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# Exponential functions

## Functions of the Form y=ab(x)+qy=ab(x)+q

Functions of the form y=ab(x)+qy=ab(x)+q are known as exponential functions. The general shape of a graph of a function of this form is shown in Figure 1.

### Investigation : Functions of the Form y=ab(x)+qy=ab(x)+q

1. On the same set of axes, plot the following graphs:
1. a(x)=-2·b(x)+1a(x)=-2·b(x)+1
2. b(x)=-1·b(x)+1b(x)=-1·b(x)+1
3. c(x)=0·b(x)+1c(x)=0·b(x)+1
4. d(x)=1·b(x)+1d(x)=1·b(x)+1
5. e(x)=2·b(x)+1e(x)=2·b(x)+1
Use your results to deduce the effect of aa.
2. On the same set of axes, plot the following graphs:
1. f(x)=1·b(x)-2f(x)=1·b(x)-2
2. g(x)=1·b(x)-1g(x)=1·b(x)-1
3. h(x)=1·b(x)+0h(x)=1·b(x)+0
4. j(x)=1·b(x)+1j(x)=1·b(x)+1
5. k(x)=1·b(x)+2k(x)=1·b(x)+2
Use your results to deduce the effect of qq.

You should have found that the value of aa affects whether the graph curves upwards (a>0a>0) or curves downwards (a<0a<0).

You should have also found that the value of qq affects the position of the yy-intercept.

These different properties are summarised in Table 1.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

### Domain and Range

For y=ab(x)+qy=ab(x)+q, the function is defined for all real values of xx. Therefore, the domain is {x:xR}{x:xR}.

The range of y=ab(x)+qy=ab(x)+q is dependent on the sign of aa.

If a>0a>0 then:

b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q
(1)

Therefore, if a>0a>0, then the range is {f(x):f(x)[q;)}{f(x):f(x)[q;)}.

If a<0a<0 then:

b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q
(2)

Therefore, if a<0a<0, then the range is {f(x):f(x)(-;q]}{f(x):f(x)(-;q]}.

For example, the domain of g(x)=3·2x+2g(x)=3·2x+2 is {x:xR}{x:xR}. For the range,

2 x 0 3 · 2 x 0 3 · 2 x + 2 2 2 x 0 3 · 2 x 0 3 · 2 x + 2 2
(3)

Therefore the range is {g(x):g(x)[2;)}{g(x):g(x)[2;)}.

### Intercepts

For functions of the form, y=ab(x)+qy=ab(x)+q, the intercepts with the xx and yy axis is calculated by setting x=0x=0 for the yy-intercept and by setting y=0y=0 for the xx-intercept.

The yy-intercept is calculated as follows:

y = a b ( x ) + q y i n t = a b ( 0 ) + q = a ( 1 ) + q = a + q y = a b ( x ) + q y i n t = a b ( 0 ) + q = a ( 1 ) + q = a + q
(4)

For example, the yy-intercept of g(x)=3·2x+2g(x)=3·2x+2 is given by setting x=0x=0 to get:

y = 3 · 2 x + 2 y i n t = 3 · 2 0 + 2 = 3 + 2 = 5 y = 3 · 2 x + 2 y i n t = 3 · 2 0 + 2 = 3 + 2 = 5
(5)

The xx-intercepts are calculated by setting y=0y=0 as follows:

y = a b ( x ) + q 0 = a b ( x i n t ) + q a b ( x i n t ) = - q b ( x i n t ) = - q a y = a b ( x ) + q 0 = a b ( x i n t ) + q a b ( x i n t ) = - q b ( x i n t ) = - q a
(6)

Which only has a real solution if either a<0a<0 or q<0q<0. Otherwise, the graph of the function of form y=ab(x)+qy=ab(x)+q does not have any xx-intercepts.

For example, the xx-intercept of g(x)=3·2x+2g(x)=3·2x+2 is given by setting y=0y=0 to get:

y = 3 · 2 x + 2 0 = 3 · 2 x i n t + 2 - 2 = 3 · 2 x i n t 2 x i n t = - 2 3 y = 3 · 2 x + 2 0 = 3 · 2 x i n t + 2 - 2 = 3 · 2 x i n t 2 x i n t = - 2 3
(7)

which has no real solution. Therefore, the graph of g(x)=3·2x+2g(x)=3·2x+2 does not have any xx-intercepts.

### Asymptotes

Functions of the form y=ab(x)+qy=ab(x)+q have a single horizontal asymptote. The asymptote can be determined by examining the range.

We have seen that the range is controlled by the value of q. If a>0a>0, then the range is {f(x):f(x)[q;)}{f(x):f(x)[q;)}.And if a>0a>0, then the range is {f(x):f(x)[q;)}{f(x):f(x)[q;)}.

This shows that the function tends towards the value of q as x x . Therefore the horizontal asymptote lies at x=qx=q.

### Sketching Graphs of the Form f(x)=ab(x)+qf(x)=ab(x)+q

In order to sketch graphs of functions of the form, f(x)=ab(x)+qf(x)=ab(x)+q, we need to determine four characteristics:

1. domain and range
2. asymptote
3. yy-intercept
4. xx-intercept

For example, sketch the graph of g(x)=3·2x+2g(x)=3·2x+2. Mark the intercepts.

We have determined the domain to be {x:xR}{x:xR} and the range to be {g(x):g(x)[2,)}{g(x):g(x)[2,)}.

The yy-intercept is yint=5yint=5 and there are no xx-intercepts.

#### Exercise 1: Drawing an exponential graph

Draw the graph of y=-2.3x+5y=-2.3x+5.

##### Solution
1. Step 1. Find the domain and range: The domain is: {x:xR}{x:xR} and the range is: {f(x):f(x)(-;5]}{f(x):f(x)(-;5]}.
2. Step 2. Calculate the asymptote: There is one asymptote for functions of this form. This occurs at y=qy=q. So the asymptote for this graph is at y=5y=5
3. Step 3. Calculate the y-intercept: The y-intercept occurs when x=0x=0.
y = -2.3x+5 y = -2.30+5 y = -2(1)+5 yint = 7 y = -2.3x+5 y = -2.30+5 y = -2(1)+5 yint = 7
(8)
So there is one y-intercept at (0,7)(0,7).
4. Step 4. Calculate the x-intercept: The x-intercept occurs when y=0y=0. Calculating the x-intercept gives:
y = -2.3x+5 0 = -2.3x+5 -5 = -2.3x 3xint = 52 xint = 0,83 y = -2.3x+5 0 = -2.3x+5 -5 = -2.3x 3xint = 52 xint = 0,83
(9)
So there is one x-intercept at (0,83,0)(0,83,0).
5. Step 5. Plot the graph: Putting all this together gives us the following graph:

#### Exponential Functions and Graphs

1. Draw the graphs of y=2xy=2x and y=(12)xy=(12)x on the same set of axes.
1. Is the xx-axis and asymptote or and axis of symmetry to both graphs ? Explain your answer.
2. Which graph is represented by the equation y=2-xy=2-x ? Explain your answer.
3. Solve the equation 2x=(12)x2x=(12)x graphically and check that your answer is correct by using substitution.
4. Predict how the graph y=2.2xy=2.2x will compare to y=2xy=2x and then draw the graph of y=2.2xy=2.2x on the same set of axes.
2. The curve of the exponential function ff in the accompanying diagram cuts the y-axis at the point A(0; 1) and B(2; 4) is on ff.
1. Determine the equation of the function ff.
2. Determine the equation of hh, the function of which the curve is the reflection of the curve of ff in the xx-axis.
3. Determine the range of hh.

## Summary

• You should know the following charecteristics of functions:
• dependent and independent variables: The given or chosen x-value is known as the independent variable, because its value can be chosen freely. The calculated y-value is known as the dependent variable, because its value depends on the chosen x-value.
• domain and range: The domain of a relation is the set of all the x values for which there exists at least one y value according to that relation. The range is the set of all the y values, which can be obtained using at least one x value.
• intercepts with axes: The intercept is the point at which a graph intersects an axis. The x-intercepts are the points at which the graph cuts the x-axis and the y-intercepts are the points at which the graph cuts the y-axis.
• turning points: Only for graphs of functions whose highest power is more than 1. There are two types of turning points: a minimal turning point and a maximal turning point. A minimal turning point is a point on the graph where the graph stops decreasing in value and starts increasing in value and a maximal turning point is a point on the graph where the graph stops increasing in value and starts decreasing.
• asymptotes: An asymptote is a straight or curved line, which the graph of a function will approach, but never touch.
• lines of symmetry: A line about which the graph is symmetric
• intervals on which the function increases/decreases: The interval on which a graph increases or decreases
• continuous nature of the function: A graph is said to be continuous if there are no breaks in the graph.
• Set notation A set of certain x values has the following form: {x : conditions, more conditions}
• Interval notation Here we write an interval in the form ’lower bracket, lower number, comma, upper number, upper bracket’
• You should know the following functions and their properties:
• Functions of the form y=ax+qy=ax+q. These are straight lines.
• Functions of the Form y=ax2+qy=ax2+q These are known as parabolic functions or parabolas.
• Functions of the Form y=ax+qy=ax+q. These are known as hyperbolic functions.
• Functions of the Form y=ab(x)+qy=ab(x)+q. These are known as exponential functions.

## End of Chapter Exercises

1. Sketch the following straight lines:
1. y=2x+4y=2x+4
2. y-x=0y-x=0
3. y=-12x+2y=-12x+2
2. Sketch the following functions:
1. y=x2+3y=x2+3
2. y=12x2+4y=12x2+4
3. y=2x2-4y=2x2-4
3. Sketch the following functions and identify the asymptotes:
1. y=3x+2y=3x+2
2. y=-4.2x+1y=-4.2x+1
3. y=2.3x-2y=2.3x-2
4. Sketch the following functions and identify the asymptotes:
1. y=3x+4y=3x+4
2. y=1xy=1x
3. y=2x-2y=2x-2
5. Determine whether the following statements are true or false. If the statement is false, give reasons why:
1. The given or chosen y-value is known as the independent variable.
2. An intercept is the point at which a graph intersects itself.
3. There are two types of turning points – minimal and maximal.
4. A graph is said to be congruent if there are no breaks in the graph.
5. Functions of the form y=ax+qy=ax+q are straight lines.
6. Functions of the form y=ax+qy=ax+q are exponential functions.
7. An asymptote is a straight or curved line which a graph will intersect once.
8. Given a function of the form y=ax+qy=ax+q , to find the y-intersect put x=0x=0 and solve for yy.
9. The graph of a straight line always has a turning point.
6. Given the functions f(x)=-2x2-18f(x)=-2x2-18 and g(x)=-2x+6g(x)=-2x+6
1. Draw ff and gg on the same set of axes.
2. Calculate the points of intersection of ff and gg.
3. Hence use your graphs and the points of intersection to solve for xx when:
1. f(x)>0f(x)>0
2. f(x)g(x)0f(x)g(x)0
4. Give the equation of the reflection of ff in the xx-axis.
7. After a ball is dropped, the rebound height of each bounce decreases. The equation y=5·(0,8)xy=5·(0,8)x shows the relationship between xx, the number of bounces, and yy, the height of the bounce, for a certain ball. What is the approximate height of the fifth bounce of this ball to the nearest tenth of a unit ?
8. Mark had 15 coins in five Rand and two Rand pieces. He had 3 more R2-coins than R5-coins. He wrote a system of equations to represent this situation, letting xx represent the number of five rand coins and yy represent the number of two rand coins. Then he solved the system by graphing.
1. Write down the system of equations.
2. Draw their graphs on the same set of axes.
3. What is the solution?

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