Rational Function Asymptotes Explorer

The graph covers −6 to 6 on both axes. The function is drawn in coral; vertical asymptotes are dashed coral lines, the horizontal or slant asymptote is a dashed teal line, zeros are solid dark dots on the x-axis, and holes are open circles.

The sim opens with $f(x) = \frac{x+2}{x-2}$. The panel shows the fraction exactly as you built it, with common factors not cancelled, and then four facts: Vertical asymptotes, End behavior (the horizontal or slant asymptote), Holes as (x, y) points, and Zeros (x-intercepts). Keep a nonzero: at a = 0 the function is just 0 (with gaps where the denominator is zero), but the panel still lists the zeros and asymptotes for the factors, so that setting does not illustrate anything useful.

Write $f(x) = \frac{p(x)}{q(x)}$ in factored form and cancel any factor shared by the numerator and denominator. Then:

End behavior depends on the degrees n (numerator) and d (denominator) and the leading coefficients: $$n < d:\ y = 0, \qquad n = d:\ y = \frac{\text{leading coefficient of } p}{\text{leading coefficient of } q}, \qquad n = d + 1:\ \text{slant asymptote}.$$ For a slant asymptote, divide p by q; the quotient $mx + c$ is the asymptote and the remainder is ignored. If $n > d + 1$, there is no horizontal or slant asymptote. In this simulator every denominator factor has coefficient 1, so when the degrees match the horizontal asymptote is simply $y = a$.

AP Precalculus describes these with limits: a vertical asymptote at $x = 3$ means $\lim_{x \to 3^+} f(x)$ or $\lim_{x \to 3^-} f(x)$ is $\pm\infty$, and a horizontal asymptote $y = 2$ means $\lim_{x \to \pm\infty} f(x) = 2$.

Step 1: cancel. $(x - 1)$ appears in both, so $f(x) = \frac{2(x+2)}{x-3}$ for $x \ne 1$.

Step 2: hole. Plug $x = 1$ into the reduced form: $\frac{2(3)}{-2} = -3$. Hole at $(1, -3)$.

Step 3: vertical asymptote. The remaining denominator factor is $(x - 3)$, so $x = 3$.

Step 4: zero and y-intercept. The remaining numerator factor gives a zero at $x = -2$. The y-intercept is $f(0) = \frac{2(2)}{-3} = -\tfrac43 \approx -1.33$.

Step 5: end behavior. Both reduced polynomials have degree 1, so the horizontal asymptote is $y = \frac{2}{1} = 2$.

Check it in the simulator: set a = 2. In the numerator, leave x+2 on and turn on x−1. In the denominator, turn off x−2 and turn on x−1 and x−3. The facts should read: vertical asymptotes x = 3; horizontal asymptote y = 2; holes (1, -3); zeros x = -2.

Follow-up (slant): set a = 1, use x and x+1 in the numerator and only x−1 in the denominator. Dividing $x^2 + x$ by $x - 1$ gives quotient $x + 2$, remainder 2, so the slant asymptote is $y = x + 2$ (the panel writes it as y = 1x + 2).

This simulator spans Topics 1.7 to 1.10 of Unit 1: end behavior, zeros, vertical asymptotes and holes of rational functions. Questions often give a rational function in factored form and ask for the limit as x approaches a value from one side, the location of a hole, or which graph matches a given set of asymptotes and intercepts.