This form of the exponential function is slightly more complex than the form studied in Grade 10.
- On the same set of axes, with
−5≤x≤3
5
x
3
and
−35≤y≤35
35
y
35
, plot the following graphs:
- f(x)=-2·2(x+1)+1f(x)=-2·2(x+1)+1
- g(x)=-1·2(x+1)+1g(x)=-1·2(x+1)+1
- h(x)=0·2(x+1)+1h(x)=0·2(x+1)+1
- j(x)=1·2(x+1)+1j(x)=1·2(x+1)+1
- k(x)=2·2(x+1)+1k(x)=2·2(x+1)+1
- On the same set of axes, with
−3≤x≤3
3
x
3
and
−5≤y≤20
5
y
20
, plot the following graphs:
- f(x)=1·2(x+1)-2f(x)=1·2(x+1)-2
- g(x)=1·2(x+1)-1g(x)=1·2(x+1)-1
- h(x)=1·2(x+1)+0h(x)=1·2(x+1)+0
- j(x)=1·2(x+1)+1j(x)=1·2(x+1)+1
- k(x)=1·2(x+1)+2k(x)=1·2(x+1)+2
- On the same set of axes, with
−5≤x≤3
5
x
3
and
−35≤y≤35
35
y
35
, plot the following graphs:
- f(x)=-2·2(x+1)+1f(x)=-2·2(x+1)+1
- g(x)=-1·2(x+1)+1g(x)=-1·2(x+1)+1
- h(x)=0·2(x+1)+1h(x)=0·2(x+1)+1
- j(x)=1·2(x+1)+1j(x)=1·2(x+1)+1
- k(x)=2·2(x+1)+1k(x)=2·2(x+1)+1
- Following the general method of the above activities, choose your own values of aa and qq to plot 5 graphs of y=ab(x+p)+qy=ab(x+p)+q on the same set of axes (choose your own limits for xx and yy carefully). Make sure that you use the same values of aa, bb and qq for each graph, and different values of pp. Use your results to understand the effect of changing the value of pp.
These different properties are summarised in Table 1.
Table 1: Table summarising general shapes and positions of functions of the form y=ab(x+p)+qy=ab(x+p)+q.
| |
p
<
0
p
<
0
|
p
>
0
p
>
0
|
| |
a
>
0
a
>
0
|
a
<
0
a
<
0
|
a
>
0
a
>
0
|
a
<
0
a
<
0
|
|
q
>
0
q
>
0
|
|
|
|
|
|
q
<
0
q
<
0
|
|
|
|
|
For y=ab(x+p)+qy=ab(x+p)+q, the function is defined for all real values of xx. Therefore, the domain is {x:x∈R}{x:x∈R}.
The range of y=ab(x+p)+qy=ab(x+p)+q is dependent on the sign of aa.
If a>0a>0 then:
b
(
x
+
p
)
>
0
a
·
b
(
x
+
p
)
>
0
a
·
b
(
x
+
p
)
+
q
>
q
f
(
x
)
>
q
b
(
x
+
p
)
>
0
a
·
b
(
x
+
p
)
>
0
a
·
b
(
x
+
p
)
+
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,∞)}. In other words
f(x)f(x) can be any real number greater than qq.
If a<0a<0 then:
b
(
x
+
p
)
>
0
a
·
b
(
x
+
p
)
<
0
a
·
b
(
x
+
p
)
+
q
<
q
f
(
x
)
<
q
b
(
x
+
p
)
>
0
a
·
b
(
x
+
p
)
<
0
a
·
b
(
x
+
p
)
+
q
<
q
f
(
x
)
<
q
(2)
Therefore, if a<0a<0, then the range is (-∞,q)(-∞,q), meaning that
f(x)f(x) can be any real number less than qq. Equivalently, one could write that the range is {y∈R:y<q}{y∈R:y<q}.
For example, the domain of g(x)=3·2x+1+2g(x)=3·2x+1+2 is {x:x∈R}{x:x∈R}.
For the range,
2
x
+
1
>
0
3
·
2
x
+
1
>
0
3
·
2
x
+
1
+
2
>
2
2
x
+
1
>
0
3
·
2
x
+
1
>
0
3
·
2
x
+
1
+
2
>
2
(3)
Therefore the range is {g(x):g(x)∈[2,∞)}{g(x):g(x)∈[2,∞)}.
- Give the domain of y=3xy=3x.
- What is the domain and range of f(x)=2xf(x)=2x ?
- Determine the domain and range of y=(1,5)x+3y=(1,5)x+3.
For functions of the form, y=ab(x+p)+qy=ab(x+p)+q, the intercepts with the xx- and yy-axis are 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
+
p
)
+
q
y
i
n
t
=
a
b
(
0
+
p
)
+
q
=
a
b
p
+
q
y
=
a
b
(
x
+
p
)
+
q
y
i
n
t
=
a
b
(
0
+
p
)
+
q
=
a
b
p
+
q
(4)
For example, the yy-intercept of g(x)=3·2x+1+2g(x)=3·2x+1+2 is given by setting x=0x=0 to get:
y
=
3
·
2
x
+
1
+
2
y
i
n
t
=
3
·
2
0
+
1
+
2
=
3
·
2
1
+
2
=
3
·
2
+
2
=
8
y
=
3
·
2
x
+
1
+
2
y
i
n
t
=
3
·
2
0
+
1
+
2
=
3
·
2
1
+
2
=
3
·
2
+
2
=
8
(5)
The xx-intercepts are calculated by setting y=0y=0 as follows:
y
=
a
b
(
x
+
p
)
+
q
0
=
a
b
(
x
i
n
t
+
p
)
+
q
a
b
(
x
i
n
t
+
p
)
=
-
q
b
(
x
i
n
t
+
p
)
=
-
q
a
y
=
a
b
(
x
+
p
)
+
q
0
=
a
b
(
x
i
n
t
+
p
)
+
q
a
b
(
x
i
n
t
+
p
)
=
-
q
b
(
x
i
n
t
+
p
)
=
-
q
a
(6)
Since b>0b>0 (this is a requirement in the original definition) and a positive number raised to any power is always positive, the last equation above only has a real solution if either a<0a<0 or q<0q<0 (but not both). Additionally, aa must not be zero for the division to be valid. If these conditions are not satisfied, the graph of the function of the form y=ab(x+p)+qy=ab(x+p)+q does not have any xx-intercepts.
For example, the xx-intercept of g(x)=3·2x+1+2g(x)=3·2x+1+2 is given by setting y=0y=0 to get:
y
=
3
·
2
x
+
1
+
2
0
=
3
·
2
x
i
n
t
+
1
+
2
-
2
=
3
·
2
x
i
n
t
+
1
2
x
i
n
t
+
1
=
-
2
2
y
=
3
·
2
x
+
1
+
2
0
=
3
·
2
x
i
n
t
+
1
+
2
-
2
=
3
·
2
x
i
n
t
+
1
2
x
i
n
t
+
1
=
-
2
2
(7)
which has no real solution. Therefore, the graph of g(x)=3·2x+1+2g(x)=3·2x+1+2 does not have a xx-intercept. You will notice that calculating g(x)g(x) for any value of xx will always give a positive number, meaning that yy will never be zero and so the graph will never intersect the xx-axis.
- Give the y-intercept of the graph of y=bx+2y=bx+2.
- Give the x- and y-intercepts of the graph of y=12(1,5)x+3-0,75y=12(1,5)x+3-0,75.
Functions of the form y=ab(x+p)+qy=ab(x+p)+q always have exactly one horizontal asymptote.
When examining the range of these functions, we saw that we always have either y<qy<q or y>qy>q for all input values of xx. Therefore the line y=qy=q is an asymptote.
For example, we saw earlier that the range of g(x)=3·2x+1+2g(x)=3·2x+1+2 is (2,∞)(2,∞) because g(x)g(x) is always greater than 2. However, the value of g(x)g(x) can get extremely close to 2, even though it never reaches it. For example, if you calculate g(-20)g(-20), the value is approximately 2.000006. Using larger negative values of xx will make g(x)g(x) even closer to 2: the value of g(-100)g(-100) is so close to 2 that the calculator is not precise enough to know the difference, and will (incorrectly) show you that it is equal to exactly 2.
From this we deduce that the line y=2y=2 is an asymptote.
- Give the equation of the asymptote of the graph of y=3x-2y=3x-2.
- What is the equation of the horizontal asymptote of the
graph of y=3(0,8)x-1-3y=3(0,8)x-1-3 ?
In order to sketch graphs of functions of the form, f(x)=ab(x+p)+qf(x)=ab(x+p)+q, we need to determine four characteristics:
- domain and range
- yy-intercept
- xx-intercept
For example, sketch the graph of g(x)=3·2x+1+2g(x)=3·2x+1+2. Mark the intercepts.
We have determined the domain to be {x:x∈R}{x:x∈R} and the range to be {g(x):g(x)∈(2,∞)}{g(x):g(x)∈(2,∞)}.
The yy-intercept is yint=8yint=8 and there is no xx-intercept.
- Draw the graphs of the following on the same set of axes. Label the horizontal asymptotes and y-intercepts clearly.
- y=bx+2y=bx+2
- y=bx+2y=bx+2
- y=2bxy=2bx
- y=2bx+2+2y=2bx+2+2
- Draw the graph of f(x)=3xf(x)=3x.
- Explain where a solution of 3x=53x=5 can be read off the graph.
"Accessible versions of this collection are available at Bookshare. DAISY and BRF provided. "