Compare each graph to f(x) = 2^{x} . Write a description of each transformation and graph each function.
Problem 1 :
Solution :
Here the horizontal asymptote is y = -3
y = 2^{x - h} + (-3) ---(1)
The exponential curve is passing through the (1, 5).
5 = 2^{1 - h} + (-3)
8 = 2^{1 - h}
2^{3} = 2^{1 - h}
1 - h = 3
h = -2
By applying the value of h in (1), we get
y = 2^{x - (-2)} + (-3)
y = 2^{x + 2} - 3
Transformations done :
Moving the parent function y = 2^{x},
2 units left and 3 units down
Problem 2 :
Solution :
Here the horizontal asymptote is y = 5. Observing the curve, it is reflected across x-axis.
y = -2^{x - h} + 5 ---(1)
The exponential curve is passing through the (2, 4).
4 = -2^{2 - h} + 5
-1 = -2^{2 - h}
-2^{0} = -2^{2 - h}
2 - h = 0
h = 2
By applying the value of h in (1), we get
y = -2^{x - 2} + 5
Transformations done :
Moving the parent function y = 2^{x},
Reflection across x-axis, 2 units right and 5 units up.
Problem 3 :
Solution :
Here the horizontal asymptote is y = -4.
y = 2^{x - h} + (-4) ---(1)
The exponential curve is passing through the (3, 0).
0 = 2^{3 - h} - 4
4 = 2^{3 - h}
2^{2} = 2^{3 - h}
3 - h = 2
h = 3 - 2
h = 1
By applying the value of h in (1), we get
y = 2^{x - 1} - 4
Transformations done :
Moving the parent function y = 2^{x},
1 unit right and 4 units down.
Problem 4 :
Solution :
Here the horizontal asymptote is y = -5.
y = 2^{x - h} + (-5) ---(1)
The exponential curve is passing through the (1, 3).
3 = 2^{1 - h} - 5
8 = 2^{1 - h}
2^{3} = 2^{1 - h}
1 - h = 3
h = 1 - 3
h = -2
By applying the value of h in (1), we get
y = 2^{x - (-2)} - 5
y = 2^{x + 2} - 5
Transformations done :
Moving the parent function y = 2^{x},
2 unit left and 5 units down.
Problem 5 :
Solution :
Here the horizontal asymptote is y = 1.
y = 2^{x - h} + 1 ---(1)
The exponential curve is passing through the (1, 2).
2 = 2^{1 - h }+ 1
1 = 2^{1 - h}
2^{0} = 2^{1 - h}
1 - h = 0
h = 1
By applying the value of h in (1), we get
y = 2^{x - 1} + 1
Transformations done :
Moving the parent function y = 2^{x},
1 unit right and 1 unit up
Problem 6 :
Solution :
Here the horizontal asymptote is y = 3. By observing the curve, it is reflected across x -axis
y = -2^{x - h} + 3 ---(1)
The exponential curve is passing through the (3, 2).
2 = -2^{3 - h }+ 3
-1 = -2^{3 - h}
-2^{0} = -2^{3}^{- h}
3 - h = 0
h = 3
By applying the value of h in (1), we get
y = -2^{x - 3} + 3
Transformations done :
Moving the parent function y = 2^{x},
Reflected across x-axis, 3 units left and 3 unit up.
Problem 7 :
Solution :
Here the horizontal asymptote is y = -3. By observing the curve, it is reflected across x -axis
y = 2^{x - h} + (-3) ---(1)
The exponential curve is passing through the (3, -2).
-2 = 2^{3 - h }- 3
1 = 2^{3 - h}
2^{0} = 2^{3}^{- h}
3 - h = 0
h = 3
By applying the value of h in (1), we get
y = 2^{x - 3} - 3
Transformations done :
Moving the parent function y = 2^{x},
3 units right and 3 unit down.
Problem 8 :
Solution :
Here the horizontal asymptote is y = 4. By observing the curve, it is reflected across x -axis
y = -2^{x - h} + 4 ---(1)
The exponential curve is passing through the (3, 3).
3 = -2^{3 - h }+ 4
-1 = -2^{3 - h}
-2^{0} = -2^{3}^{- h}
3 - h = 0
h = 3
By applying the value of h in (1), we get
y = -2^{x - 3} + 4
Transformations done :
Moving the parent function y = 2^{x},
3 units right and 4 units up.
Problem 9 :
Solution :
Here the horizontal asymptote is y = 2. By observing the curve, it is reflected across x -axis
y = -2^{x - h} + 2 ---(1)
The exponential curve is passing through the (1, 1).
1 = -2^{1 - h }+ 2
-1 = -2^{1 - h}
-2^{0} = -2^{1}^{- h}
1 - h = 0
h = 1
By applying the value of h in (1), we get
y = -2^{x - 1} + 2
Transformations done :
Moving the parent function y = 2^{x},
Reflected across x-axis,1 unit right and 2 units up.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM