DWG NO. 2.6 — Lesson 6 of 6

Graphing Rational Functions

Unit 2: Polynomial and Rational Functions · ~30 min

Objective Graph a rational function by combining its intercepts, asymptotes, holes, and a sign analysis of the intervals between them.

Everything from the last lesson — vertical asymptotes, holes, horizontal and slant asymptotes — describes the skeleton of a rational function's graph. This lesson fills it in with the two remaining pieces: intercepts, and which side of the x-axis the graph sits on in each region.

The full workflow

1. Factor and simplify. Cancel any common factors first — they become holes, not part of the graph's main shape.
2. Find the vertical asymptote(s). Zeros of the simplified denominator.
3. Find the horizontal or slant asymptote. Compare degrees, as in 2.5.
4. Find the intercepts. x-intercepts come from zeros of the simplified numerator; the y-intercept is f(0).
5. Run a sign analysis. The vertical asymptotes and x-intercepts split the x-axis into intervals. The function's sign can only change at those boundary points, so test one point in each interval to find whether the graph sits above or below the x-axis there.
6. Sketch. Draw the asymptotes as dashed guide lines, plot the intercepts, then draw each branch approaching its asymptotes in the direction the sign analysis and asymptote behavior require.

f(x) = \(\frac{x-1}{x^{2}-4}\) — vertical asymptotes x = ±2, horizontal asymptote y = 0, x-intercept x = 1

Worked Example 1 · Full graph, matching the diagram above

ProblemGraph f(x) = \(\frac{x-1}{x^{2}-4}\). Denominator factors as (x−2)(x+2); nothing cancels with the numerator, so no holes.

1Vertical asymptotes: x = 2 and x = −2. Horizontal asymptote: numerator degree 1 < denominator degree 2, so y = 0.

2x-intercept: x−1 = 0 → x = 1. y-intercept: f(0) = (−1)/(−4) = 0.25.

3The asymptotes and x-intercept split the axis into four intervals. Testing one point in each: f(−3) < 0, f(0) > 0, f(1.5) < 0, f(3) > 0.

Below axis on (−∞,−2) and (1,2); above axis on (−2,1) and (2,∞) — matching the four branches in the diagram

Worked Example 2 · A simpler single-asymptote graph

ProblemGraph f(x) = \(\frac{2x}{x-3}\). Vertical asymptote: x = 3. Horizontal asymptote: equal degrees, ratio of leading coefficients 2/1 = 2, so y = 2.

1x-intercept: 2x = 0 → x = 0, which is also the y-intercept (they coincide here).

2Test the two intervals split by x = 0 and x = 3: f(−1) = −2/−4 = 0.5 > 0; f(1) = 2/−2 = −1 < 0; f(4) = 8/1 = 8 > 0.

Above axis on (−∞,0) and (3,∞); below axis on (0,3)

Worked Example 3 · A rational function with no vertical asymptote

ProblemGraph f(x) = \(\frac{1}{x^{2}+1}\). The denominator x²+1 is never 0 for any real x (it's always ≥ 1), so there is no vertical asymptote and no break in the graph at all.

1Horizontal asymptote: numerator degree 0 < denominator degree 2, so y = 0.

2The numerator is the constant 1, never 0, so there's no x-intercept. y-intercept: f(0) = 1.

No vertical asymptote; horizontal asymptote y = 0; the graph is entirely above the x-axis, peaking at (0, 1)

Guided practice

1.List the vertical asymptote(s), any holes, and the horizontal asymptote of f(x) = \(\frac{x+1}{x^{2}-1}\).
Show answer

x²−1 = (x−1)(x+1); the (x+1) factor cancels, leaving 1/(x−1). Hole at x = −1 · Vertical asymptote: x = 1 · Horizontal asymptote: y = 0
2.Find the x- and y-intercepts of f(x) = \(\frac{x-4}{x+2}\).
Show answer

x-intercept: x−4=0 → x = 4. y-intercept: f(0) = −4/2 = −2.
3.Determine the sign of f(x) = \(\frac{x+3}{x-1}\) on the interval (−3, 1) by testing a point.
Show answer

Test x = 0: f(0) = 3/(−1) = −3. f(x) is negative on (−3, 1).
4.State the vertical asymptote, horizontal asymptote, and any intercepts of f(x) = \(\frac{3}{x+2}\).
Show answer

Vertical asymptote: x = −2. Numerator degree 0 < denominator degree 1, so horizontal asymptote: y = 0. The constant numerator is never 0, so no x-intercept; y-intercept: f(0) = 3/2.
5.Explain why f(x) = \(\frac{1}{x^{2}+4}\) has no vertical asymptote, and state its horizontal asymptote.
Show answer

x²+4 is always at least 4 for real x, so it's never 0 — no vertical asymptote. Numerator degree 0 < denominator degree 2, so the horizontal asymptote is y = 0.

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