Any transformed exponential can be written in the form
y = ab^{x - h} + k
where y = k is the horizontal asymptote.
Note that, due to the shift, the vertical intercept is shifted to (0, a+k)
Find the exponential function represented by the graph given below.
Problem 1 :
Solution :
y = ab^{x} + k
Here, the Horizontal asymptote is y = -1. So, k = -1
y = ab^{x} - 1 ------(1)
Tracing two points from the graph of the given function given above.
(0, 2) and (-5, -1)
Applying the point (0, 2) in (1), we get
2 = ab^{0} - 1
2 = a(1) - 1
a = 2 + 1
a = 3
Applying the point (1, 5) in (1), we get
5 = ab^{1} - 1
6 = 3(b^{1})
b = 6/3
b = 2
So, the required exponential function is
y = 3(2)^{x} - 1
Problem 2 :
Solution :
y = ab^{x} + k
Here, the Horizontal asymptote is y = 5. So, k = 5
y = ab^{x} + 5 ------(1)
Tracing two points from the graph of the given function given above.
(0, 2) and (-1, 3)
Applying the point (0, 2) in (1), we get
2 = ab^{0} + 5
2 - 5 = a(1)
a = -3
Applying the point (-1, 3) in (1), we get
3 = (-3)b^{-1} + 5
3 - 5 = -3(1/b)
-2 = -3/b
b = 3/2
b = 1.5
So, the required exponential function is
y = -3(1.5)^{x} + 5
Problem 3 :
Solution :
y = ab^{x} + k
Here, the Horizontal asymptote is y = 1. So, k = 1
y = ab^{x} + 1 ------(1)
Tracing two points from the graph of the given function given above.
(-2, 0) and (0, -3)
Applying the point (-2, 0) in (1), we get
0 = ab^{-2} + 1
-1 = a/b^{2}
Applying the point (0, -3) in (1), we get
-3 = ab^{0} + 1
-3 - 1 = a(1)
a = -4
Applying the value of a, we get
-4/b^{2} = -1
b^{2} = 4
b = 2
So, the required exponential function is
y = -4(1.5)^{x} + 5
Problem 4 :
Solution :
y = ab^{x} + k
Here, the Horizontal asymptote is y = 3. So, k = 3
y = ab^{x} + 3 ------(1)
Tracing two points from the graph of the given function given above.
(0, 2) and (1, 1)
Applying the point (0, 2) in (1), we get
2 = ab^{0} + 3
-1 = a(1)
a = -1
Applying the point (1, 1) in (1), we get
1 = (-1)b^{1} + 3
1 - 3 = -b
b = 2
So, the required exponential function is
y = -1(2)^{x} + 3
Problem 5 :
Solution :
y = ab^{x} + k
Here, the Horizontal asymptote is y = 3. So, k = 3
y = ab^{x} + 3 ------(1)
Tracing two points from the graph of the given function given above.
(0, 1) and (-1, -1)
Applying the point (0, 1) in (1), we get
1 = ab^{0} + 3
-2 = a(1)
a = -2
Applying the point (-1, -1) in (1), we get
-1 = (-2)b^{-1} + 3
-4/(-2) = 1/b
b = 1/2
So, the required exponential function is
y = -2(1/2)^{x} + 3
Problem 6 :
Solution :
y = ab^{x} + k
Here, the Horizontal asymptote is y = -2. So, k = -2
y = ab^{x} - 2------(1)
Tracing two points from the graph of the given function given above.
(0, 1) and (-1, 4)
Applying the point (0, 1) in (1), we get
1 = ab^{0} - 2
3 = a(1)
a = 3
Applying the point (-1, 4) in (1), we get
4 = (2)b^{-1} - 2
4 + 2 = 2/b
6b = 2
b = 1/3
So, the required exponential function is
y = 3(1/3)^{x} - 2
Problem 7 :
Solution :
y = ab^{x} + k
Here, the Horizontal asymptote is y = 7. So, k = 7
y = ab^{x} + 7------(1)
Tracing two points from the graph of the given function given above.
(0, 5) and (1, 1)
Applying the point (0, 5) in (1), we get
5 = ab^{0} + 7
5 - 7 = a(1)
a = -2
Applying the point (1, 1) in (1), we get
1 = (-2)b^{1} + 7
-6 = -2b
b = 3
So, the required exponential function is
y = -2(3)^{x} + 7
Problem 8 :
Solution :
y = ab^{x} + k
Here, the Horizontal asymptote is y = 5. So, k = 5
y = ab^{x} + 5------(1)
Tracing two points from the graph of the given function given above.
(0, 3) and (-1, -1)
Applying the point (0, 3) in (1), we get
3 = ab^{0} + 5
-2 = a(1)
a = -2
Applying the point (-1, -1) in (1), we get
-1 = (-2)b^{-1} + 5
-6 = -2/b
b = 1/3
So, the required exponential function is
y = -2(1/3)^{x} + 5
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM