H Algebra 2/Trig · Unit 7
For an exponential, write the horizontal asymptote and the y-intercept. For a log, write the vertical asymptote and the x-intercept.
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Exponential y-intercept
For f(x) = a·bx + c, the asymptote is y = c.
Substitute x = 0. Because b0 = 1, the y-intercept is (0, a + c).
Log asymptote
Set the inside equal to 0 and solve. That x-value is the vertical asymptote.
Factor a common factor first when the inside has one. log4(8(x − 2)) is still 0 only when x − 2 = 0.
Log x-intercept
Set y = 0. Undo the added constant, then divide by the coefficient of the log.
Rewrite the log equation as an exponential equation, then solve for x.
The inside has to stay positive at that x-value.
Exponential: asymptote and y-intercept
f(x) = 9(13)x − 4
Horizontal asymptote of y = a·bx + c:
y = c
y = −4
The y-intercept is the point where x = 0.
f(0) = 9(13)⁰ − 4
(13)⁰ = 1
f(0) = 9 · 1 − 4
f(0) = 5
(0, 5)
Answer: Asymptote y = −4; y-intercept (0, 5)
f(x) = 4(2x) − 7
Horizontal asymptote of y = a·bx + c:
y = c
y = −7
The y-intercept is the point where x = 0.
f(0) = 4(2⁰) − 7
2⁰ = 1
f(0) = 4 · 1 − 7
f(0) = −3
(0, −3)
Answer: Asymptote y = −7; y-intercept (0, −3)
f(x) = 6(12)x + 3
Horizontal asymptote of y = a·bx + c:
y = c
y = 3
The y-intercept is the point where x = 0.
f(0) = 6(12)⁰ + 3
(12)⁰ = 1
f(0) = 6 · 1 + 3
f(0) = 9
(0, 9)
Answer: Asymptote y = 3; y-intercept (0, 9)
f(x) = 5(3x) − 1
Horizontal asymptote of y = a·bx + c:
y = c
y = −1
The y-intercept is the point where x = 0.
f(0) = 5(3⁰) − 1
3⁰ = 1
f(0) = 5 · 1 − 1
f(0) = 4
(0, 4)
Answer: Asymptote y = −1; y-intercept (0, 4)
f(x) = 8(14)x − 5
Horizontal asymptote of y = a·bx + c:
y = c
y = −5
The y-intercept is the point where x = 0.
f(0) = 8(14)⁰ − 5
(14)⁰ = 1
f(0) = 8 · 1 − 5
f(0) = 3
(0, 3)
Answer: Asymptote y = −5; y-intercept (0, 3)
f(x) = 2(7x) + 6
Horizontal asymptote of y = a·bx + c:
y = c
y = 6
The y-intercept is the point where x = 0.
f(0) = 2(7⁰) + 6
7⁰ = 1
f(0) = 2 · 1 + 6
f(0) = 8
(0, 8)
Answer: Asymptote y = 6; y-intercept (0, 8)
Log of (x − h): asymptote and x-intercept
g(x) = 2 · log3 (x + 1) − 6
The vertical asymptote is where the inside equals 0.
x + 1 = 0
x = −1
Asymptote: x = −1
The x-intercept is where y = 0.
2 · log3 (x + 1) − 6 = 0
2 · log3 (x + 1) = 6
log3 (x + 1) = 3
If log3(inside) = c, then the inside equals 3 raised to c.
(x + 1) = 3³
x = 26
(26, 0)
Answer: Asymptote x = −1; x-intercept (26, 0)
g(x) = log2 (x − 3) − 4
The vertical asymptote is where the inside equals 0.
x − 3 = 0
x = 3
Asymptote: x = 3
The x-intercept is where y = 0.
log2 (x − 3) − 4 = 0
log2 (x − 3) = 4
If log2(inside) = c, then the inside equals 2 raised to c.
(x − 3) = 2⁴
x = 19
(19, 0)
Answer: Asymptote x = 3; x-intercept (19, 0)
g(x) = 3 · log5 (x + 2) − 3
The vertical asymptote is where the inside equals 0.
x + 2 = 0
x = −2
Asymptote: x = −2
The x-intercept is where y = 0.
3 · log5 (x + 2) − 3 = 0
3 · log5 (x + 2) = 3
log5 (x + 2) = 1
If log5(inside) = c, then the inside equals 5 raised to c.
(x + 2) = 5¹
x = 3
(3, 0)
Answer: Asymptote x = −2; x-intercept (3, 0)
g(x) = −2 · log4 (x − 1) + 2
The vertical asymptote is where the inside equals 0.
x − 1 = 0
x = 1
Asymptote: x = 1
The x-intercept is where y = 0.
−2 · log4 (x − 1) + 2 = 0
−2 · log4 (x − 1) = −2
log4 (x − 1) = 1
If log4(inside) = c, then the inside equals 4 raised to c.
(x − 1) = 4¹
x = 5
(5, 0)
Answer: Asymptote x = 1; x-intercept (5, 0)
g(x) = log3 (x + 4) − 2
The vertical asymptote is where the inside equals 0.
x + 4 = 0
x = −4
Asymptote: x = −4
The x-intercept is where y = 0.
log3 (x + 4) − 2 = 0
log3 (x + 4) = 2
If log3(inside) = c, then the inside equals 3 raised to c.
(x + 4) = 3²
x = 5
(5, 0)
Answer: Asymptote x = −4; x-intercept (5, 0)
g(x) = 4 · log2 (x − 5) − 8
The vertical asymptote is where the inside equals 0.
x − 5 = 0
x = 5
Asymptote: x = 5
The x-intercept is where y = 0.
4 · log2 (x − 5) − 8 = 0
4 · log2 (x − 5) = 8
log2 (x − 5) = 2
If log2(inside) = c, then the inside equals 2 raised to c.
(x − 5) = 2²
x = 9
(9, 0)
Answer: Asymptote x = 5; x-intercept (9, 0)
A coefficient inside the log
g(x) = 3 · log4 (8x − 16) − 6
The vertical asymptote is where the inside equals 0.
8x − 16 = 0
8x = 16
x = 2
Asymptote: x = 2
Factor the inside before you solve.
8x − 16 = 8(x − 2)
The x-intercept is where y = 0.
3 · log4 (8x − 16) − 6 = 0
3 · log4 (8x − 16) = 6
log4 (8x − 16) = 2
If log4(inside) = c, then the inside equals 4 raised to c.
(8x − 16) = 4²
8x = 32
x = 4
(4, 0)
Answer: Asymptote x = 2; x-intercept (4, 0)
g(x) = 2 · log2 (4x − 20) − 4
The vertical asymptote is where the inside equals 0.
4x − 20 = 0
4x = 20
x = 5
Asymptote: x = 5
Factor the inside before you solve.
4x − 20 = 4(x − 5)
The x-intercept is where y = 0.
2 · log2 (4x − 20) − 4 = 0
2 · log2 (4x − 20) = 4
log2 (4x − 20) = 2
If log2(inside) = c, then the inside equals 2 raised to c.
(4x − 20) = 2²
4x = 24
x = 6
(6, 0)
Answer: Asymptote x = 5; x-intercept (6, 0)
g(x) = 3 · log5 (5x + 10) − 6
The vertical asymptote is where the inside equals 0.
5x + 10 = 0
5x = −10
x = −2
Asymptote: x = −2
Factor the inside before you solve.
5x + 10 = 5(x + 2)
The x-intercept is where y = 0.
3 · log5 (5x + 10) − 6 = 0
3 · log5 (5x + 10) = 6
log5 (5x + 10) = 2
If log5(inside) = c, then the inside equals 5 raised to c.
(5x + 10) = 5²
5x = 15
x = 3
(3, 0)
Answer: Asymptote x = −2; x-intercept (3, 0)
g(x) = −2 · log3 (3x − 6) + 6
The vertical asymptote is where the inside equals 0.
3x − 6 = 0
3x = 6
x = 2
Asymptote: x = 2
Factor the inside before you solve.
3x − 6 = 3(x − 2)
The x-intercept is where y = 0.
−2 · log3 (3x − 6) + 6 = 0
−2 · log3 (3x − 6) = −6
log3 (3x − 6) = 3
If log3(inside) = c, then the inside equals 3 raised to c.
(3x − 6) = 3³
3x = 33
x = 11
(11, 0)
Answer: Asymptote x = 2; x-intercept (11, 0)
g(x) = 4 · log2 (4x + 4) − 12
The vertical asymptote is where the inside equals 0.
4x + 4 = 0
4x = −4
x = −1
Asymptote: x = −1
Factor the inside before you solve.
4x + 4 = 4(x + 1)
The x-intercept is where y = 0.
4 · log2 (4x + 4) − 12 = 0
4 · log2 (4x + 4) = 12
log2 (4x + 4) = 3
If log2(inside) = c, then the inside equals 2 raised to c.
(4x + 4) = 2³
4x = 4
x = 1
(1, 0)
Answer: Asymptote x = −1; x-intercept (1, 0)
g(x) = 2 · log9 (3x − 15) − 1
The vertical asymptote is where the inside equals 0.
3x − 15 = 0
3x = 15
x = 5
Asymptote: x = 5
Factor the inside before you solve.
3x − 15 = 3(x − 5)
The x-intercept is where y = 0.
2 · log9 (3x − 15) − 1 = 0
2 · log9 (3x − 15) = 1
log9 (3x − 15) = 12
If log9(inside) = c, then the inside equals 9 raised to c.
(3x − 15) =
= 3
(3x − 15) = 3
3x = 18
x = 6
(6, 0)
Answer: Asymptote x = 5; x-intercept (6, 0)
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