Zeros and Vertical Asymptotes

A rational function

A rational function is one polynomial divided by another, such as $f\left(x\right)=\frac{\left(x-1\right)}{\left(x-3\right)}$f(x) = (x − 1)/(x − 3). The polynomial on top is the numerator and the one underneath is the denominator. The reciprocal graph $y=\frac{6}{x}$y = 6/x is the simplest example, with numerator 6 and denominator x.

Two questions decide most of the shape of its graph: where is the function zero, and where does it have no value at all? The numerator answers the first, and the denominator answers the second.

Zeros come from the numerator

A fraction is zero exactly when its numerator is zero and its denominator is not. A denominator can make a fraction very small, as 6 ÷ 1000 = 0.006, but never zero: for $6/d$6 / d to be 0, d × 0 would have to be 6.

The numerator x − 1 is zero at x = 1, and there the denominator is 1 − 3 = −2, not zero. So $f\left(1\right)=0/\left(-2\right)=0$f(1) = 0 / (−2) = 0, and the graph crosses the x-axis at (1, 0). That is the only zero of f.

The graph crosses the y-axis where x = 0: $f\left(0\right)=\frac{\left(-1\right)}{\left(-3\right)}=\frac{1}{3}$f(0) = (−1)/(−3) = 1/3.

Where the denominator is zero

The denominator x − 3 is zero at x = 3. There the numerator is 3 − 1 = 2, so f(3) would be 2 ÷ 0. Division by zero is undefined: no number multiplied by 0 gives 2. So f has no value at x = 3, and its graph has no point there.

Close to x = 3

Try values of x just above 3. At x = 3.1 the numerator is 2.1 and the denominator is 0.1, so $f\left(3.1\right)=2.1/0.1=21$f(3.1) = 2.1 / 0.1 = 21. At x = 3.01, $f=2.01/0.01=201$f = 2.01 / 0.01 = 201, and at x = 3.001, f = 2001.

The numerator stays close to 2 while the denominator shrinks toward 0, and dividing by a smaller and smaller number gives a larger and larger answer. So as x comes down toward 3, the graph climbs without bound. It gets closer and closer to the vertical line x = 3 and never reaches it. That line is a vertical asymptote.

The sign on each side

Now try values just below 3. At x = 2.9 the numerator is 1.9 but the denominator is 2.9 − 3 = −0.1, so $f\left(2.9\right)=1.9/\left(-0.1\right)=-19$f(2.9) = 1.9 / (−0.1) = −19. At x = 2.99, f = −199. The denominator is negative on this side, so the values are large and negative.

In symbols: as $x\to 3$x → 3 from above, $f\left(x\right)\to +\infty$f(x) → +∞, and as $x\to 3$x → 3 from below, $f\left(x\right)\to -\infty$f(x) → −∞. The curve runs up on the right of the asymptote and down on the left. To find which way each side goes, work out the sign of the numerator and of the denominator at a number just left and a number just right of the asymptote.

Values of x just below and just above 3. The closer x is to 3, the larger f(x) is in size, negative on the left and positive on the right.

The graph of $f\left(x\right)=\frac{\left(x-1\right)}{\left(x-3\right)}$f(x) = (x − 1)/(x − 3). It crosses the x-axis at (1, 0), where the numerator is 0. The dashed line x = 3 is the vertical asymptote: the curve falls on its left and rises on its right. Far out on both sides the curve levels off toward the height 1, a horizontal asymptote, which is the subject of a later lesson.

The curve is $y=\frac{1}{\left(x-2\right)}$y = 1/(x − 2), with its asymptote dashed at x = 2. Drag x to 1.9 and the value is −10; drag it to 2.1 and it is 10. At x = 2 there is no value at all.

Two asymptotes and a sign test

Take $g\left(x\right)=\frac{\left(x+2\right)}{\left(\left(x-1\right)\left(x-4\right)\right)}$g(x) = (x + 2)/((x − 1)(x − 4)). The numerator is zero at x = −2, and the denominator is not zero there, so the graph crosses the x-axis at (−2, 0). The denominator is zero at x = 1 and at x = 4, and the numerator is not zero at either, so there are two vertical asymptotes, x = 1 and x = 4.

These three values cut the number line into four parts, and g can change sign only at one of them. Test one number in each part, the sign test used for Solving a Rational Inequality. $g\left(-3\right)=\frac{\left(-1\right)}{\left(\left(-4\right)\times \left(-7\right)\right)}=-\frac{1}{28}$g(−3) = (−1)/((−4) × (−7)) = −1/28, negative. $g\left(0\right)=\frac{2}{\left(\left(-1\right)\times \left(-4\right)\right)}=\frac{1}{2}$g(0) = 2/((−1) × (−4)) = 1/2, positive. $g\left(2\right)=\frac{4}{\left(1\times \left(-2\right)\right)}=-2$g(2) = 4/(1 × (−2)) = −2, negative. $g\left(5\right)=\frac{7}{\left(4\times 1\right)}=\frac{7}{4}$g(5) = 7/(4 × 1) = 7/4, positive.

So near x = 1 the curve rises on the left and falls on the right, and near x = 4 it falls on the left and rises on the right. Between the two asymptotes the whole curve is below the x-axis.

The graph of $g\left(x\right)=\frac{\left(x+2\right)}{\left(\left(x-1\right)\left(x-4\right)\right)}$g(x) = (x + 2)/((x − 1)(x − 4)), with asymptotes dashed at x = 1 and x = 4. It crosses the x-axis at (−2, 0), passes through $\left(0,\frac{1}{2}\right)$(0, 1/2), and stays below the axis between the asymptotes, through (2, −2).

When the sign does not change

Not every vertical asymptote has the curve going opposite ways on its two sides. Take $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$h(x) = 1/(x − 2)². The denominator is zero at x = 2, so x = 2 is a vertical asymptote. But the denominator is a square, and a square is never negative.

At x = 1.9, $h=1/{\left(-0.1\right)}^2=1/0.01=100$h = 1 / (−0.1)² = 1 / 0.01 = 100. At x = 2.1, $h=1/0.1^2=100$h = 1 / 0.1² = 100 as well. On both sides of x = 2 the values are large and positive, so the curve rises on both sides. This is why each side is tested with a number rather than assumed to have the opposite sign.

The graph of $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$h(x) = 1/(x − 2)². The dashed line x = 2 is its vertical asymptote, and the curve rises on both sides of it.

When the numerator is zero too

Both rules need the other part of the fraction to be nonzero. In $\frac{\left(x-3\right)}{\left(\left(x-3\right)\left(x+1\right)\right)}$(x − 3)/((x − 3)(x + 1)), the numerator and the denominator are both zero at x = 3, and 0 ÷ 0 is neither a zero of the function nor a vertical asymptote. What happens there, a single missing point called a hole, is the subject of the next lesson.

The usual mistakes

Taking the zero of the denominator as a zero of the function. At x = 3, $\frac{\left(x-1\right)}{\left(x-3\right)}$(x − 1)/(x − 3) has no value at all; the function is zero where the numerator is zero, at x = 1.

Taking the y-intercept as the zero. $f\left(0\right)=\frac{1}{3}$f(0) = 1/3, which is where the curve crosses the y-axis; it crosses the x-axis where the numerator is zero, at x = 1.

Getting the sign of the asymptote wrong. The denominator x + 3 is zero at x = −3, so the asymptote of $\frac{1}{\left(x+3\right)}$1/(x + 3) is x = −3, not x = 3.

Assuming the curve always goes up on one side and down on the other. For $\frac{1}{{\left(x-2\right)}^2}$1/(x − 2)² it goes up on both sides. Test a number on each side.

Writing f(3) = 0 or $f\left(3\right)=\infty$f(3) = ∞. f(3) is undefined: there is no point on the graph at x = 3.