Section 6.3: Exponential functions
1) let f(x) = 2 x
1a) make a table of values and sketch a graph x f(x)
2 2 2 = 4 (2,4)
1 2 1 = 2 (1,2)
0 2 0 = 1 (0,1)
-1 2 -1 = ½ (-1, ½)
-2 2 -2 = ¼ (-2, ¼)
1b) Find the domain of f(x) the graph needs to be extended all the way to the left and right edge of the x-axis.
Answer: domain (−∞, ∞)
1c) Find the range of f(x)
The graph is supposed to always be just slightly above the x-axis. So the bottom y for the range is 0. I use a round bracket as the graph doesn’t touch the x-axis. The graph goes up forever, so the end of the range is ∞
Answer: 𝒓𝒂𝒏𝒈𝒆 (𝟎, ∞)
1d) Find the horizontal asymptote
The x-axis is the horizontal asymptote
Answer: y = 0

3) let f(x) = 3 x
3a) make a table of values and sketch a graph x f(x)
2 3 2 = 9 (2,9)
1 3 1 = 3 (1,3)
0 3 0 = 1 (0,1)
-1 3 -1 = 1/3 (-1, 1/3)
-2 3 -2 = 1/9 (-2, 1/9)
3b) Find the domain of f(x) the graph needs to be extended all the way to the left and right edges of the x-axis.
Answer: domain (−∞, ∞)
3c) Find the range of f(x)
The graph is supposed to always be just slightly above the x-axis. So the bottom y for the range is 0. I use a round bracket as the graph doesn’t touch the x-axis. The graph goes up forever, so the end of the range is ∞
Answer: 𝒓𝒂𝒏𝒈𝒆 (𝟎, ∞)
3d) Find the horizontal asymptote
The x-axis is the horizontal asymptote
Answer: y = 0

5) f(x) = e x-5
5a) Make a table of values and sketch a graph change the sign of the number in the exponent, put that number in the middle of the x-column x f(x)
7 e 7-5 = e 2 = 7.39 (7, 7.39)
6 e 6-5 = e = 2.72 (6, 2.72)
5 e 5-5 = e 0 = 1 (5,1)
4 e 4-5 = e -1 = .37 (4, .37)
3 e 3-5 = e -2 = .14 (3, .14)
5b) Find the domain of f(x) the graph needs to be extended all the way to the left and right edges of the x-axis.
Answer: domain (−∞, ∞)
5c) Find the range of f(x)
The graph is supposed to always be just slightly above the x-axis. So the bottom y for the range is 0. I use a round bracket as the graph doesn’t touch the x-axis. The graph goes up forever, so the end of the range is ∞
Answer: 𝒓𝒂𝒏𝒈𝒆 (𝟎, ∞)
5d) Find the horizontal asymptote
The x-axis is the horizontal asymptote
Answer: y = 0
5e) Describe the transformation from f(x) = e x
Answer: The graph is the same, but shifted 5 units to the right.

7) f(x) = e x+4
7a) Make a table of values and sketch a graph change the sign of the number in the exponent, put that number in the middle of the x-column x f(x)
-2 e -2+4 = e 2 = 7.39 (-2, 7.39)
-3 e -3+4 = e = 2.72 (-3, 2.72)
-4 e -4+4 = e 0 = 1 (-4,1)
-5 e -5+4 = e -1 = .37 (-5, .37)
-6 e -6+4 = e -2 = .14 (-6, .14)
7b) Find the domain of f(x) the graph needs to be extended all the way to the left and right edges of the x-axis.
Answer: domain (−∞, ∞)
7c) Find the range of f(x)
The graph is supposed to always be just slightly above the x-axis. So the bottom y for the range is 0. I use a round bracket as the graph doesn’t touch the x-axis. The graph goes up forever, so the end of the range is ∞
Answer: 𝒓𝒂𝒏𝒈𝒆 (𝟎, ∞)
7d) Find the horizontal asymptote
The x-axis is the horizontal asymptote
Answer: y = 0
7e) Describe the transformation from f(x) = e x
Answer: The graph is the same, but shifted 4 units to the left.

9) f(x) = e -x
9a) Make a table of values and sketch a graph as there is no number after the x in the exponent, this table will do x f(x)
2 e -2 = .14 (2, .14)
1 e -4 = .37 (1, .37)
0 e -0 = e 0 = 1 (0,1)
-1 e -(-1) = e 1 = 2.72 (-1, 2.72)
-2 e -(-2) = e 2 = 7.39 (-2, 7.39)
9b) Find the domain of f(x) the graph needs to be extended all the way to the left and right edges of the x-axis.
Answer: domain (−∞, ∞)
9c) Find the range of f(x)
The graph is supposed to always be just slightly above the x-axis. So the bottom y for the range is 0. I use a round bracket as the graph doesn’t touch the x-axis. The graph goes up forever, so the end of the range is ∞
Answer: 𝒓𝒂𝒏𝒈𝒆 (𝟎, ∞)
9d) Find the horizontal asymptote
The x-axis is the horizontal asymptote
Answer: y = 0
9e) Describe the transformation from f(x) = e x
Answer: The graph is the same, but reflected about the y-axis

11) f(x) = -e -x
11a) As there is no number in after the x in the exponent, this table will do x f(x)
2 -e -2 = -.14 (2, -.14)
1 -e -1 = -.37 (1, -.37)
0 -e 0 = 1
-1 -e -(-1) = -2.72 (-1, -2.72)
-2 -e -(-2) = -7.39 (-2, -7.39)
11b) Find the domain of f(x) the graph needs to be extended all the way to the left and right edges of the x-axis.
Answer: domain (−∞, ∞)
11c) Find the range of f(x)
The graph is supposed to always be just slightly below the x-axis. So the top y for the range is 0. I use a round bracket as the graph doesn’t touch the x-axis. The graph goes up forever, so the end of the range is ∞
Answer: 𝒓𝒂𝒏𝒈𝒆 (−∞, 𝟎)
11d) Find the horizontal asymptote, the x-axis is the horizontal asymptote
Answer: y = 0
11e) the graph is reflected about the x-axis and the y-axis

13) f(x) = 5 + e x
13a) Make a table of values and sketch a graph change the sign of the number in the exponent, put that number in the middle of the x-column x f(x)
2 5 +e 6-4 = 5 + e 2 = 12.39 (2, 12.39)
1 5 + e 5-4 = 5 + e = 7.72 (1, 7.72)
0 5 + e 4-4 = 5 + e 0 = 6 (0, 6)
-1 5 + e 3-4 = 5 + e -1 = 5.37 (-1, 5.37)
-2 5 + e -2-4 = 5 + e -2 = 5.14 (-2, 5.14)
13b) Find the domain of f(x) the graph needs to be extended all the way to the left and right edges of the x-axis.
Answer: domain (−∞, ∞)
13c) Find the range of f(x)
The graph is supposed to always be just slightly above the line y = 5. So the bottom y for the range is 5.
I use a round bracket as the graph doesn’t touch the x-axis. The graph goes up forever, so the end of the range is ∞
Answer: 𝒓𝒂𝒏𝒈𝒆 (𝟓, ∞)
13d) Find the horizontal asymptote the + 5 moves the horizontal asymptote up 5 units.
Answer: y = 5
13e) Describe the transformation from f(x) = e x same shape as f(x) = e x , except shifted up 5 units.

15) Describe the transformation of the graph of g(x) as compared to the graph of f(x) = 3 x .
15a) g(x) = 3 x+5 the + 5 in the exponent will shift the graph left 5 units
Answer: left 5 units
15b) g(x) = 3 x+3 the + 3 in the exponent will shift the graph left units
Answer: left 3 units
15c) g(x) = 3 x-4 the -4 in the exponent will move the graph right 4 units
Answer: right 4 units
15d) g(x) = 3 x-6 the -6 in the exponent will move the graph right 6 units
Answer: right 6 units
15e) g(x) = 3 x + 2 The +2 will move the graph up 2 units
Answer: up 2 units
15f) g(x) = 3 x +13 The +13 will move the graph up 13 units
Answer: up 13 units
15g) g(x) = 3 x - 7 The -7 will move the graph down 7 units
Answer: down 7 units
15h) g(x) = 3 x – 2 The -2 will move the graph down 2 units
Answer: down 2 units
15i) g(x) = -3 x The negative in front of the 3 will reflect the graph over the x-axis
Answer: reflected over x-axis
15j) g(x) = 3 -x The negative in front of the x will reflect the graph over the y-axis
Answer: reflect over y-axis
15k) g(x) = 3 x+1 – 4 the +1 in the exponent moves left 1, the -4 moves down 4
Answer: left 1 unit, down 4 units
15l) g(x) = 3 x-3 + 2 the -3 in the exponent moves right 3 units, the +2 moves up 2 units
Answer: right 3 units, up 2 units

15m) g(x) = -3 x-3 the – in front of the 3 reflects over the x-axis. The -3 in the exponent moves right 3.
Answer: reflected over x-axis, moved right 3 units
15n) g(x) = 3 -x – 1 the –in front of the x reflects over y-axis. The -1 moves down 1 unit
Answer: reflects over y-axis moves down 1 unit.
15o) g(x) = -3 -x The – in front of the 3 reflects over the x-axis. The – in front of the x reflects over y =axis
Answer: reflects over x axis and reflects over y-axis
15p) g(x) = 5 -3 x-4 The 5 shifts up 5 units, the – in front of 3 reflects over x-axis, the -4 in the exponent shifts right 4 units.
Answer: reflected over x-axis, moved up 5 units and right 4 units
17) 2 x+5 = 16
2 x+5 = 2 4 x + 5 = 4
Answer: x = 1
19) 6 -x = 216
6 -x = 6 3
-x = 3
Answer: x = -3
21) (
1
3
) 𝑥−4
=
1
27
(
1
3
) 𝑥−4
= (
1
3
)
3 x– 4 = 3
Answer: x = 7
1 this is the algebra for the right side
27
=
1
3 3
=
1 3
3 3
= (
1
3
)
3

2x + 1 = 3x - 6
7 = x
Answer: x = 5
27) 7 4-5x = 49 x+1
7 4-5x = (7 2 ) x+1
7 4-5x = 7 2x+2
4 – 5x = 2x + 2
2 = 7x
Answer: x = 2/7
29) e x-5 = e 2x-4 x – 5 = 2x - 4
-1 = x
Answer: x = -1
23) (
1
2
)
3𝑥−4
= 32
(
1
2
)
3𝑥−4
= 2
5
( I need to change ½ to 2 -1 , as they are equal and this will make both sides have a base of 2)
(2 −1 ) 3𝑥−4
= 2
5
(now multiply exponents)
2 −3𝑥+4 = 2 5
-3x+4 = 5
-3x= 1
Answer: x = -1/3
25) 2 2x+1 = 8 x-2
2
2𝑥+1
= (2
3 ) 𝑥−2
(now multiply exponents)
2 2x+1 = 2 3x - 6

31) 𝑒 𝑥−4
∗ 𝑒
3𝑥
= 𝑒
2
(add exponents on left side) e 4x – 4 = e 2
4x – 4 = 2
4x = 6 x = 6/4
Answer: x = 3/2
33) 3 𝑥−2
∗ 3
4
= 3
5
(add exponents on left side)
3 x+2 = 3 5 x+2 = 5
Answer: x = 3