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Learn to plot log functions by finding the vertical asymptote, domain, intercepts, and key points, then check your graph with a simple table.
Graphing a logarithmic function is mostly about location: where is the vertical asymptote, where does the curve cross the x-axis, and which key points show the curve's slow growth. This lesson uses the parent form y = log_b(x - h) + k and a consistent plotting convention. You will build a small value table, mark the asymptote as a dashed vertical line, plot the intercept and two or three key points, then connect them with a smooth curve that never touches the asymptote. The same steps work for any base b > 1, and the check method uses arithmetic you can do by hand.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
For y = log_b(x - h) + k with b > 1, the argument x - h must be positive, so the domain is x > h. The vertical asymptote is the line x = h. The graph approaches this line but never crosses it.
Write the asymptote as a dashed vertical line on the grid before plotting any points. Label it with its equation. This single line controls the shape: the curve rises slowly to the right of it and drops steeply as x gets close to h from the right.
Pick x-values that make x - h equal to simple powers of the base b. For base 2, use x - h = 1, 2, 4, 8 because log_2(1)=0, log_2(2)=1, log_2(4)=2, log_2(8)=3. Then add k to each result. This keeps the arithmetic checkable without a calculator.
Plot at least three points: the x-intercept when it exists, and two points on either side of it. The x-intercept occurs when log_b(x - h) + k = 0, so x - h = b^(-k) and x = h + b^(-k).
Plot the intercept and key points, then draw a smooth curve through them that approaches the asymptote without touching it. The curve should be steep near the asymptote and flatten as x increases. Do not connect points with straight segments; the graph is curved.
Check by reading one point back from your graph and confirming it satisfies the equation. Also confirm that no plotted point has x ≤ h. If a point falls left of the asymptote, it is invalid for this function.
Graph y = log_2(x - 1) + 1. State the domain, asymptote, x-intercept, and three key points.
Domain: x > 1. Asymptote: x = 1 (dashed vertical line). x-intercept: solve log_2(x - 1) + 1 = 0, so log_2(x - 1) = -1, x - 1 = 1/2, x = 1.5; intercept (1.5, 0). Key points: x - 1 = 1 gives x = 2, y = 0 + 1 = 1, point (2, 1); x - 1 = 2 gives x = 3, y = 1 + 1 = 2, point (3, 2); x - 1 = 4 gives x = 5, y = 2 + 1 = 3, point (5, 3). Plot these, draw a smooth increasing curve approaching x = 1 from the right, and check (3, 2): log_2(2) + 1 = 1 + 1 = 2.
Graph y = log_3(x - 2). State the domain, asymptote, x-intercept, and two key points.
Domain x > 2; asymptote x = 2; x-intercept (3, 0); key points (5, 1) and (11, 2).
Why: The argument x - 2 must be positive, so x > 2 and the asymptote is x = 2. Set log_3(x - 2) = 0, so x - 2 = 1 and x = 3. For x - 2 = 3, x = 5 and y = 1. For x - 2 = 9, x = 11 and y = 2.
Check: Substitute (5, 1): log_3(3) = 1. Confirm no plotted x-value is 2 or less.
Graph y = log_2(x + 1) - 2. State the domain, asymptote, x-intercept, and two key points.
Domain x > -1; asymptote x = -1; x-intercept (3, 0); key points (1, -1) and (7, 1).
Why: Here h = -1, so x + 1 > 0 gives x > -1 and the asymptote is x = -1. Solve log_2(x + 1) - 2 = 0, so log_2(x + 1) = 2, x + 1 = 4, x = 3. For x + 1 = 2, x = 1 and y = 1 - 2 = -1. For x + 1 = 8, x = 7 and y = 3 - 2 = 1.
Check: Substitute (7, 1): log_2(8) - 2 = 3 - 2 = 1. Confirm all x-values exceed -1.
Graph y = log_5(x - 4) + 1. State the domain, asymptote, x-intercept, and two key points.
Domain x > 4; asymptote x = 4; x-intercept (4.2, 0); key points (9, 2) and (29, 3).
Why: The domain is x > 4 and the asymptote is x = 4. Solve log_5(x - 4) + 1 = 0, so log_5(x - 4) = -1, x - 4 = 1/5, x = 4.2. For x - 4 = 5, x = 9 and y = 1 + 1 = 2. For x - 4 = 25, x = 29 and y = 2 + 1 = 3.
Check: Substitute (9, 2): log_5(5) + 1 = 1 + 1 = 2. Confirm the intercept 4.2 is greater than 4.
Graph y = log_2(x - 3) + 2. State the domain, asymptote, whether an x-intercept exists, and two key points.
Domain x > 3; asymptote x = 3; an x-intercept exists at (3.25, 0) because log_2(x - 3) + 2 = 0 gives log_2(x - 3) = -2, so x - 3 = 1/4 and x = 3.25. Key points (4, 2) and (5, 3).
Why: The domain is x > 3 and the asymptote is x = 3. Solving log_2(x - 3) + 2 = 0 gives log_2(x - 3) = -2, x - 3 = 1/4, x = 3.25, so an intercept does exist at (3.25, 0). For x - 3 = 1, x = 4 and y = 0 + 2 = 2. For x - 3 = 2, x = 5 and y = 1 + 2 = 3.
Check: Substitute (4, 2): log_2(1) + 2 = 0 + 2 = 2. Confirm 3.25 > 3 and that the curve stays above the asymptote's x-value.
No. If k makes the curve shift entirely above or below the axis, there is no x-intercept. Solve log_b(x - h) + k = 0 to check.
The argument x - h approaches zero from the right, and log_b of a number near zero becomes very negative, so the curve drops without bound near x = h.
At least three well-chosen points plus the intercept when it exists. More points help, but easy powers of the base keep the arithmetic simple.
Yes, but the curve then decreases instead of increases. This lesson uses b > 1 so the shape is consistently increasing.