Holes Where a Factor Cancels

Factor both parts first

A rational function is zero where its numerator is zero, and it has a vertical asymptote where its denominator is zero. Both rules assume that the numerator and the denominator are not zero at the same x. When they are, factor both before reading anything.

Take $y=\frac{\left(x^2-4\right)}{\left(x-2\right)}$y = (x² − 4)/(x − 2). The numerator is a difference of two squares, $x^2-4=\left(x-2\right)\left(x+2\right)$x² − 4 = (x − 2)(x + 2), so $y=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)}$y = (x − 2)(x + 2)/(x − 2). The same bracket, x − 2, appears in the numerator and in the denominator.

Cancel, except at x = 2

Canceling the bracket divides the numerator and the denominator by x − 2. That is allowed whenever x − 2 is not zero, so for every x except 2 the function is y = x + 2.

At x = 2 the original function reads (4 − 4) ÷ (2 − 2) = 0 ÷ 0, and 0 ÷ 0 has no value. A quotient 0 ÷ 0 would have to be a number c with c × 0 = 0, and every number does that, so no single answer exists. The original function is undefined at x = 2, even though x + 2 on its own would give 4 there.

So the simplified form comes with a condition: y = x + 2, for $x\ne 2$x ≠ 2. Without the condition it describes a different function.

A line with one point missing

The graph is the straight line y = x + 2 with the single point at x = 2 taken out. The height of the missing point comes from the simplified form: 2 + 2 = 4. So there is a gap at (2, 4), and it is drawn as a hollow dot, an open circle, which is the notation for a point that is not on the graph. The gap is called a hole.

Values near the hole stay close to 4. At x = 1.9 the function is 3.9, at x = 2.1 it is 4.1, and at x = 1.99 it is 3.99. From both sides they close in on 4. A vertical asymptote is the opposite: there the values grow without bound.

The graph of $y=\frac{\left(x^2-4\right)}{\left(x-2\right)}$y = (x² − 4)/(x − 2): the line y = x + 2 with a hole at (2, 4), drawn as a hollow dot.

Values of $y=\frac{\left(x^2-4\right)}{\left(x-2\right)}$y = (x² − 4)/(x − 2) for x just below and just above 2. They close in on 4 from both sides, and at x = 2 itself there is no value.

Change the denominator: nothing cancels

Now take $y=\frac{\left(x^2-4\right)}{\left(x-3\right)}$y = (x² − 4)/(x − 3). The numerator factors as before, (x − 2)(x + 2), but the denominator is x − 3, and it is not a factor of the numerator. There is no shared bracket, so nothing cancels.

Then the curve runs away

At x = 3 the numerator is 9 − 4 = 5 and the denominator is 0. That is 5 ÷ 0, not 0 ÷ 0, and the values near it are not close to anything. At x = 3.1, $y=5.61/0.1=56.1$y = 5.61 / 0.1 = 56.1, and at x = 2.9, $y=4.41/\left(-0.1\right)=-44.1$y = 4.41 / (−0.1) = −44.1. The graph has a vertical asymptote at x = 3. Its zeros are where the numerator is zero, at x = −2 and x = 2.

So the same zero in the denominator gives two different graphs. If its factor cancels with the numerator, one point is missing: a hole. If it does not, the curve runs off along an asymptote.

The graph of $y=\frac{\left(x^2-4\right)}{\left(x-3\right)}$y = (x² − 4)/(x − 3). It crosses the x-axis at (−2, 0) and (2, 0), and the dashed line x = 3 is a vertical asymptote: nothing canceled.

A hole and an asymptote together

One function can have both. Take $y=\frac{\left(x^2-x-6\right)}{\left(x^2-9\right)}$y = (x² − x − 6)/(x² − 9). Factor both parts: the numerator is (x − 3)(x + 2) and the denominator is (x − 3)(x + 3). The factor x − 3 is shared, so for $x\ne 3$x ≠ 3, $y=\frac{\left(x+2\right)}{\left(x+3\right)}$y = (x + 2)/(x + 3).

Read everything from the factors. The factor that canceled, x − 3, gives a hole at x = 3, and its height comes from the simplified form: $\frac{\left(3+2\right)}{\left(3+3\right)}=\frac{5}{6}$(3 + 2)/(3 + 3) = 5/6. The factor left in the denominator, x + 3, gives a vertical asymptote at x = −3. The factor left in the numerator, x + 2, gives a zero at x = −2.

The order matters: cancel the common factors first, then read the zeros and the asymptotes from what is left.

The graph of $y=\frac{\left(x^2-x-6\right)}{\left(x^2-9\right)}$y = (x² − x − 6)/(x² − 9). The dashed line x = −3 is a vertical asymptote, the curve crosses the x-axis at (−2, 0), and the hollow dot at $\left(3,\frac{5}{6}\right)$(3, 5/6) is a hole.

A factor that does not cancel completely

In $y=\frac{\left(x-2\right)}{{\left(x-2\right)}^2}$y = (x − 2)/(x − 2)², the numerator has one factor of x − 2 and the denominator has two. Canceling one leaves $y=\frac{1}{\left(x-2\right)}$y = 1/(x − 2), for $x\ne 2$x ≠ 2, and a factor x − 2 is still in the denominator. So x = 2 is a vertical asymptote, not a hole.

A hole needs the factor to disappear from the denominator completely. If any of it is left underneath, the curve still runs away there.

The usual mistakes

Calling x = 2 an asymptote because the denominator is zero there. The numerator is zero too, and the factor x − 2 cancels, so the values near x = 2 stay close to 4.

Drawing the point (2, 4) as a solid dot. The original function has no value at x = 2, so the point is hollow.

Reading the height of the hole from the original numerator, or multiplying the numbers. The numerator of $\frac{\left(x^2-4\right)}{\left(x-2\right)}$(x² − 4)/(x − 2) is 0 at x = 2, but the hole is at height 2 + 2 = 4, from the simplified form. For $\frac{\left(x-3\right)\left(x+1\right)}{\left(x-3\right)}$(x − 3)(x + 1)/(x − 3) the height is 3 + 1 = 4, from x + 1, not 3 × 1 = 3.

Dropping the condition $x\ne 2$x ≠ 2. y = x + 2 on its own has a value at 2; the original function does not.

Canceling terms instead of factors. In $\frac{\left(x+2\right)}{\left(x+3\right)}$(x + 2)/(x + 3) nothing cancels: striking out the x to leave $\frac{2}{3}$2/3 is not allowed, because x is a term of the sum, not a factor of the whole numerator.