Unit outcomes

By the end of this unit you will be able to:

• Sketch and find the equation of the graph [latex]\scriptsize y=a.{{b}^{x}}+q,b \gt 0[/latex].
• Investigate and generalise the impact of [latex]\scriptsize a[/latex] and [latex]\scriptsize q[/latex] on [latex]\scriptsize y=a.{{b}^{x}}+q,b \gt 0[/latex].

What you should know

Before you start this unit, make sure you can:

• Determine the domain and range of a function by looking at its graph. Go over Units 1, 2 and 3 in this subject outcome if you need help with this.
• Manipulate and simplify algebraic expressions. Go over Subject outcome 2.2, Unit 1: Simplifying algebraic expressions if you need more help with the basics.
• Solve exponential equations. Go over Subject outcome 2.3, Unit 2: Solve exponential and literal equations if you need more help with the basics.
• Plot points on the Cartesian plane. If you do not know how to plot points onto the Cartesian plane, then you should do Subject outcome 3.3, Unit 1: Plotting points on the Cartesian plane before continuing with this unit.
• Solve linear inequalities. Go over Subject outcome 2.3, Unit 3: Solve algebraic inequalities if you need more help with this.
• Solve simultaneous equations. Go over Subject outcome 2.3, Unit 4: Solve simultaneous equations if you need more help with this.
• Work with the exponential function. This is briefly introduced in Subject outcome 1.2, Unit 2: Introduction to exponents. You should complete this unit before continuing.

Introduction

You have probably heard the term “exponential growth” before. However, the word ‘exponential’ is often used incorrectly to refer to something that happens very quickly. In Mathematics, ‘exponential’ has a very specific meaning and exponential functions are specific kinds of functions.

The good news is that these functions behave quite similarly to the other functions we have explored so far. Therefore, much of what we know about linear, quadratic and hyperbolic functions can be applied to exponential functions as well.

There are many processes in nature and finance where the relationship between the inputs and outputs is exponential. One example is how bacteria reproduce. Bacteria reproduce by simply splitting into two new bacteria. No sexual reproduction takes place.

So, if one bacterium splits into two and then each of these splits into two and then each of these splits into two, etc., then pretty soon we have a whole lot of bacteria! How many bacteria? Let’s find out.

The exponential function

Activity 4.1: Investigate the exponential function

Time required: 30 minutes

What you need:

• a pen or pencil
• a calculator
• blank paper or a notebook

What to do:

For this activity, we will assume that bacteria multiply once every day. When they multiply, each one splits into two new bacteria. If we start with one bacterium, the progression for the first three days will look like Figure 1.

1. Use and extend Figure 1 to create a table of values of the number of bacteria after each day for eight days.
2. Plot these points and join them with a smooth curve. To help you plot the graph use a scale of [latex]\scriptsize 1[/latex], [latex]\scriptsize 2[/latex], [latex]\scriptsize 3[/latex]… on the x-axis and [latex]\scriptsize 10[/latex], [latex]\scriptsize 20[/latex], [latex]\scriptsize 30[/latex]… on the y-axis.
3. Write an expression that describes the relationship between the inputs (the number of days) and the outputs (the number of bacteria). Hint: the outputs increase in powers of 2.
4. How many bacteria are there after two weeks?
5. Where will the graph cut the y-axis? Add this point to your plot.
6. What happens when [latex]\scriptsize x \lt 0[/latex]? Try a few values of [latex]\scriptsize x[/latex] to find out and add these to your sketch.
7. Does the graph have a horizontal asymptote? If so, what is this asymptote?
8. Does the graph have a vertical asymptote? If so, what is this asymptote?
9. What is the domain and range of this graph?
10. Is the relationship between [latex]\scriptsize x[/latex] and [latex]\scriptsize y[/latex] a function? Why?

What did you find?

1. Here is the completed table of values.
   

   Days ([latex]\scriptsize x[/latex])	[latex]\scriptsize 1[/latex]	[latex]\scriptsize 2[/latex]	[latex]\scriptsize 3[/latex]	[latex]\scriptsize 4[/latex]	[latex]\scriptsize 5[/latex]	[latex]\scriptsize 6[/latex]	[latex]\scriptsize 7[/latex]	[latex]\scriptsize 8[/latex]
   Number of bacteria ([latex]\scriptsize y[/latex])	[latex]\scriptsize 2[/latex]	[latex]\scriptsize 4[/latex]	[latex]\scriptsize 8[/latex]	[latex]\scriptsize 16[/latex]	[latex]\scriptsize 32[/latex]	[latex]\scriptsize 64[/latex]	[latex]\scriptsize 128[/latex]	[latex]\scriptsize 256[/latex]

2. If we plot these points and join them with a smooth curve, we get the graph in Figure 2.
   

   

3. We can see that the y-values increase by powers of [latex]\scriptsize 2[/latex]. Here is the same table of values but with the y-values expressed as powers of [latex]\scriptsize 2[/latex].
   

   Days ([latex]\scriptsize x[/latex])	[latex]\scriptsize 1[/latex]	[latex]\scriptsize 2[/latex]	[latex]\scriptsize 3[/latex]	[latex]\scriptsize 4[/latex]	[latex]\scriptsize 5[/latex]	[latex]\scriptsize 6[/latex]	[latex]\scriptsize 7[/latex]	[latex]\scriptsize 8[/latex]
   Number of bacteria ([latex]\scriptsize y[/latex])	[latex]\scriptsize 2^1[/latex]	[latex]\scriptsize 2^2[/latex]	[latex]\scriptsize 2^3[/latex]	[latex]\scriptsize 2^4[/latex]	[latex]\scriptsize 2^5[/latex]	[latex]\scriptsize 2^6[/latex]	[latex]\scriptsize 2^7[/latex]	[latex]\scriptsize 2^8[/latex]

   We can see that the exponent in each case matches the x-value. Therefore, we can say that the equation of the graph is [latex]\scriptsize y=2^x[/latex].

4. [latex]\scriptsize 2[/latex] weeks is [latex]\scriptsize 14[/latex] days. If we use our equation, we get [latex]\scriptsize y=2^{14}=16~384[/latex]. That is a lot of bacteria!
5. The y-intercept is when [latex]\scriptsize x=0[/latex]. Therefore, [latex]\scriptsize y=2^0=1[/latex]. So the y-intercept is the point [latex]\scriptsize (0,1)[/latex].
6. You can try any values but let’s use [latex]\scriptsize -1[/latex], [latex]\scriptsize -2[/latex] and [latex]\scriptsize -3[/latex].
   [latex]\scriptsize y=2^{-1}=\displaystyle \frac{1}{2}[/latex]
   [latex]\scriptsize y=2^{-2}=\displaystyle \frac{1}{4}[/latex]
   [latex]\scriptsize y=2^{-3}=\displaystyle \frac{1}{8}[/latex]
   If we add these to our sketch, we get the graph in Figure 3. We have zoomed into this part of the graph to make things clearer.

   

7. As the value of [latex]\scriptsize x[/latex] decreases, the value of [latex]\scriptsize y[/latex] will get closer and closer to zero but will never equal zero and will never cross the x-axis to become negative. Therefore the x-axis (the line [latex]\scriptsize y=0[/latex]) is a horizontal asymptote.
8. As [latex]\scriptsize x[/latex] increases, [latex]\scriptsize y[/latex] will get bigger and bigger but there are no values that [latex]\scriptsize x[/latex] gets close to but never reaches. Therefore there is no vertical asymptote.
9. [latex]\scriptsize x[/latex] can be any real number. Therefore the domain is [latex]\scriptsize x\in \mathbb{R}[/latex]. We have already seen that the line [latex]\scriptsize y=0[/latex] is a horizontal asymptote and that [latex]\scriptsize y[/latex] is never negative. Therefore the range is [latex]\scriptsize \{y\ |\ y\in \mathbb{R},y \gt 0\}[/latex].
10. Using the vertical line test (see Figure 4), we can see that for every input value we only ever get one output value. Therefore [latex]\scriptsize y={{2}^{x}}[/latex] is a function.
    

    

The effect of [latex]\scriptsize \large b[/latex] on [latex]\scriptsize \large y={{b}^{x}}[/latex]

From Activity 4.1 we learnt that equations like [latex]\scriptsize y={{2}^{x}}[/latex] (general form [latex]\scriptsize y={{b}^{x}}[/latex]) are functions, and that they have a horizontal asymptote and a y-intercept. But what happens if we have the function [latex]\scriptsize g(x)={{3}^{x}}[/latex] or [latex]\scriptsize h(x)={{\left( \displaystyle \frac{1}{3} \right)}^{x}}[/latex]? Can [latex]\scriptsize b \gt 0[/latex]?

Activity 4.2: Investigate the effect of b on y= x²

Time required: 30 minutes

What you need:

• a pen or pencil
• a calculator
• blank paper or a notebook
• an internet connection

What to do:

Part A

1. Given [latex]\scriptsize f(x)={{2}^{x}}[/latex], [latex]\scriptsize g(x)={{3}^{x}}[/latex] , [latex]\scriptsize h(x)={{\left( \displaystyle \frac{1}{3} \right)}^{x}}[/latex] and [latex]\scriptsize y={{(-2)}^{x}}[/latex], create and complete the following table of values. Note: You are given [latex]\scriptsize y={{(-2)}^{x}}[/latex]. The negative sign is included in the base. If you were given [latex]\scriptsize y=-{{2}^{x}}[/latex], you would assume that only [latex]\scriptsize 2[/latex] is in the base and the negative sign (or [latex]\scriptsize -1[/latex]) is out front and is not raised to the exponent.
   

   [latex]\scriptsize x[/latex]	[latex]\scriptsize -2[/latex]	[latex]\scriptsize -1[/latex]	[latex]\scriptsize 0[/latex]	[latex]\scriptsize 1[/latex]	[latex]\scriptsize 2[/latex]
   [latex]\scriptsize f(x)[/latex]
   [latex]\scriptsize g(x)[/latex]
   [latex]\scriptsize h(x)[/latex]
   [latex]\scriptsize y=(-2)^x[/latex]

2. Sketch the graphs on the same set of axes.
3. What is the difference between [latex]\scriptsize f(x)[/latex] and [latex]\scriptsize g(x)[/latex]?
4. What is the difference between [latex]\scriptsize f(x)[/latex] and [latex]\scriptsize h(x)[/latex]?
5. What are the y-intercepts of [latex]\scriptsize f(x)[/latex], [latex]\scriptsize g(x)[/latex] and [latex]\scriptsize h(x)[/latex]?
6. Why do the graphs all have the same y-intercept?
7. Can you plot [latex]\scriptsize y=-2^x[/latex]. If not, why not?

Part B

When you have access to the Internet, visit this exponential function interactive simulation.

Here you will find an exponential function of the form [latex]\scriptsize y=b^x[/latex] with a slider to change the values of [latex]\scriptsize b[/latex].

1. What happens when [latex]\scriptsize b[/latex] changes from being greater than one to less than one?
2. What happens when [latex]\scriptsize b=1[/latex]? Why is this?
3. What happens when [latex]\scriptsize b \lt 0[/latex]? Why is this?

What did you find?

Part A

1. Here is the completed table of values.
   

   ([latex]\scriptsize x[/latex])	[latex]\scriptsize -2[/latex]	[latex]\scriptsize -1[/latex]	[latex]\scriptsize 0[/latex]	[latex]\scriptsize 1[/latex]	[latex]\scriptsize 2[/latex]
   ([latex]\scriptsize f(x)[/latex])	[latex]\scriptsize \displaystyle \frac{1}{4}[/latex]	[latex]\scriptsize \displaystyle \frac{1}{2}[/latex]	[latex]\scriptsize 1[/latex]	[latex]\scriptsize 2[/latex]	[latex]\scriptsize 4[/latex]
   ([latex]\scriptsize g(x)[/latex])	[latex]\scriptsize \displaystyle \frac{1}{9}[/latex]	[latex]\scriptsize \displaystyle \frac{1}{3}[/latex]	[latex]\scriptsize 1[/latex]	[latex]\scriptsize 3[/latex]	[latex]\scriptsize 9[/latex]
   ([latex]\scriptsize h(x)[/latex])	[latex]\scriptsize 9[/latex]	[latex]\scriptsize 3[/latex]	[latex]\scriptsize 1[/latex]	[latex]\scriptsize \displaystyle \frac{1}{3}[/latex]	[latex]\scriptsize \displaystyle \frac{1}{9}[/latex]
   [latex]\scriptsize y=(-2)^x[/latex]	[latex]\scriptsize \displaystyle \frac{1}{4}[/latex]	[latex]\scriptsize -\displaystyle \frac{1}{2}[/latex]	[latex]\scriptsize 1[/latex]	[latex]\scriptsize -2[/latex]	[latex]\scriptsize 4[/latex]

    

2. The graphs of each of these functions is shown in Figure 5. The different points generated by [latex]\scriptsize y=-2^x[/latex] are also shown.
   

   

3. The graph of [latex]\scriptsize g(x)[/latex] is steeper than [latex]\scriptsize f(x)[/latex]. The y-values increase faster. This makes sense because for [latex]\scriptsize g(x)[/latex] we have powers of [latex]\scriptsize 3[/latex] not [latex]\scriptsize 2[/latex].
4. While [latex]\scriptsize f(x)[/latex] increases for positive [latex]\scriptsize x[/latex] values and approaches the asymptote for negative [latex]\scriptsize x[/latex] values, [latex]\scriptsize h(x)[/latex] does the opposite. If you look carefully you can see that [latex]\scriptsize g(x)=3^x[/latex] and [latex]\scriptsize h(x)=\left(\displaystyle \frac{1}{3}\right)^x[/latex] are mirror images of each other. They are symmetrical about the y-axis.
5.  (and 6). The y-intercept of all three functions is [latex]\scriptsize (0,1)[/latex]. This is because any base to the power of zero is one.

7. We cannot plot [latex]\scriptsize y=-2^x[/latex] because the points jump around. They don’t all lie on a smooth curve. This makes sense because if we raise a negative number to an even power, the answer will be positive. But if we raise it to an odd power, the answer will be negative.

Part B

1. As soon as [latex]\scriptsize b[/latex] changes from greater than one to less than one, the graph bends in the opposite direction. When [latex]\scriptsize b \gt 0[/latex], the graph increases as [latex]\scriptsize x[/latex] get more positive. When [latex]\scriptsize 0 \lt b \lt 1[/latex], the graph increases as [latex]\scriptsize x[/latex] gets more negative.
2. When [latex]\scriptsize b=1[/latex], the graph is a horizontal line. This is because one raised to any power is always equal to one. So no matter what the [latex]\scriptsize x[/latex] value is, the function value will always be one.
3. When [latex]\scriptsize b \lt 0[/latex], the input/output points do not all line on the same smooth curve so we cannot draw the graph.

From Activity 4.2 we learnt that as [latex]\scriptsize b[/latex] in [latex]\scriptsize y=b^x[/latex] increases, the graph gets steeper as the function values increase more quickly. When [latex]\scriptsize 0 \lt b \lt 1[/latex], the graph flips over to be symmetrical about the y-axis. This means that [latex]\scriptsize y={{b}^{x}}[/latex] and [latex]\scriptsize y={{\left( \displaystyle \frac{1}{b} \right)}^{x}}[/latex] are symmetrical to each other about the y-axis. They are mirror images of each other if we treat the y-axis as the mirror (see Figure 6).

We also saw that [latex]\scriptsize b[/latex] cannot be negative. This is why we always write the basic exponential function as [latex]\scriptsize y=b^x, b \gt 0[/latex].

Example 4.1

Given [latex]\scriptsize j(x)={{4}^{x}}[/latex] and [latex]\scriptsize l(x)={{\left( \displaystyle \frac{1}{4} \right)}^{x}}[/latex].

1. Use a table of values to plot each function on the same set of axes.
2. State the domain and range of each function.
3. About what line are these two graphs symmetrical?

Solutions

1. The graphs are shown in Figure 6.
   

   

2. [latex]\scriptsize j(x)[/latex]:
   Domain: [latex]\scriptsize \{x\ |\ x\in \mathbb{R}\}[/latex]
   Range: [latex]\scriptsize \{y\ |\ y\in \mathbb{R},y \lt 0\}[/latex]
   [latex]\scriptsize l(x)[/latex]:
   Domain: [latex]\scriptsize \{x\ |\ x\in \mathbb{R}\}[/latex]
   Range: [latex]\scriptsize \{y\ |\ y\in \mathbb{R},y \lt 0\}[/latex]
3. The graphs are symmetrical about the y-axis, the line [latex]\scriptsize x=0[/latex].

The effect of [latex]\scriptsize \large q[/latex] on [latex]\scriptsize \large y={{b}^{x}}+q[/latex]

So far, we have looked at the graphs of functions like [latex]\scriptsize y={{b}^{x}}[/latex]. But what about graphs of functions like [latex]\scriptsize y={{b}^{x}}+q[/latex]? How do you think the value of [latex]\scriptsize q[/latex] will affect the graph? Study the next example to see.

Example 4.2

1. Using a table of values, sketch the graphs of [latex]\scriptsize d(x)={{2}^{x}}+2[/latex] and [latex]\scriptsize e(x)={{2}^{x}}-1[/latex].
2. What is the y-intercept of each graph?
3. What are the asymptotes of each graph? How do these compare to the values of [latex]\scriptsize q[/latex] in each function?
4. What is the domain and range of each graph?

Solutions

1. You could have used any values you liked. Here is a table of values that can be used to sketch each function (Figure 8).
   

   [latex]\scriptsize x[/latex]	[latex]\scriptsize -2[/latex]	[latex]\scriptsize -1[/latex]	[latex]\scriptsize 0[/latex]	[latex]\scriptsize 1[/latex]	[latex]\scriptsize 2[/latex]
   [latex]\scriptsize d(x)[/latex]	[latex]\scriptsize \displaystyle \frac{9}{4}[/latex]	[latex]\scriptsize \displaystyle \frac{5}{2}[/latex]	[latex]\scriptsize 3[/latex]	[latex]\scriptsize 4[/latex]	[latex]\scriptsize 6[/latex]
   [latex]\scriptsize e(x)[/latex]	[latex]\scriptsize -\displaystyle \frac{3}{4}[/latex]	[latex]\scriptsize -\displaystyle \frac{1}{2}[/latex]	[latex]\scriptsize 0[/latex]	[latex]\scriptsize 1[/latex]	[latex]\scriptsize 3[/latex]

   

2. The y-intercept of [latex]\scriptsize d(x)[/latex] is [latex]\scriptsize (0,3)[/latex] and the y-intercept of [latex]\scriptsize e(x)[/latex] is [latex]\scriptsize (0,0)[/latex].
3. The horizontal asymptote of [latex]\scriptsize d(x)[/latex] is [latex]y=2[/latex]. The horizontal asymptote of [latex]\scriptsize e(x)[/latex] is [latex]\scriptsize y=-1[/latex]. In each case, the horizontal asymptote is the same as the value of [latex]\scriptsize q[/latex].
4. [latex]\scriptsize d(x)[/latex]:
   Domain: [latex]\scriptsize \{x:x\in \mathbb{R}\}[/latex]
   Range: [latex]\scriptsize \{y:y\in \mathbb{R},y \gt 2\}[/latex]
   [latex]\scriptsize e(x)[/latex]:
   Domain: [latex]\scriptsize \{x:x\in \mathbb{R}\}[/latex]
   Range: [latex]\scriptsize \{y:y\in \mathbb{R},y \gt -1\}[/latex]

In Example 4.2 we saw that changing the value of [latex]\scriptsize q[/latex] shifts the whole graph vertically up or down. This is the same effect changing the value of [latex]\scriptsize q[/latex] has on the linear function ([latex]\scriptsize y=ax+q[/latex]), the quadratic function ([latex]\scriptsize y=a{{x}^{2}}+q[/latex]) and the hyperbolic function ([latex]\scriptsize y=\displaystyle \frac{a}{x}+q[/latex]).

In the exponential function, the horizontal asymptote is the line [latex]\scriptsize y=q[/latex]. This is the same as for the hyperbolic function.

The effect of [latex]\scriptsize \large a[/latex] on [latex]\scriptsize \large y=a.{{b}^{x}}+q[/latex]

Finally, we need to have a look at exponential functions of the form [latex]\scriptsize y=a.{{b}^{x}}+q[/latex]. The functions [latex]\scriptsize j(x)={{2}^{x}}[/latex], [latex]\scriptsize k(x)={{2.2}^{x}}[/latex], and [latex]\scriptsize l(x)=-{{2.2}^{x}}[/latex] are shown in Figure 9.

Firstly, what is the difference between [latex]\scriptsize j(x)={{2}^{x}}[/latex], [latex]\scriptsize k(x)={{2.2}^{x}}[/latex]? Can you see that the graph of [latex]\scriptsize k(x)[/latex] is stretched up so that it intersects the y-axis at [latex]\scriptsize (0,2)[/latex] instead of [latex]\scriptsize (0,1)[/latex]? Every point on the graph of [latex]\scriptsize j(x)={{2}^{x}}[/latex] has been multiplied by two to make the graph of [latex]\scriptsize k(x)={{2.2}^{x}}[/latex].

Secondly, we can also see that the graph of [latex]\scriptsize l(x)=-{{2.2}^{x}}[/latex] is the same shape as [latex]\scriptsize k(x)={{2.2}^{x}}[/latex] except that it has been flipped over horizontally. Instead of going to positive infinity as [latex]\scriptsize x[/latex] gets bigger, the graph goes to negative infinity. Each function value is multiplied by negative two instead of positive two. The function [latex]\scriptsize l(x)=-{{2.2}^{x}}[/latex] still has a horizontal asymptote at [latex]\scriptsize y=0[/latex] (remember that the value of [latex]\scriptsize q[/latex] in the function is zero) but this asymptote now represents a maximum value for the graph rather than a minimum value. The graphs of [latex]\scriptsize k(x)={{2.2}^{x}}[/latex] and [latex]\scriptsize l(x)=-{{2.2}^{x}}[/latex] are mirror images of each other if the mirror is the x-axis. In general, we say that [latex]\scriptsize y=a.{{b}^{x}}+q[/latex] and [latex]\scriptsize y=-a.{{b}^{x}}q[/latex] are symmetrical about the x-axis.

The domains of each of these functions are all still [latex]\scriptsize \{x\ |\ x\in \mathbb{R}\}[/latex]. However, the ranges are different and depend on whether [latex]\scriptsize a \gt 0[/latex] or [latex]\scriptsize a \lt 0[/latex].

When [latex]\scriptsize a \gt 0[/latex], the range of the function is [latex]\scriptsize \{y\ |\ y\in \mathbb{R},y \gt q\}[/latex].
When [latex]\scriptsize a \lt 0[/latex], the range of the function is [latex]\scriptsize \{y\ |\ y\in \mathbb{R},y \lt q\}[/latex].

Figures 10 and 11 show a summary of what we know about the exponential function so far.

Before looking at the next example, get online and visit this interactive simulation.

Here you will find an exponential function of the form [latex]\scriptsize y=a.{{b}^{x}}+q,b \lt 0[/latex] with sliders to change the values of [latex]\scriptsize a[/latex], [latex]\scriptsize b[/latex] and [latex]\scriptsize q[/latex]. Spend some time playing with the graph to make sure that you understand how changing the values of [latex]\scriptsize a[/latex], [latex]\scriptsize b[/latex] and [latex]\scriptsize q[/latex] affects the shape and position of the exponential graph of the form [latex]\scriptsize y=a.{{b}^{x}}+q,b \lt 0[/latex].

Example 4.3

1. Sketch the graph of [latex]\scriptsize t(x)={{3.2}^{2}}+2[/latex]. Mark the intercepts and asymptote.
2. Find the domain and range of [latex]\scriptsize v(x)=-{{5.3}^{x}}-1[/latex].

Solutions

1. To sketch an exponential function, we need to know the sign of [latex]\scriptsize b[/latex], the sign of [latex]\scriptsize a[/latex], the y-intercept, the x-intercept (if one exists) and the horizontal asymptote.
   .
   In [latex]\scriptsize t(x)={{3.2}^{2}}+2[/latex], we can see that [latex]\scriptsize b \gt 1[/latex] and [latex]\scriptsize a \gt 0[/latex]. This means that the graph will grow to positive infinity as [latex]\scriptsize x[/latex] gets larger. [latex]\scriptsize q=2[/latex], which means that the graph will be shifted two units up.
   .
   y-intercept (let [latex]\scriptsize x=0[/latex]):
   [latex]\scriptsize y={{3.2}^{0}}+2=3.1+2=5[/latex]
   The y-intercept is [latex]\scriptsize (0,5)[/latex].
   .
   x-intercept (let [latex]\scriptsize y=0[/latex]):
   [latex]\scriptsize \begin{align} & 0={{3.2}^{x}}+2 \\ & \therefore {{3.2}^{x}}=-2 \\ & \therefore {{2}^{x}}=-\displaystyle \frac{2}{3} \\ \end{align}[/latex]
   Therefore, there is no real solution and hence, no x-intercept.
   .
   The horizontal axis is the line [latex]\scriptsize y=2[/latex]. We plot the y-intercept and the asymptote and draw the graph as shown in Figure 12.

   

2. We need to find the domain and range of [latex]\scriptsize v(x)=-{{5.3}^{x}}-1[/latex]. The domain is all real values. So the domain is [latex]\scriptsize \{x\ |\ x\in \mathbb{R}\}[/latex].
   .
   We can see that [latex]\scriptsize a \lt 0[/latex] and [latex]\scriptsize b \gt 1[/latex]. This means that the graph will decrease infinitely as [latex]\scriptsize x[/latex] gets larger and the horizontal asymptote (the line [latex]\scriptsize y=-1[/latex]) is an upper limit on the range. The range is [latex]\scriptsize y \lt q[/latex] and so is [latex]\scriptsize \{v(x)\ |\ v(x)\in \mathbb{R},v(x) \lt -1\}[/latex].
   .
   Alternatively, you could use inequalities to determine the range.
   [latex]\scriptsize \begin{align} & -{{5.3}^{x}} \lt 0 \\ & \therefore -{{5.3}^{x}}-1 \lt -1 \\ \end{align}[/latex]
   Therefore the range will be all real values less than negative one: [latex]\scriptsize \{v(x)\ |\ v(x)\in \mathbb{R},v(x) \lt -1\}[/latex]

Exercise 4.1

1. Given [latex]\scriptsize f(x)=-\displaystyle \frac{3}{4}.4^{x}+3[/latex]:
   1. Calculate the y-intercept.
   2. Calculate the x-intercept.
   3. Calculate the equation of the horizontal asymptote.
   4. Calculate the domain and range of [latex]\scriptsize f(x)[/latex].
   5. Make a sketch of [latex]\scriptsize f(x)[/latex] indicating the intercepts and equation of the asymptote.
2. Sketch the graph of [latex]\scriptsize y=-2\times {{4}^{x}}+4[/latex] indicating the intercepts and asymptote.
3. Make a neat sketch of [latex]\scriptsize g(x)=-3.{{\left( \displaystyle \frac{1}{3} \right)}^{x}}+3[/latex] showing the intercepts and horizontal asymptote.

The full solutions are at the end of the unit.

Find the equations of exponential functions

Example 4.4

1. Figure 13 shows the graph of an exponential function of the form [latex]\scriptsize y=a\times {{\left( \displaystyle \frac{1}{2} \right)}^{x}}+q[/latex] which cuts the y-axis at [latex]\scriptsize (0,2)[/latex] and the x-axis at [latex]\scriptsize (-1,0)[/latex]. Find the values of [latex]\scriptsize a[/latex] and [latex]\scriptsize q[/latex] and hence, the equation of the function.
   

   

2. An exponential function with an asymptote of [latex]\scriptsize y=0[/latex] passes through [latex]\scriptsize \text{A}(0,1)[/latex] and [latex]\scriptsize \text{B}(2,9)[/latex]. What is the equation of the function?

Solutions

1. We are told that the function is of the form [latex]\scriptsize y=a\times {{\left( \displaystyle \frac{1}{2} \right)}^{x}}+q[/latex]. Therefore, we already know that [latex]\scriptsize b=\displaystyle \frac{1}{2}[/latex]. With the two intercepts, we can set up a system of two simultaneous equations as follows:
   .
   Using [latex]\scriptsize (-1,0)[/latex]:
   [latex]\scriptsize \begin{align} 0 & =a{{\left( \displaystyle \frac{1}{2} \right)}^{-1}}+q \\ \therefore 2a+q & =0 \\ \therefore 2a & =-q\quad \text{Equation 1} \end{align}[/latex]
   .
   Using [latex]\scriptsize (0,2)[/latex]:
   [latex]\scriptsize \begin{align} 2 & =a\times {{\left( \displaystyle \frac{1}{2} \right)}^{0}}+q \\ \therefore 2 & =a+q \\ \therefore a & =2-q\quad \text{Equation 2} \end{align}[/latex]
   .
   Substitute equation 2 into equation 1:
   .
   [latex]\scriptsize \begin{align} 2(2-q) & =-q \\ \therefore 4-2q & =-q \\ \therefore q & =4 \end{align}[/latex]
   .
   Substitute [latex]\scriptsize q=4[/latex] into equation 2:
   [latex]\scriptsize \begin{align} a & =2-4 \\ \therefore a & =-2 \end{align}[/latex]
   Therefore, the equation of the function is [latex]\scriptsize y=-2\times {{\left( \displaystyle \frac{1}{2} \right)}^{x}}+4[/latex].
2. We are told that the asymptote of the exponential function is the line [latex]\scriptsize y=0[/latex]. Therefore, [latex]\scriptsize q=0[/latex] and the function is of the form [latex]\scriptsize y=a.{{b}^{x}}[/latex]. Using [latex]\scriptsize \text{A}(0,1)[/latex], we get [latex]\scriptsize 1=a.{{b}^{0}}=a[/latex]. Now we can use [latex]\scriptsize \text{B}(2,9)[/latex] to find the value of [latex]\scriptsize b[/latex].
   [latex]\scriptsize \begin{align} 9 & ={{b}^{2}} \\ \therefore b & =\pm \text{3} \end{align}[/latex]But we know that [latex]\scriptsize b\ge 0[/latex]. Therefore, we need to exclude the answer [latex]\scriptsize b=-3[/latex].
   Therefore, the equation of the function is [latex]\scriptsize y={{3}^{x}}[/latex].

Exercise 4.2

1. What is the equation of the exponential function of the form [latex]\scriptsize y=a.{{b}^{x}}+q[/latex] represented in the graph?
   
2. An exponential function of the form [latex]\scriptsize y=a.{{b}^{x}}+q[/latex] has a range of [latex]\scriptsize \{y\ |\ y\in \mathbb{R},y \lt-2\}[/latex] and a y-intercept of [latex]\scriptsize (0,-\displaystyle \frac{5}{2})[/latex]. If it passes through the point [latex]\scriptsize (2,-10)[/latex], what is the equation of the function?

The full solutions are at the end of the unit.

Summary

In this unit you have learnt the following:

• How to sketch exponential functions of the form [latex]\scriptsize y=a.{{b}^{x}}+q,b \gt 0[/latex].
• How to find the equation of graphs of the form [latex]\scriptsize y=a.{{b}^{x}}+q,b \gt 0[/latex].
• The effect of [latex]\scriptsize a[/latex], [latex]\scriptsize b[/latex] and [latex]\scriptsize q[/latex] on the shape and position of [latex]\scriptsize y=a.{{b}^{x}}+q,b \gt 0[/latex].
• How to find the asymptote of an exponential function of the form [latex]\scriptsize y=a.{{b}^{x}}+q,b \gt 0[/latex].
• How to determine the domain and range of an exponential function of the form [latex]\scriptsize y=a.{{b}^{x}}+q,b \gt 0[/latex].

Unit 4: Assessment

Suggested time to complete: 60 minutes

1. Pictured below are the following functions:
   [latex]\scriptsize y=-{{2.3}^{x}}+2[/latex]
   [latex]\scriptsize y=3{{\left( \displaystyle \frac{1}{2} \right)}^{x}}-1[/latex]
   [latex]\scriptsize y={{2}^{x}}+1[/latex]
   [latex]\scriptsize y={{2.2}^{x}}-3[/latex]
   [latex]\scriptsize y=-3{{\left( \displaystyle \frac{2}{3} \right)}^{x}}-1[/latex]
   
   Match each one to its graph.
2. Given the functions [latex]\scriptsize q(x)={{3}^{x}}+1[/latex] and [latex]\scriptsize r(x)={{\left( \displaystyle \frac{1}{3} \right)}^{x}}+1[/latex].
   1. Draw the graphs on the same set of axes.
   2. Is the line [latex]\scriptsize y=1[/latex] an asymptote or axis of symmetry of each graph?
   3. Solve the equation [latex]\scriptsize {{3}^{x}}+1={{\left( \displaystyle \frac{1}{3} \right)}^{x}}+1[/latex] graphically.
   4. About what line are these two functions symmetrical?
3. Sketched below are the functions [latex]\scriptsize v(x)=-\displaystyle \frac{3}{x}[/latex] and [latex]\scriptsize w(x)={{3}^{x}}[/latex].
   
   1. What feature(s) or characteristic(s) do the graphs share?
   2. Evaluate [latex]\scriptsize w(1)-v(1)[/latex].
   3. Describe what the expression [latex]\scriptsize w(1)-v(1)[/latex] means graphically.
4. The function [latex]\scriptsize f(x)=-\displaystyle \frac{1}{2}{{\left( \displaystyle \frac{1}{4} \right)}^{x}}+2[/latex] is shown in the diagram.
   
   1. What is the domain and range of [latex]\scriptsize f(x)[/latex]?
   2. What is the equation of the function [latex]\scriptsize g(x)[/latex], [latex]\scriptsize f(x)[/latex] shifted 3 units down?
   3. What is the equation of [latex]\scriptsize h(x)[/latex], the reflection of [latex]\scriptsize g(x)[/latex] about the y-axis?

The full solutions are at the end of the unit.

Unit 4: Solutions

Exercise 4.1

1. .
   1. y-intercept (let [latex]\scriptsize x=0[/latex]):
      [latex]\scriptsize \begin{align} f(0) & =-\displaystyle \frac{3}{4}{{.4}^{0}}+3 \\ \therefore f(0) & =-\displaystyle \frac{3}{4}+3 \\ \therefore f(0) & =\displaystyle \frac{9}{4} \end{align}[/latex]
      Therefore the y-intercept is [latex]\scriptsize (0,\displaystyle \frac{9}{4})[/latex].
   2. x-intercept (let [latex]\scriptsize y=0[/latex]):
      [latex]\scriptsize \begin{align} 0 & =-\displaystyle \frac{3}{4}{{4}^{x}}+3 \\ \therefore \displaystyle \frac{3}{4}{{4}^{x}} & =3 \\ \therefore {{4}^{x}} & =4 \\ \therefore x & =1 \end{align}[/latex]
      Therefore the x-intercept is [latex]\scriptsize (1,0)[/latex].
   3. [latex]\scriptsize q=3[/latex] therefore the horizontal asymptote is [latex]\scriptsize y=3[/latex].
   4. Domain: [latex]\scriptsize \{x\ |\ x\in \mathbb{R}\}[/latex]
      Range:
      [latex]\scriptsize a \lt 0[/latex] therefore
      [latex]\scriptsize \begin{align} -\displaystyle \frac{3}{4}{{4}^{x}} \lt 0 \\ \therefore -\displaystyle \frac{3}{4}{{4}^{x}}+3 \lt 3 \\ \end{align}[/latex]
      So the range is [latex]\scriptsize \{y\ |\ y\in \mathbb{R},y \lt 3\}[/latex]
   5. .
      
2. y-intercept (let [latex]\scriptsize x=0[/latex]):
   [latex]\scriptsize \begin{align} y & =-2\times {{4}^{0}}+4 \\ \therefore y & =2 \end{align}[/latex]
   The y-intercept is [latex]\scriptsize (0,2)[/latex].
   .
   x-intercept (let [latex]\scriptsize y=0[/latex]):
   [latex]\scriptsize \begin{align} 0 & =-2\times {{4}^{x}}+4 \\ \therefore 2\times {{4}^{x}} & =4 \\ \therefore {{4}^{x}} & =2 \\ \therefore x & =\displaystyle \frac{1}{2} \end{align}[/latex]
   The x-intercept is [latex]\scriptsize (\displaystyle \frac{1}{2},0)[/latex].
   .
   The asymptote is [latex]\scriptsize y=4[/latex].
   
3. y-intercept (let [latex]\scriptsize x=0[/latex]):
   [latex]\scriptsize \begin{align} g(x) & =-3.{{\left( \displaystyle \frac{1}{3} \right)}^{0}}+3 \\ \therefore g(x) & =0 \end{align}[/latex]The y-intercept is [latex]\scriptsize (0,0)[/latex]. Therefore the graph passes through the origin and the x- and y-intercepts are not different points.
   .
   The asymptote is [latex]\scriptsize y=3[/latex].
   .
   [latex]\scriptsize a \lt 0[/latex] and [latex]\scriptsize 0 \lt b \lt 1[/latex]. Therefore the graph goes to negative infinity as [latex]\scriptsize x[/latex] gets more negative.
   

Back to Exercise 4.1

Exercise 4.2

1. From the graph, the horizontal asymptote is [latex]\scriptsize y=1[/latex]. Therefore [latex]\scriptsize q=1[/latex]. Therefore the function is of the form [latex]\scriptsize y=a.{{b}^{x}}+1[/latex].Substitute [latex]\scriptsize (0,-2)[/latex]:
   [latex]\scriptsize \begin{align} -2 & =a.{{b}^{0}}+1 \\ \therefore -2 & =a+1 \\ \therefore a & =-3 \end{align}[/latex]Substitute [latex]\scriptsize (1,0)[/latex]:
   [latex]\scriptsize \begin{align} 0=-3\times {{b}^{1}}+1 \\ \therefore 3b=1 \\ \therefore b=\displaystyle \frac{1}{3} \\ \end{align}[/latex]Therefore the equation of the function is [latex]\scriptsize y=-3{{\left( \displaystyle \frac{1}{3} \right)}^{x}}+1[/latex].
2. We are told that the range of the function is [latex]\scriptsize y \lt -2[/latex]. Therefore [latex]\scriptsize y=-2[/latex] is the asymptote and [latex]\scriptsize q=-2[/latex].We are told that the y-intercept is [latex]\scriptsize (0,-\displaystyle \frac{5}{2})[/latex] which is below the asymptote. Therefore the graph curves down and so:Substitute [latex]\scriptsize (0,-\displaystyle \frac{5}{2})[/latex] into [latex]\scriptsize y=a.{{b}^{x}}-2[/latex]:
   [latex]\scriptsize \begin{align} -\displaystyle \frac{5}{2} & =a.{{b}^{0}}-2 \\ \therefore a & =-\displaystyle \frac{1}{2} \end{align}[/latex]Now substitute [latex]\scriptsize (2,-10)[/latex] into [latex]\scriptsize y=-\displaystyle \frac{1}{2}.{{b}^{x}}-2[/latex]:
   [latex]\scriptsize \begin{align} -10 & =-\displaystyle \frac{1}{2}{{b}^{2}}-2 \\ \therefore -8 & =-\displaystyle \frac{1}{2}{{b}^{2}} \\ \therefore {{b}^{2}} & =16 \\ \therefore b & =\pm \text{4} \end{align}[/latex]But for an exponential function we know that [latex]\scriptsize b \gt 0[/latex]. Therefore [latex]\scriptsize b=4[/latex] and the equation of the function is [latex]\scriptsize y=-\displaystyle \frac{1}{2}{{4}^{x}}-2[/latex].

Back to Exercise 4.2

Unit 4: Assessment

1. [latex]\scriptsize h(x)=-{{2.3}^{x}}+2[/latex]
   [latex]\scriptsize k(x)=3{{\left( \displaystyle \frac{1}{2} \right)}^{x}}-1[/latex]
   [latex]\scriptsize f(x)={{2}^{x}}+1[/latex]
   [latex]\scriptsize g(x)={{2.2}^{x}}-3[/latex]
   [latex]\scriptsize j(x)=-3{{\left( \displaystyle \frac{2}{3} \right)}^{x}}-1[/latex]
2. .
   1. .
      
   2. The line [latex]\scriptsize y=x[/latex] is an asymptote of each graph.
   3. The solution to [latex]\scriptsize {{3}^{x}}+1={{\left( \displaystyle \frac{1}{3} \right)}^{x}}+1[/latex] is the point where the graphs intersect each other. From the graph, this is the point [latex]\scriptsize (0,2)[/latex].
   4. The functions are symmetrical about the y-axis.
3. .
   1. Both graphs have the line [latex]\scriptsize y=0[/latex] as a horizontal asymptote.
   2. .
      [latex]\scriptsize \begin{align*} w(1)-v(1)&={{3}^{1}}-\left( -\displaystyle \frac{3}{1} \right) \\ &=3+3\\ & =6 \end{align*}[/latex]
   3. [latex]\scriptsize w(1)[/latex] and [latex]\scriptsize v(1)[/latex] are the y-values of the functions when [latex]\scriptsize x=1[/latex]. Therefore [latex]\scriptsize w(1)-v(1)[/latex] is the difference in or distance between these y-values. Therefore the graphs are six units apart vertically when [latex]\scriptsize x=1[/latex].
4. .
   1. Domain: [latex]\scriptsize \{x\ |\ x\in \mathbb{R}\}[/latex]
      Range:
      [latex]\scriptsize q=2[/latex] and [latex]\scriptsize a \lt 0[/latex]
      Range is [latex]\scriptsize \{f(x)\ |\ f(x)\in \mathbb{R},f(x) \lt 2\}[/latex]
   2. .
      [latex]\scriptsize \begin{align} g(x) & =f(x)-3 \\ \therefore g(x) & =-\displaystyle \frac{1}{2}{{\left( \displaystyle \frac{1}{4} \right)}^{x}}+2-3 \\ \therefore g(x) & =-\displaystyle \frac{1}{2}{{\left( \displaystyle \frac{1}{4} \right)}^{x}}-1 \end{align}[/latex]
   3. We know that [latex]\scriptsize g(x)=-\displaystyle \frac{1}{2}{{\left( \displaystyle \frac{1}{4} \right)}^{x}}-1[/latex]. We also know that [latex]\scriptsize y=a.{{b}^{x}}+q[/latex] and [latex]\scriptsize y=a.{{\left( \displaystyle \frac{1}{b} \right)}^{x}}+q[/latex] are symmetrical about the y-axis. Therefore, if [latex]\scriptsize b=\displaystyle \frac{1}{4}[/latex] in [latex]\scriptsize g(x)[/latex], the value of [latex]\scriptsize b[/latex] for the graph symmetrical about the y-axis to [latex]\scriptsize g(x)[/latex] will be [latex]\scriptsize 4[/latex]. Therefore [latex]\scriptsize h(x)=-\displaystyle \frac{1}{2}\times {{4}^{x}}-1[/latex].

Back to Unit 4: Assessment