4.4Graphs of Logarithmic Functions
Graph f(x) = a·log_b(x−h)+k, apply transformations, identify the vertical asymptote x = h, and determine domain and range.
Logarithmic graphs appear in every field that uses scales — sound intensity, earthquake magnitude, information entropy. Mastering their transformations, especially how horizontal shifts move the vertical asymptote, is essential for modeling and calculus.
Essential Question
How do transformations of f(x) = log_b(x) move the vertical asymptote and change the domain — and how is this different from what happens when you transform an exponential function?
Lesson Overview
Every logarithmic graph f(x) = a·log_b(x−h) + k is a transformation of the parent y = log_b(x). Unlike exponential functions whose asymptote is horizontal, logarithmic functions have a vertical asymptote at x = h. Horizontal shifts move this asymptote and change the domain — the key difference to master in this chapter.
Parent Function Properties
- Domain: (0, ∞) — argument must be positive
- Range: (−∞, ∞) — all real numbers
- X-intercept: (1, 0) — since log_b(1) = 0
- Vertical asymptote: x = 0 (y-axis)
- Passes through: (b, 1) — since log_b(b) = 1
General Form
f(x) = a · log_b(x − h) + k
- a: vertical stretch; if a < 0, reflects over x-axis
- h: horizontal shift → asymptote moves to x = h
- k: vertical shift (does NOT move asymptote)
- Domain: (h, ∞) when b > 1 and a > 0
- Range: always (−∞, ∞)
log₂x (increasing, b > 1) vs log₁/₂x (decreasing, 0 < b < 1) — both pass through (1, 0)
Effect of each transformation on asymptote, domain, and range
Worked Examples
Graph f(x) = log₂(x) + 3. State the asymptote, domain, range, and x-intercept.
Start with parent y = log₂(x). The "+3" shifts the graph UP 3 units.
Vertical shift does NOT move the vertical asymptote: x = 0.
Domain: (0, ∞) — unchanged by vertical shift.
Range: (−∞, ∞) — always all reals for any log function.
X-intercept: set f(x) = 0 → log₂(x) + 3 = 0 → log₂(x) = −3 → x = 2⁻³ = 1/8.
Key points: f(1) = 0+3 = 3; f(2) = 1+3 = 4; f(4) = 2+3 = 5.
Graph g(x) = log₂(x − 3). State the asymptote, domain, range, and x-intercept.
The "(x − 3)" shifts the graph RIGHT 3 units.
Vertical asymptote moves right 3: x = 3.
Domain: (3, ∞) — argument (x−3) must be positive, so x > 3.
Range: (−∞, ∞).
X-intercept: log₂(x−3) = 0 → x−3 = 2⁰ = 1 → x = 4. X-int: (4, 0).
Key points: g(4) = log₂(1) = 0; g(5) = log₂(2) = 1; g(7) = log₂(4) = 2.
Graph h(x) = −log₃(x). State the asymptote, domain, range, and x-intercept.
The negative sign reflects the parent y = log₃(x) over the x-axis.
Asymptote: x = 0 (reflection does not move the vertical asymptote).
Domain: (0, ∞) — unchanged.
Range: (−∞, ∞) — still all reals after reflection.
X-intercept: −log₃(x) = 0 → log₃(x) = 0 → x = 1. X-int: (1, 0).
The graph is now decreasing (falls left to right) instead of increasing.
Describe all transformations of f(x) = 2·log₅(x + 4) − 1 and state the asymptote and domain.
Parent: y = log₅(x).
a = 2: vertical stretch by factor 2.
(x + 4): horizontal shift LEFT 4 units.
− 1: vertical shift DOWN 1 unit.
Asymptote: x = −4 (horizontal shift moves asymptote left 4).
Domain: (−4, ∞) — argument (x+4) > 0 → x > −4.
X-intercept: 2·log₅(x+4) − 1 = 0 → log₅(x+4) = 1/2 → x+4 = 5^(1/2) = √5 → x = √5 − 4 ≈ −1.76.
Write the equation of a logarithmic function with base 3, shifted right 2 and up 5, reflected over the x-axis.
Start with parent: y = log₃(x).
Shift right 2: replace x with (x−2) → y = log₃(x−2).
Reflect over x-axis: multiply by −1 → y = −log₃(x−2).
Shift up 5: add 5 → y = −log₃(x−2) + 5.
Asymptote: x = 2 (from the horizontal shift).
Domain: (2, ∞).
Guided Practice
For f(x) = log₃(x) − 4, state the asymptote, domain, range, and x-intercept.
Hint: Vertical shift does not move the asymptote. Find x-intercept by setting f(x) = 0.
For g(x) = log₂(x + 5), state the asymptote, domain, range, and x-intercept.
Hint: (x+5) shifts left 5. Asymptote moves to x = −5. Domain: x > −5.
Describe all transformations of h(x) = −3·log₂(x − 1) + 2.
Hint: Identify a, h, k from f(x) = a·log_b(x−h)+k. List each transformation separately.
Write the equation of a log function with base 2, shifted left 3 and down 1.
Hint: Left shift: replace x with (x+3). Down shift: subtract 1 outside the log.
For f(x) = log₄(x + 2) − 3, find the domain and x-intercept.
Hint: Domain: set argument > 0. X-intercept: set f(x) = 0 and convert to exponential form.
Key Vocabulary
Vertical Asymptote
A vertical line x = h that the logarithmic graph approaches but never crosses. Determined by the horizontal shift h in f(x) = a·log_b(x−h)+k.
Domain of a Log Function
For f(x) = a·log_b(x−h)+k, the domain is (h, ∞). The argument (x−h) must be strictly positive.
Range of a Log Function
Always (−∞, ∞) for any logarithmic function, regardless of transformations. The range never changes.
X-intercept
Found by setting f(x) = 0 and solving. For log functions, convert to exponential form after isolating the log.
Increasing vs Decreasing
If b > 1: log_b(x) is increasing (rises left to right). If 0 < b < 1: log_b(x) is decreasing (falls left to right).
Vertical Stretch/Compress
Multiplying by |a| > 1 stretches the graph away from the x-axis; 0 < |a| < 1 compresses it. Does not change asymptote or domain.
Horizontal Shift
Replacing x with (x−h) shifts the graph right h units (h > 0) or left |h| units (h < 0). Moves the vertical asymptote to x = h.
Reflection
Multiplying by −1 outside (−log) reflects over x-axis; replacing x with −x reflects over y-axis (changes domain to (−∞, 0)).
Check Your Understanding
Interactive Practice — 5 Questions
What is the vertical asymptote of f(x) = log₂(x − 5)?
What is the domain of g(x) = log₃(x + 4)?
What is the range of h(x) = −2·log₅(x − 1) + 7?
Find the x-intercept of f(x) = log₂(x − 1).
Which function has vertical asymptote x = −3?
Independent Practice
Independent Practice
For f(x) = log₂(x) + 5, state the asymptote, domain, range, and x-intercept.
For g(x) = log₃(x − 4), state the asymptote, domain, range, and x-intercept.
Describe all transformations of h(x) = −4·log₂(x + 3) − 2 and state the asymptote and domain.
Write the equation of a log function with base 5, shifted right 1 and up 3, reflected over the x-axis.
For f(x) = 2·log₄(x + 1) − 3, find: (a) asymptote (b) domain (c) x-intercept.
Vertical shift: log₂x vs log₂x+3 (asymptote stays at x=0)
Horizontal shift: log₂x vs log₂(x−2) (asymptote moves to x=2)
Reflection over x-axis: log₂x vs −log₂x (x-intercept stays at (1,0))
Common Mistakes
Saying the asymptote of f(x) = log(x) + 5 is x = 5 (confusing vertical shift with horizontal shift).
Adding outside the log shifts the graph vertically — the asymptote stays at x = 0. Only (x − h) inside the log moves the asymptote to x = h.
Writing the domain of f(x) = log₂(x − 3) as (0, ∞) instead of (3, ∞).
The domain is determined by the argument: (x − 3) > 0 → x > 3. Domain: (3, ∞).
Saying the range of f(x) = −log(x) + 7 is (−∞, 7) because of the reflection and shift.
The range of every logarithmic function is always (−∞, ∞). Transformations never restrict the range.
Thinking f(x) = log(x + 3) shifts the graph RIGHT 3.
f(x+3) shifts LEFT 3. To shift right, use f(x−3) = log(x−3). The sign inside is counterintuitive — same rule as exponentials.
Math Tips
The vertical asymptote of f(x) = a·log_b(x−h)+k is always x = h. Set the argument equal to zero and solve.
Domain rule: set the argument > 0 and solve for x. The domain is that solution set as an interval.
Range is always (−∞, ∞) for any log function — no exceptions. You never need to calculate it.
To find the x-intercept: set f(x) = 0, isolate the log, then convert to exponential form.
Sketch checklist: (1) find asymptote x = h, (2) plot x-intercept, (3) plot (h+b, 1) — the point one unit above the x-axis, (4) draw smooth curve approaching asymptote.