(x\)-values that can be selected.

However, for the square root function, it is important to select appropriate \(x\)-values. There are many \(x\)-values that will not work.

E.g., \(f(-2)\) is undefined, since the square root of a negative number is not a real number.

In fact, \(f(x)\) is undefined for all \(x \lt 0\). We need to select \(x \geq 0\).

For \(f(x)=\sqrt x\),

Here I've selected the \(x\)-values \(0\), \(1\), \(2\), \(3\) and \(4\).

\(x\) \(f(x)=\sqrt{x}\)
\(0\)  
\(1\)  
\(2\)  
\(3\)  
​\(4\)  

The second column of the table is completed by evaluating the function at each of these \(x\)-values.

Sample Calculation:

When \(x=0\), evaluate \(f(0)\). 

\(\begin{align*}f(0) &= \sqrt0 \\ &= 0\end{align*}\)

The rest of the table can be completed similarly.

\(x\) \(f(x)=\sqrt{x}\)
\(0\) \(0\)
\(1\) \(1\)
\(2\) \(1.41\)
\(3\) \(1.73\)
​​​​​\(4\) \(2\)

Note that \(\sqrt{2}\) and \(\sqrt{3}\) have been rounded to two decimal places.

We can now sketch the graph of \(y = f(x)\) by plotting the five points in the table and joining them with a smooth curve.

The graph of the function, f of x equals the square root of x.

This curve has a definite end point at \((0, 0)\) and extends in the direction of positive \(x\)- and \(y\)-values only.

It's important to mark the end point clearly and to show that the graph continues in the other direction, either by drawing an arrow on the end of the curve or by extending the curve to the edge of the grid.

It's important to notice that the points we have chosen here only form a small part of the curve and two of them are difficult to draw accurately by hand.

We are going to recreate the table and redraw the graph using \(x\)-values that are perfect squares, \(0\), \(1\), \(4\), \(9\), and \(16\).

These \(x\)-values result in whole number of values after the square root is taken and so the points will be easier to plot accurately.

You may not be used to selecting \(x\)-values that increase inconsistently in your tables. However, this would only be a problem if we were trying to check first or second differences.

As before, take the square root of each \(x\)-coordinate to find the corresponding \(y\)-coordinates, which are \(0\), \(1\), \(2\), \(3\), and \(4\).

\(x\) \(f(x)= \sqrt x\)
\(0\) \(0\)
\(1\) \(1\)
\(4\) \(2\)
\(9\) \(3\)
​​​​​\(16\) \(4\)

We can plot the five points and join them with a curve.

The graph of the square root function.

Again, make sure that the end point is clearly marked at \((0, 0)\) and the curve extends in the other direction.

With these five points, we get a more extended look at the graph. We can see that the curve becomes less and less steep as it gets further from the origin.

Having a graph of a function can make it easier to describe its domain and range.

The graph of the square root function.

Recall

The domain of \(f(x) = \sqrt x\) is \(\{x \in \mathbb{R}\mid x \geq 0\}\).

We now know this for two reasons.

  1. The square root of a negative number is not a real number, so \(x \ge 0\).
  2. We can see on the graph that there is a defined end point where \(x = 0\), and that this represents the minimum \(x\)-value. All of the points on the curve have \(x\)-coordinates greater than or equal to \(0\).

Recall

The range of \(f(x) = \sqrt x\) is \(\{y \in \mathbb{R} \mid y \geq 0\}\).

This is also for two reasons.

  1. The \(\sqrt{x}\) refers to the positive square root only, so \(f(x) \ge 0\).
  2. We can see that the graph has a defined end point where \(y = 0\), and that this is the minimum possible \(y\)-value. All of the points on the curve have \(y\)-coordinates greater than or equal to \(0\).

Check Your Understanding 1

Question

Which of the following points lies on the graph of \(f(x)=\sqrt{x}\)?

  1. \((-100, -10)\)
  2. \((100, -10)\)
  3. \((100, 10)\)
  4. \((-100, 10)\)

Answer

  1. \((100, 10)\)

Feedback

The point \((100, 10)\) lies on the graph of \(f(x)=\sqrt{x}\).

For \(f(x)=\sqrt{x}\) remember that

  • the square root of a negative number is not a real number, so \(x\) cannot be negative.
  • \(\sqrt{x}\) is referring to the positive square root of \(x\), so \(f(x)\) or \(y\) will always be greater than or equal to zero.

Check Your Understanding 2

Question

Which of the following graphs represents the function \(f(x)=\sqrt{x}\)?

Graph A

A parabola turned on its side, opening to the right. The points (0, 0), (1, 1), (1, negative 1), (4, 2), and (4, negative 2) are identifiable.

Graph B

A function originating at the point (0,0) and passing through the point (16, 2).

Graph C

A curve originating at the point (0,0) and passing through the points (4, 2) and (16, 4).

Answer

Graph C

Feedback

Graph A

A parabola turned on its side, opening to the right. The points (0, 0), (1, 1), (1, negative 1), (4, 2), and (4, negative 2) are identifiable.

This graph does not represent a function. It does not pass the vertical line test. Therefore, this cannot be the graph of \(f(x)=\sqrt{x}\).

Graph B

A function originating at the point (0,0) and passing through the point (16, 2).

Based on some of the points on this curve,

  • when \(x=4\), \(f(x)\approx 1.4\)
  • when \(x=9\), \(f(x)\approx 1.7\)

  • when \(x=16\), \(f(x)=2\)

\(\sqrt{4}\ne 1.4\), \(\sqrt{9}\ne 1.7\), and \(\sqrt{16}\ne 2\).

This is not the graph of \(f(x)=\sqrt{x}\).

Graph C

A curve originating at the point (0,0) and passing through the points (4, 2) and (16, 4).

Based on some of the points on this curve,

  • when \(x=1\), \(f(x)=1\)
  • when \(x=4\), \(f(x)=2\)
  • when \(x=9\), \(f(x)=3\)

  • when \(x=16\), \(f(x)=4\)

\(\sqrt{1}=1\), \(\sqrt{4}=2\), \(\sqrt{9}=3\), and \(\sqrt{16}=4\).

This is the graph of \(f(x)=\sqrt{x}\).


The Reciprocal Function


The reciprocal of a real number \(k\), for \(k \neq 0\), is the number that, when multiplied by \(k\), gives a product equal to \(1\).

The reciprocal of a real number \(k\) can be obtained by dividing \(1\) by \(k\). That is, the reciprocal of \(k\) is \(\dfrac{1}{k}\)   (which can be written as \(k^{-1}\)) since:  

\(\begin{align*} k \left( \dfrac1k \right)&= \dfrac kk \\ &= 1 \end{align*}\)

and

\(\begin{align*} k\times k^{-1}&= k^0 \\ &= 1\end{align*} \)

The product is \(1\), as required.

Did You Know?

Since \(k\left(\dfrac1k \right)=1\) and \(\dfrac1k(k)=1\) for \(k \neq 0\),

\(\dfrac1k\) is the reciprocal of \(k\)
and
\(k\) is the reciprocal of \(\dfrac1k\).

Some examples of reciprocals:

  1. \(\)Since \(2\left(\dfrac12\right)=1\) (which also means that \( \dfrac12(2)=1\)),

    \(\dfrac12\) is the reciprocal of \(2\)
    and
    \(2\) is the reciprocal of \(\dfrac12\).

  2. Since \(-\dfrac16(-6)=1\) ,

    \(-6\) is the reciprocal of \(-\dfrac16\)
    and
    \(-\dfrac16\) is the reciprocal of \(-6\).

  3. Since \(\dfrac45\left(\dfrac54\right)=\dfrac{20}{20} = 1\),

    \(\dfrac54 \) is the reciprocal of \(\dfrac45\)
    and
    \(\dfrac45\) is the reciprocal of \(\dfrac54\).

  4. Since \(-0.002(-500)=1\),

    \(-500\) is the reciprocal of \(-0.002\)
    and
    \(-0.002\) is the reciprocal of \(-500\).

It is also possible to take the reciprocal of irrational numbers (such as \(\sqrt 2\)), although we will not need to do that in this lesson.

The number \(0\) does not have a reciprocal, since there is no real number that multiplies with \(0\) to give a product that is equal to \(1\).


The ​​Reciprocal Function

Let's sketch the graph of another common function.

The function \(f(x)=\dfrac1x\) is called the reciprocal function.

The output, \(\dfrac1x\), is the reciprocal of the input, \(x\). 

Math in Action

The function \(f(x)=\dfrac1x\) can be used to represent the time (in minutes) that it takes to walk \(1\) km at a speed of \(x\) km/min.

As with other functions, it is possible to graph this function by creating a table of values.

Graphing the Reciprocal Function

To create a table, we select various values of \(x\) and evaluate the function at these values.

Let's start with the integer values \(-3\), \(-2\), \(-1\), \(0\), \(1\), \(2\), and \(3\).

\(x\) \(f(x)=\dfrac1x\)
\(-3\)  
\(-2\)  
\(-1\)  
\(0\)  
\(1\)  
\(2\)  
\(3\)  

\(f(-3) = -\dfrac{1}{3}\), since the reciprocal of \(-3\) is \(-\dfrac{1}{3}\), or because \(1 \div -3 = -\dfrac{1}{3}\).

Similarly, the reciprocal of \(-2\) is \(-\dfrac{1}{2}\) and the reciprocal of \(-1\) is \(-1\), since \(1  \div -1 = -1\).

When we come to \(x = 0\), the reciprocal is undefined, since it's not possible to divide \(1\) by \(0\).

This means that the graph will not contain any points with an \(x\)-coordinate of \(0\).

\(x\) \(f(x)=\dfrac1x\)
\(-3\) \(-\frac13 \)
\(-2\) \(-\frac12\)
\(-1\) \(-1\)
\(0\) undefined
\(1\)  
\(2\)  
\(3\)  

Moving on, we can find that the reciprocal of \(1\) is \(1\), the reciprocal of \(2\) is \(\dfrac{1}{2}\), and the reciprocal of \(3\) is \(\dfrac{1}{3}\).

There are now \(6\) points that we can plot from the values in the table.

\(x\) \(f(x)=\dfrac1x\)
\(-3\) \(-\frac13 \)
\(-2\) \(-\frac12\)
\(-1\) \(-1\)
\(0\) undefined
\(1\) \(1\)
\(2\) \(\frac12\)
\(3\) \(\frac13 \)

It appears that the graph of this function is in two pieces.

So let's join the three points in quadrant \(\text{I}\) and join the three points in quadrant \(\text{III}\).

At this point, we still don't have a great sense of what this function truly looks like. It will help to consider two more \(x\)-values, \(-\dfrac{1}{2}\) and \(\dfrac{1}{2}\).

Since the reciprocal of \(-2\) is \(-\dfrac{1}{2}\). The reciprocal of \(-\dfrac{1}{2}\) is \(-2\).

That is \(f\left(-\dfrac{1}{2}\right) = -2\).

Similarly, \(f\left(\dfrac{1}{2}\right) = 2\).

\(x\) \(f(x)=\dfrac1x\)
\(-3\) \(-\frac13 \)
\(-2\) \(-\frac12\)
\(-1\) \(-1\)
\(-\frac12\) \(-2\)
\(0\) undefined
\(\frac12\) \(2\)
\(1\) \(1\)
\(2\) \(\frac12\)
\(3\) \(\frac13 \)

If we add these two points to our graph and extend the curves to join, the shape of the function becomes a little more clear.

However, we still need to investigate what happens at large positive and large negative \(x\)-values as well as what happens for \(x\)-values close to \(0\). That is, points that are closer to the \(y\)-axis.

Let's start by considering larger positive \(x\)-values.

We can add to our existing points by finding \(f(4)\), \(f(5)\), and \(f(10)\).

  • \(f(4) = \dfrac{1}{4}\)
  • \(f(5) = \dfrac{1}{5}\)
  • \(f(10) = \dfrac{1}{10}\)

If we plot these three additional points, we can see that they are getting closer and closer to the \(x\)-axis, but are not quite touching it.

This pattern continues for even larger values of \(x\).

\(f(100) = \dfrac{1}{100}\), which is still not quite zero.

\(x\) \(f(x)=\dfrac1x\)
\(\frac12\) \(2\)
\(1\) \(1\)
\(2\) \(\frac12\)
\(3\) \(\frac13 \)
\(4\) \(\frac14\)
\(5\) \(\frac15\)
\(\vdots\) \(\vdots\)
\(10\) \(\frac1{10}\)
\(\vdots\) \(\vdots\)
\(100\) \(\frac1{100}\)

Even if we have \(x = 1~000~000\), the output of the function will be \(\dfrac{1}{1~000~000} = 0.000001\). Much closer, but still not equal to zero.

No matter how large of a positive \(x\)-value we input, the output of the function will not exactly equal zero. 

So this portion of the graph will not cross the \(x\)-axis. Since all of the values for \(f(x)\) are positive here, the curve will stay above the \(x\)-axis and get closer and closer to it, without ever touching it.

We find a similar result for larger negative \(x\)-values such \(f(-4)\), \(f(-5)\), and \(f(-10)\).

If we add them to the graph, we can see that these points are also getting closer to the \(x\)-axis, although this time they are below the axis.

\(f(-100) = - \dfrac{1}{100}\), which is closer to zero, but still not quite there.

\(x\) \(f(x)=\dfrac1x\)
\(-\frac12\) \(-2\)
\(-1\) \(-1\)
\(-2\) \(-\frac12\)
\(-3\) \(-\frac13 \)
\(-4\) \(-\frac14\)
\(-5\) \(-\frac15\)
\(\vdots\) \(\vdots\)
\(-10\) \(-\frac1{10}\)
\(\vdots\) \(\vdots\)
\(-100\) \(-\frac1{100}\)

As the \(x\)-coordinate of the points on this portion of the graph get more and more negative, \(f(x)\) gets closer and closer to zero, but will not equal zero exactly.

So this portion of the curve doesn't touch the \(x\)-axis, either. It stays below the \(x\)-axis.

It turns out that there are no points on the function \(f(x) = \dfrac{1}{x}\) that touch the \(x\)-axis.

Now, let's look at some \(x\)-values that are closer to zero.

Starting with positive \(x\)-values, we find that

  • \(f\left(\dfrac{1}{3}\right) = 3\), since \(\dfrac{1}{3}\) and \(3\) are reciprocals, 
  • \(f\left(\dfrac{1}{4}\right) = 4\), and
  • \(f\left(\dfrac{1}{5}\right) = 5\).

Adding these points to the graph, we can see that the output of the function or \(y\)-value is getting significantly larger as the \(x\)-values get closer to zero.

This is confirmed when we check the value of \(f\left(\dfrac{1}{10}\right)\), which is \(10\).

This pattern would continue, \(f\left(\dfrac{1}{100}\right) = 100\). So that point would have a \(y\)-coordinate of \(100\). The value, \(f(x)\), becomes a larger and larger positive number as \(x\) gets closer to zero on the positive side. However, we know that the graph will not touch the \(y\)-axis since \(f(0)\) is undefined.

Something similar happens for negative \(x\)-values.

  • \(f\left(-\dfrac{1}{3}\right) = -3\)
  • \(f\left(-\dfrac{1}{4}\right) = 4\)
  • \(f\left(-\dfrac{1}{5}\right) = -5\), and
  • \(f\left(-\dfrac{1}{10}\right) = -10\).

The output of the function becomes a larger and larger negative number, the closer the \(x\)-value gets to zero on the negative side. However this portion of the graph doesn't touch the \(y\)-axis either.

We now have a much better sense of the shape of this curve. The graph of the function \(f(x) = \dfrac{1}{x}\) consists of two parts or branches. It never crosses the \(x\)-axis or the \(y\)-axis, although the curve does get quite close to both.

Asymptotes

One important feature of the reciprocal function is that it has what are known as asymptotes. 

A line is called an asymptote if, as a function \(y=f(x)\) heads towards infinity, the function continually gets closer to (but does not touch) that line.

There are three kinds of asymptote: horizontal, vertical, and slant/oblique.

A slant/oblique asymptote refers to a line that is not vertical or horizontal, but we will not be seeing any of those in this lesson.

Note: A function can cross an asymptote.

For the functions that we typically consider in high school, such asymptote crossings (if they occur) usually happen a finite number of times (and for values of \(x\) that aren’t too far from zero).

More complicated functions exist that do cross their asymptote(s) an infinite number of times.

For the function \(f(x)=\dfrac1x\) has two asymptotes,

  • The graph of \(y=f(x)\) approaches but does not touch the \(x\)-axis (the line \(y=0\)). So there is a horizontal asymptote at \(y=0\).

  • The graph of \(y=f(x)\) approaches but does not touch the \(y\)-axis (the line \(x=0\)). So there is a vertical asymptote at \(x=0\).

These asymptotes are features that we will want to include when sketching the graph.

Drawing Asymptotes

Did You Know?

When graphing the reciprocal function \(f(x)=\dfrac1x\) by hand, the asymptotes should be drawn with dotted lines (since they are not actually part of the curve).

This differentiates the asymptotes from the function, which is drawn as a solid curve.

A Sketch of the Reciprocal Function

In our exploration of the graph of the reciprocal function, we found the coordinates of many different points on the function.

When sketching a graph of this function by hand, it isn't necessary to include this many points.

Instead we can use the asymptotes to help us correctly approximate the shape of the curve.

Here are some recommended steps for graphing by hand.

To sketch \(y=f(x)\) for \(f(x)=\dfrac1x\):

  • Plot some of the points on each "branch" of the curve
    \(x\) \(f(x)=\dfrac1x\)
    \(-\frac12\) \(-2\)
    \(-1\) \(-1\)
    \(-2\) \(-\frac12\)
    \(x\) \(f(x)=\dfrac1x\)
    \(\frac12\) \(2\)
    \(1\) \(1\)
    \(2\) \(\frac12\)

    This gives us a pretty good sense of the shape of the curves.

  • Mark the horizontal asymptote \(y=0\) and the vertical asymptote \(x=0\).

    Recall

    When graphing this function by hand, the asymptotes should be drawn with dotted lines (since they are not actually part of the curve).

  • Join the points and extend the curve so it approaches both asymptotes

    It is important to ensure that your graph does not ever cross either of the asymptotes.

  • Also make sure to indicate that each part of the curve extends past what has been drawn, either by adding arrows on the ends of the curve or by extending both branches of the curve to the edge of the grid as has been done here.

Domain and Range

Now, that we have become more familiar with the reciprocal function and its graph, let's discuss its domain and range.

Recall

The domain of \(f(x) = \dfrac1x\) is \(\{x \in \mathbb{R} \mid x \neq 0\}\).

We know this for two reasons.

  1. \(1\) divided by \(0\) is undefined. So \(0\) is not a possible input for the function. In fact, it is the only input that doesn't work.
  2. The graph does not cross the \(y\)-axis. So there are no points with \(x = 0\).

Recall

The range of \(f(x) = \dfrac 1x\) is \(\{y \in \mathbb{R} \mid y \neq 0 \}\).

Again, there are two reasons for this.

  1. \(1\) divided by any real value of \(x\) can never equal \(0\). So the output of the function can never be \(0\).
  2. The graph doesn't touch the \(x\)-axis so there are no points with \(y = 0\).

Check Your Understanding 3

Question

Which of the following points lies on the graph of \(f(x)=\dfrac{1}{x}\)?

  1. \(\left(\dfrac{1}{5}, 5\right)\)
  2. \(\left(0, -200\right)\)
  3. \(\left(10, 0\right)\)
  4. \(\left(9, -\dfrac{1}{9}\right)\)

Answer

  1. \(\left(\dfrac{1}{5}, 5\right)\)

Feedback

The point \(\left(\dfrac{1}{5}, 5\right)\) lies on the graph of \(f(x)=\dfrac{1}{x}\).

For \(f(x)=\dfrac{1}{x}\) remember that

  • \(1\) divided by a number can never equal \(0\), so \(f(x)\) or \(y\) will never be \(0\).
  • division by \(0\) is undefined, so \(x\) cannot be \(0\).

  • \(x\) multiplied by its reciprocal, \(f(x)\) or \(y\), will always equal \(1\).


Check Your Understanding 4

Question

Which of the following graphs represents the function \(f(x)=\dfrac{1}{x}\)?

Graph A

A function with a horizontal asymptote at y equals 0 and a vertical asymptote at x equals 0. The curve goes through the points at (negative 2, negative 1 over 8), (negative 1, negative 1), (negative 1 over 2, negative 8), (1 over 2, 8), (1, 1), and (2, 1 over 8).

Graph B

A function with a horizontal asymptote at y equals 0 and a vertical asymptote at x equals 0. The curve goes through the points at (negative 2, 1 over 8), (negative 1, 1), (negative 1 over 2, 8), (1 over 2, negative 8), (1, negative 1), and (2, negative 1 over 8).

Graph C

A function with a horizontal asymptote at y equals 0 and a vertical asymptote at x equals 0. The curve goes through the points at (negative 2, negative 1 over 2), (negative 1, negative 1), (negative 1 over 2, negative 2), (1 over 2, 2), (1, 1), and (2, 1 over 2).

Graph D

A function with a horizontal asymptote at y equals 0 and a vertical asymptote at x equals 0. The curve goes through the points at (negative 2, 1 over 2), (negative 1, 1), (negative 1 over 2, 2), (1 over 2, 2), (1, 1), and (2, 1 over 2).

Answer

Graph C

Feedback

Graph A

A function with a horizontal asymptote at y equals 0 and a vertical asymptote at x equals 0. The curve goes through the points at (negative 2, negative 1 over 8), (negative 1, negative 1), (negative 1 over 2, negative 8), (1 over 2, 8), (1, 1), and (2, 1 over 8).

On this graph we can see that when \(x=\dfrac{1}{2}\), \(f(x)=8\).

\(8\) is not the reciprocal of \(\dfrac{1}{2}\).

This is not the graph of \(f(x)=\dfrac{1}{x}\).

Graph B

A function with a horizontal asymptote at y equals 0 and a vertical asymptote at x equals 0. The curve goes through the points at (negative 2, 1 over 8), (negative 1, 1), (negative 1 over 2, 8), (1 over 2, negative 8), (1, negative 1), and (2, negative 1 over 8).

On this graph

  • when \(x\) is positive, \(f(x)\) is negative.
  • when \(x\) is negative, \(f(x)\) is positive.

However, the reciprocal of a positive number is always positive, and the reciprocal of a negative number is always negative.

This is not the graph of \(f(x)=\dfrac{1}{x}\).

Graph C

A function with a horizontal asymptote at y equals 0 and a vertical asymptote at x equals 0. The curve goes through the points at (negative 2, negative 1 over 2), (negative 1, negative 1), (negative 1 over 2, negative 2), (1 over 2, 2), (1, 1), and (2, 1 over 2).

On this graph

  • when \(x=\dfrac{1}{2}\), \(f(x)=2\).
  • when \(x=1\), \(f(x)=1\).

  • when \(x=2\), \(f(x)=\dfrac{1}{2}\).

\(1\) is the reciprocal of \(1\) and \(\dfrac{1}{2}\) is the reciprocal of \(2\).

Also, \(-1\) is the reciprocal of \(-1\) and \(-\dfrac{1}{2}\) is the reciprocal of \(-2\).

This is the graph of \(f(x)=\dfrac{1}{x}\).

Graph D

A function with a horizontal asymptote at y equals 0 and a vertical asymptote at x equals 0. The curve goes through the points at (negative 2, 1 over 2), (negative 1, 1), (negative 1 over 2, 2), (1 over 2, 2), (1, 1), and (2, 1 over 2).

For positive values of \(x\), \(f(x)\) represents the correct reciprocal of \(x\).

However, for negative values of \(x\), \(f(x)\) is still positive.

Negative numbers do not have positive reciprocals.

This is not the graph of \(f(x)=\dfrac{1}{x}\).


Summary of Functions


  Quadratic Function, \(f(x)=x^2\) Square Root Function, \(f(x)=\sqrt x\) Reciprocal Function, \(f(x)= \dfrac1x\)
Graph

Graph of f(x)=x^2

Graph of square root function f(x)=square root of x

Graph of the reciprocal function f(x) = 1 over x, asymptotes defined at x=0 and y=0
Domain \(\{ x\in \mathbb{R}\}\) \(\{x \in \mathbb{R} \mid x \geq 0\}\) \(\{x \in \mathbb{R} \mid x \neq 0\}\)
Range \(\{y \in \mathbb{R} \mid y \geq 0\}\) \(\{y \in \mathbb{R} \mid y \geq 0\}\) \(\{y \in \mathbb{R} \mid y \neq 0\}\)

The following Check Your Understanding is a slideshow without audio. Advance through the slides at your own pace.


Wrap-Up


Lesson Summary

  • In this lesson, we sketched three common functions:
    • Here is the quadratic function \(f(x)=x^2\) with some key points marked:

      Graph of quadratic function y equals x squared, with points marked at (negative 3,9), (negative 2, 4),(negative 1,1), (0,0), (1,1), (2,4) and (3,9).

    • Here is the graph of the square root function \(f(x) = \sqrt x\) with some key points marked:

      Graph of the square root function, with points marked at (0,0),(1,1),(4,2) and (9,3)

    • Here is the graph of the reciprocal function \(f(x)= \dfrac1x\) with some key points and both asymptotes:

      The graph of y equals 1 over x with the points (negative 2, negative 1 over 2), (negative 1, negative 1), (negative 1 over 2, negative 2), (1 over 2, 2), (1, 1), and (2, 1 over 2).

  • Based on the above sketches, we also identified the domain and the range for each of the functions:

    • The domain of \(f(x)=x^2\)is \( \mathbb{R}\) and the range is \(\{y \in \mathbb{R} \mid y\geq 0\}\). 

    • The domain of \(f(x) = \sqrt x\) is \(\{x \in \mathbb{R} \mid x \geq 0\}\) and the range is \(\{y \in \mathbb{R} \mid y\geq 0\}\).

    • The domain of \(f(x) = \dfrac1x\) is \(\{x \in \mathbb{R} \mid x \neq 0\}\) and the range is \(\{y \in \mathbb{R} \mid y\neq 0\}\). 

  • Some functions, such as the reciprocal function \(f(x)= \dfrac1x\), have one or more vertical and/or horizontal asymptotes. The curve \(f(x)= \dfrac1x\) approaches, but does not touch, its asymptotes. 

Take It With You

Consider the functions \(f(x)=\dfrac1x-5\) and \(g(x)=\dfrac{1}{x+3}\).

What are the domain and range of each of these functions? Where would you need to draw asymptotes for these functions?

Hint: Consider which values of \(x\) would not be allowable as inputs, and which values are impossible to achieve as outputs of \(f(x)\) or \(g(x)\).