H Algebra 2/Trig · Unit 7
Sketch a log graph from at least three exact points, then name the asymptote, domain, range, and end behavior.
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The parent graph
y = log2 x passes through (1, 0), (2, 1), and (4, 2).
The vertical asymptote is x = 0. The domain is x > 0. The range is all real numbers.
Pick friendly inputs
Set the inside equal to 1, then the base, then the base squared. Those logs are 0, 1, and 2.
Asymptote and domain
The asymptote is the line where the inside is 0. The domain is the side where the inside is positive.
Neither growth nor decay
Growth and decay describe exponential functions. A log graph is neither. Watch the sign of a for the end behavior.
log of (x − h), plus a vertical move
f(x) = log2 (x − 1) + 3
The vertical asymptote is where the inside equals 0.
x − 1 = 0
x = 1
Asymptote: x = 1
The inside must be positive.
x − 1 > 0
x > 1
Domain: (1, ∞)
Set the inside equal to 1, because log2 1 = 0.
x − 1 = 1
x = 2
f(2) = 0 + 3
f(2) = 3
(2, 3)
Set the inside equal to 2, because log2 2 = 1.
x − 1 = 2
x = 3
f(3) = 1 + 3
f(3) = 4
(3, 4)
Set the inside equal to 4, because log2 4 = 2.
x − 1 = 4
x = 5
f(5) = 2 + 3
f(5) = 5
(5, 5)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 1 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = 1; domain (1, ∞); range (−∞, ∞); As x → 1 from the right, y → −∞; As x → ∞, y → ∞
f(x) = log2 (x + 4) − 1
The vertical asymptote is where the inside equals 0.
x + 4 = 0
x = −4
Asymptote: x = −4
The inside must be positive.
x + 4 > 0
x > −4
Domain: (−4, ∞)
Set the inside equal to 1, because log2 1 = 0.
x + 4 = 1
x = −3
f(−3) = 0 − 1
f(−3) = −1
(−3, −1)
Set the inside equal to 2, because log2 2 = 1.
x + 4 = 2
x = −2
f(−2) = 1 − 1
f(−2) = 0
(−2, 0)
Set the inside equal to 4, because log2 4 = 2.
x + 4 = 4
x = 0
f(0) = 2 − 1
f(0) = 1
(0, 1)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → −4 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = −4; domain (−4, ∞); range (−∞, ∞); As x → −4 from the right, y → −∞; As x → ∞, y → ∞
f(x) = log3 (x − 2) + 2
The vertical asymptote is where the inside equals 0.
x − 2 = 0
x = 2
Asymptote: x = 2
The inside must be positive.
x − 2 > 0
x > 2
Domain: (2, ∞)
Set the inside equal to 1, because log3 1 = 0.
x − 2 = 1
x = 3
f(3) = 0 + 2
f(3) = 2
(3, 2)
Set the inside equal to 3, because log3 3 = 1.
x − 2 = 3
x = 5
f(5) = 1 + 2
f(5) = 3
(5, 3)
Set the inside equal to 9, because log3 9 = 2.
x − 2 = 9
x = 11
f(11) = 2 + 2
f(11) = 4
(11, 4)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 2 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = 2; domain (2, ∞); range (−∞, ∞); As x → 2 from the right, y → −∞; As x → ∞, y → ∞
f(x) = log5 (x + 1) − 3
The vertical asymptote is where the inside equals 0.
x + 1 = 0
x = −1
Asymptote: x = −1
The inside must be positive.
x + 1 > 0
x > −1
Domain: (−1, ∞)
Set the inside equal to 1, because log5 1 = 0.
x + 1 = 1
x = 0
f(0) = 0 − 3
f(0) = −3
(0, −3)
Set the inside equal to 5, because log5 5 = 1.
x + 1 = 5
x = 4
f(4) = 1 − 3
f(4) = −2
(4, −2)
Set the inside equal to 25, because log5 25 = 2.
x + 1 = 25
x = 24
f(24) = 2 − 3
f(24) = −1
(24, −1)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → −1 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = −1; domain (−1, ∞); range (−∞, ∞); As x → −1 from the right, y → −∞; As x → ∞, y → ∞
f(x) = log4 (x − 5)
The vertical asymptote is where the inside equals 0.
x − 5 = 0
x = 5
Asymptote: x = 5
The inside must be positive.
x − 5 > 0
x > 5
Domain: (5, ∞)
Set the inside equal to 1, because log4 1 = 0.
x − 5 = 1
x = 6
f(6) = 0
f(6) = 0
(6, 0)
Set the inside equal to 4, because log4 4 = 1.
x − 5 = 4
x = 9
f(9) = 1
f(9) = 1
(9, 1)
Set the inside equal to 16, because log4 16 = 2.
x − 5 = 16
x = 21
f(21) = 2
f(21) = 2
(21, 2)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 5 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = 5; domain (5, ∞); range (−∞, ∞); As x → 5 from the right, y → −∞; As x → ∞, y → ∞
f(x) = log2 (x + 6) + 4
The vertical asymptote is where the inside equals 0.
x + 6 = 0
x = −6
Asymptote: x = −6
The inside must be positive.
x + 6 > 0
x > −6
Domain: (−6, ∞)
Set the inside equal to 1, because log2 1 = 0.
x + 6 = 1
x = −5
f(−5) = 0 + 4
f(−5) = 4
(−5, 4)
Set the inside equal to 2, because log2 2 = 1.
x + 6 = 2
x = −4
f(−4) = 1 + 4
f(−4) = 5
(−4, 5)
Set the inside equal to 4, because log2 4 = 2.
x + 6 = 4
x = −2
f(−2) = 2 + 4
f(−2) = 6
(−2, 6)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → −6 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = −6; domain (−6, ∞); range (−∞, ∞); As x → −6 from the right, y → −∞; As x → ∞, y → ∞
A coefficient in front of the log
f(x) = 2 · log3 (x − 1) − 4
The vertical asymptote is where the inside equals 0.
x − 1 = 0
x = 1
Asymptote: x = 1
The inside must be positive.
x − 1 > 0
x > 1
Domain: (1, ∞)
Set the inside equal to 1, because log3 1 = 0.
x − 1 = 1
x = 2
f(2) = 2 · 0 − 4
f(2) = −4
(2, −4)
Set the inside equal to 3, because log3 3 = 1.
x − 1 = 3
x = 4
f(4) = 2 · 1 − 4
f(4) = −2
(4, −2)
Set the inside equal to 9, because log3 9 = 2.
x − 1 = 9
x = 10
f(10) = 2 · 2 − 4
f(10) = 0
(10, 0)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 1 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = 1; domain (1, ∞); range (−∞, ∞); As x → 1 from the right, y → −∞; As x → ∞, y → ∞
f(x) = −log2 (x + 3) + 1
The vertical asymptote is where the inside equals 0.
x + 3 = 0
x = −3
Asymptote: x = −3
The inside must be positive.
x + 3 > 0
x > −3
Domain: (−3, ∞)
Set the inside equal to 1, because log2 1 = 0.
x + 3 = 1
x = −2
f(−2) = −1 · 0 + 1
f(−2) = 1
(−2, 1)
Set the inside equal to 2, because log2 2 = 1.
x + 3 = 2
x = −1
f(−1) = −1 · 1 + 1
f(−1) = 0
(−1, 0)
Set the inside equal to 4, because log2 4 = 2.
x + 3 = 4
x = 1
f(1) = −1 · 2 + 1
f(1) = −1
(1, −1)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → −3 from the right, y → ∞
As x → ∞, y → −∞
Answer: Neither; asymptote x = −3; domain (−3, ∞); range (−∞, ∞); As x → −3 from the right, y → ∞; As x → ∞, y → −∞
f(x) = 3 · log4 (x − 2) + 1
The vertical asymptote is where the inside equals 0.
x − 2 = 0
x = 2
Asymptote: x = 2
The inside must be positive.
x − 2 > 0
x > 2
Domain: (2, ∞)
Set the inside equal to 1, because log4 1 = 0.
x − 2 = 1
x = 3
f(3) = 3 · 0 + 1
f(3) = 1
(3, 1)
Set the inside equal to 4, because log4 4 = 1.
x − 2 = 4
x = 6
f(6) = 3 · 1 + 1
f(6) = 4
(6, 4)
Set the inside equal to 16, because log4 16 = 2.
x − 2 = 16
x = 18
f(18) = 3 · 2 + 1
f(18) = 7
(18, 7)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 2 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = 2; domain (2, ∞); range (−∞, ∞); As x → 2 from the right, y → −∞; As x → ∞, y → ∞
f(x) = −2 · log5 (x + 2) − 3
The vertical asymptote is where the inside equals 0.
x + 2 = 0
x = −2
Asymptote: x = −2
The inside must be positive.
x + 2 > 0
x > −2
Domain: (−2, ∞)
Set the inside equal to 1, because log5 1 = 0.
x + 2 = 1
x = −1
f(−1) = −2 · 0 − 3
f(−1) = −3
(−1, −3)
Set the inside equal to 5, because log5 5 = 1.
x + 2 = 5
x = 3
f(3) = −2 · 1 − 3
f(3) = −5
(3, −5)
Set the inside equal to 25, because log5 25 = 2.
x + 2 = 25
x = 23
f(23) = −2 · 2 − 3
f(23) = −7
(23, −7)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → −2 from the right, y → ∞
As x → ∞, y → −∞
Answer: Neither; asymptote x = −2; domain (−2, ∞); range (−∞, ∞); As x → −2 from the right, y → ∞; As x → ∞, y → −∞
f(x) = 2 · log2 (x − 4) − 6
The vertical asymptote is where the inside equals 0.
x − 4 = 0
x = 4
Asymptote: x = 4
The inside must be positive.
x − 4 > 0
x > 4
Domain: (4, ∞)
Set the inside equal to 1, because log2 1 = 0.
x − 4 = 1
x = 5
f(5) = 2 · 0 − 6
f(5) = −6
(5, −6)
Set the inside equal to 2, because log2 2 = 1.
x − 4 = 2
x = 6
f(6) = 2 · 1 − 6
f(6) = −4
(6, −4)
Set the inside equal to 4, because log2 4 = 2.
x − 4 = 4
x = 8
f(8) = 2 · 2 − 6
f(8) = −2
(8, −2)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 4 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = 4; domain (4, ∞); range (−∞, ∞); As x → 4 from the right, y → −∞; As x → ∞, y → ∞
f(x) = −3 · log3 (x − 1) + 2
The vertical asymptote is where the inside equals 0.
x − 1 = 0
x = 1
Asymptote: x = 1
The inside must be positive.
x − 1 > 0
x > 1
Domain: (1, ∞)
Set the inside equal to 1, because log3 1 = 0.
x − 1 = 1
x = 2
f(2) = −3 · 0 + 2
f(2) = 2
(2, 2)
Set the inside equal to 3, because log3 3 = 1.
x − 1 = 3
x = 4
f(4) = −3 · 1 + 2
f(4) = −1
(4, −1)
Set the inside equal to 9, because log3 9 = 2.
x − 1 = 9
x = 10
f(10) = −3 · 2 + 2
f(10) = −4
(10, −4)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 1 from the right, y → ∞
As x → ∞, y → −∞
Answer: Neither; asymptote x = 1; domain (1, ∞); range (−∞, ∞); As x → 1 from the right, y → ∞; As x → ∞, y → −∞
A coefficient inside the log
f(x) = log2 (6 − 2x) + 1
The vertical asymptote is where the inside equals 0.
6 − 2x = 0
6 = 2x
x = 3
Asymptote: x = 3
The inside must be positive.
6 − 2x > 0
x < 3
Domain: (−∞, 3)
Set the inside equal to 2, because log2 2 = 1.
6 − 2x = 2
4 = 2x
x = 2
f(2) = 1 + 1
f(2) = 2
(2, 2)
Set the inside equal to 4, because log2 4 = 2.
6 − 2x = 4
2 = 2x
x = 1
f(1) = 2 + 1
f(1) = 3
(1, 3)
Set the inside equal to 8, because log2 8 = 3.
6 − 2x = 8
−2 = 2x
x = −1
f(−1) = 3 + 1
f(−1) = 4
(−1, 4)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 3 from the left, y → −∞
As x → −∞, y → ∞
Answer: Neither; asymptote x = 3; domain (−∞, 3); range (−∞, ∞); As x → 3 from the left, y → −∞; As x → −∞, y → ∞
f(x) = log2 (4x − 12) − 3
The vertical asymptote is where the inside equals 0.
4x − 12 = 0
4x = 12
x = 3
Asymptote: x = 3
The inside must be positive.
4x − 12 > 0
x > 3
Domain: (3, ∞)
Set the inside equal to 4, because log2 4 = 2.
4x − 12 = 4
4x = 16
x = 4
f(4) = 2 − 3
f(4) = −1
(4, −1)
Set the inside equal to 8, because log2 8 = 3.
4x − 12 = 8
4x = 20
x = 5
f(5) = 3 − 3
f(5) = 0
(5, 0)
Set the inside equal to 16, because log2 16 = 4.
4x − 12 = 16
4x = 28
x = 7
f(7) = 4 − 3
f(7) = 1
(7, 1)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 3 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = 3; domain (3, ∞); range (−∞, ∞); As x → 3 from the right, y → −∞; As x → ∞, y → ∞
f(x) = log3 (3x + 9) − 1
The vertical asymptote is where the inside equals 0.
3x + 9 = 0
3x = −9
x = −3
Asymptote: x = −3
The inside must be positive.
3x + 9 > 0
x > −3
Domain: (−3, ∞)
Set the inside equal to 3, because log3 3 = 1.
3x + 9 = 3
3x = −6
x = −2
f(−2) = 1 − 1
f(−2) = 0
(−2, 0)
Set the inside equal to 9, because log3 9 = 2.
3x + 9 = 9
3x = 0
x = 0
f(0) = 2 − 1
f(0) = 1
(0, 1)
Set the inside equal to 27, because log3 27 = 3.
3x + 9 = 27
3x = 18
x = 6
f(6) = 3 − 1
f(6) = 2
(6, 2)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → −3 from the right, y → −∞
As x → ∞, y → ∞
Answer: Neither; asymptote x = −3; domain (−3, ∞); range (−∞, ∞); As x → −3 from the right, y → −∞; As x → ∞, y → ∞
f(x) = −log2 (2x + 4) + 5
The vertical asymptote is where the inside equals 0.
2x + 4 = 0
2x = −4
x = −2
Asymptote: x = −2
The inside must be positive.
2x + 4 > 0
x > −2
Domain: (−2, ∞)
Set the inside equal to 2, because log2 2 = 1.
2x + 4 = 2
2x = −2
x = −1
f(−1) = −1 · 1 + 5
f(−1) = 4
(−1, 4)
Set the inside equal to 4, because log2 4 = 2.
2x + 4 = 4
2x = 0
x = 0
f(0) = −1 · 2 + 5
f(0) = 3
(0, 3)
Set the inside equal to 8, because log2 8 = 3.
2x + 4 = 8
2x = 4
x = 2
f(2) = −1 · 3 + 5
f(2) = 2
(2, 2)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → −2 from the right, y → ∞
As x → ∞, y → −∞
Answer: Neither; asymptote x = −2; domain (−2, ∞); range (−∞, ∞); As x → −2 from the right, y → ∞; As x → ∞, y → −∞
f(x) = log3 (9 − 3x) + 2
The vertical asymptote is where the inside equals 0.
9 − 3x = 0
9 = 3x
x = 3
Asymptote: x = 3
The inside must be positive.
9 − 3x > 0
x < 3
Domain: (−∞, 3)
Set the inside equal to 3, because log3 3 = 1.
9 − 3x = 3
6 = 3x
x = 2
f(2) = 1 + 2
f(2) = 3
(2, 3)
Set the inside equal to 9, because log3 9 = 2.
9 − 3x = 9
0 = 3x
x = 0
f(0) = 2 + 2
f(0) = 4
(0, 4)
Set the inside equal to 27, because log3 27 = 3.
9 − 3x = 27
−18 = 3x
x = −6
f(−6) = 3 + 2
f(−6) = 5
(−6, 5)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 3 from the left, y → −∞
As x → −∞, y → ∞
Answer: Neither; asymptote x = 3; domain (−∞, 3); range (−∞, ∞); As x → 3 from the left, y → −∞; As x → −∞, y → ∞
f(x) = 2 · log3 (6 − x) − 4
The vertical asymptote is where the inside equals 0.
6 − x = 0
6 = 1x
x = 6
Asymptote: x = 6
The inside must be positive.
6 − x > 0
x < 6
Domain: (−∞, 6)
Set the inside equal to 1, because log3 1 = 0.
6 − x = 1
5 = 1x
x = 5
f(5) = 2 · 0 − 4
f(5) = −4
(5, −4)
Set the inside equal to 3, because log3 3 = 1.
6 − x = 3
3 = 1x
x = 3
f(3) = 2 · 1 − 4
f(3) = −2
(3, −2)
Set the inside equal to 9, because log3 9 = 2.
6 − x = 9
−3 = 1x
x = −3
f(−3) = 2 · 2 − 4
f(−3) = 0
(−3, 0)
A logarithm is neither exponential growth nor exponential decay.
Neither
Range of a log function, when the coefficient of the log is not 0:
(−∞, ∞)
As x → 6 from the left, y → −∞
As x → −∞, y → ∞
Answer: Neither; asymptote x = 6; domain (−∞, 6); range (−∞, ∞); As x → 6 from the left, y → −∞; As x → −∞, y → ∞
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