FIND VERTICAL ASYMPTOTE OF LOG FUNCTION

Find the vertical asymptote of each of the following logarithmic functions.

Problem 1:

f_(x) = log₅ x + 2

Solution :

f_(x) = log₅ x + 2

x = 0

So, equation of vertical asymptote is x = 0.

Problem 2 :

f_(x) = log₃ (4x - 1) - 2

Solution :

f_(x) = log₃ (4x - 1) - 2

4x - 1 = 0

4x = 1

x = 1/4

So, equation of vertical asymptote is x = 1/4.

Problem 3 :

f_(x) = -log₂ 3x

Solution :

f_(x) = -log₂ 3x

-3x = 0

-x = 0

x = 0

So, equation of vertical asymptote is x = 0.

Problem 4 :

f_(x) = log₂ (5 - x)

Solution :

f_(x) = log₂ (5 - x)

5 - x = 0

-x = -5

x = 5

So, equation of vertical asymptote is x = 5.

Problem 5 :

f_(x) = log₅ (-x) + 5

Solution :

f_(x) = log₅ (-x) + 5

-x = 0

x = 0

So, equation of vertical asymptote is x = 0.

Problem 6 :

f_(x) = ln x - 4

Solution :

f_(x) = ln x - 4

x = 0

So, equation of vertical asymptote is x = 0.

Problem 7 :

f_(x) = ln (2 - 3x)

Solution :

f_(x) = ln (2 - 3x)

2 - 3x = 0

-3x = -2

x = 2/3

So, equation of vertical asymptote is x = 2/3.

Problem 8 :

f_(x) = log₃ (x + 5) + 1

Solution :

f_(x) = log₃ (x + 5) + 1

x + 5 = 0

x = -5

So, equation of vertical asymptote is x = -5.

Problem 9 :

f_(x) = ln (x - 3)

Solution :

f_(x) = ln (x - 3)

x - 3 = 0

x = 3

So, equation of vertical asymptote is x = 3.

Problem 10 :

Determine whether each statement is always, sometimes, or never true. Explain your reasoning.

a) A vertical translation of the graph of f (x) = log x changes the equation of the asymptote.

b) A horizontal shrink of the graph of f (x) = log x does not change the domain.

Solution :

a) Considering the example,

y = log x and y = (log x) - 3

for these two functions, the vertical asymptote is y-axis or x = 0.

Vertical translation will move the graph towards up or down. It is not affecting the vertical asymptote. Then, the vertical translation never change the equation of vertical asymptote. So, it is always true.

b) Considering the functions,

f(x) = log x

Domain is x > 0

After horizontal shrink, 

f(x) = log (2x)

Domain is x > 0

Then horizontal shrink dose not change the domain. So, it is always true.

Problem 11 :

Describe the transformation of the graph of f represented by the graph of g. Then give an equation of the asymptote.

a) f(x) = ln x, g(x) = ln(x + 6)

b) f(x) = log1/5 x, g(x) = log1/5 x + 13

Solution :

a) f(x) = ln x, g(x) = ln(x + 6)

Describing the transformation of the function g(x) :

g(x) = ln (x - (-6))

Horizontal shift of 6 units to the left.

Equation of vertical asymptote is x = 0 or y-axis.

b) f(x) = log1/5 x, g(x) = log1/5 x + 13

Describing the transformation of the function f(x) :

f(x) = log1/5 x

Horizontal stretch of 1/5.

Equation of horizontal asymptote is x = 0 or y-axis.

Describing the transformation of the function f(x) :

g(x) =  log1/5 x + 13

Horizontal stretch of 1/5.

Vertical translation of 13 units up

Equation of vertical asymptote is x = 0 or y-axis.

Problem 12 :

Sketch the graph of the transformation of f(x) =  log₅ x onto the graph. Label the graph.

a) g(x) = 3 log₅(x + 2) - 4

a) g(x) = 3 log₅(3 - x) + 1

Solution :

a) g(x) = 3 log₅(x + 2) - 4

Describing the transformation :

g(x) = 3 log₅(x - (-2)) - 4

• Vertical stretch of 3 units
• Horizontal translation of 2 units to the left and
• Vertical translation of 4 units down.

Finding vertical asymptote :

x + 2 = 0

x = -2

b) g(x) = 3 log₅(3 - x) + 1

Describing the transformation :

g(x) = 3 log₅(-(x - 3)) + 1

• Reflection across x-axis.
• Vertical stretch of 3 units
• Horizontal translation of 3 units to the right
• Vertical translation of 1 unit up.

Finding vertical asymptote :

x - 3 = 0

x = 3