Values a Rational Function Cannot Take

A band the curve skips

$y=\frac{x^2+1}{x}$y = (x² + 1)/x is the same as $y=x+\frac{1}{x}$y = x + 1/x. For x > 0 the curve comes down from the asymptote x = 0, turns at (1, 2) and rises again. For x < 0 it is the same shape turned upside down, with its highest point at (−1, −2).

So no point of the curve has a height between −2 and 2. The question is how to find that band without drawing the graph.

Turn it into a quadratic in x

A height y is reached when some x gives it. Treat y as a fixed number and ask whether the equation $y=\frac{x^2+1}{x}$y = (x² + 1)/x can be solved for x.

Multiply both sides by x: yx $=x^2+1$= x² + 1. Collect everything on one side: $x^2-$x² − yx + 1 = 0. This is a quadratic in x, with a = 1, b = −y and c = 1.

The discriminant decides

The quadratic has a real root exactly when $b^2-4ac\ge 0$b² − 4ac ≥ 0. Here that is ${\left(-y\right)}^2-4\times 1\times 1\ge 0$(−y)² − 4 × 1 × 1 ≥ 0, so $y^2\ge 4$y² ≥ 4.

$y^2\ge 4$y² ≥ 4 means $y\le -2$y ≤ −2 or $y\ge 2$y ≥ 2. Those are the heights the curve reaches, and the values it cannot take are −2 < y < 2.

The edges are the turning points

At y = 2 the discriminant is exactly 0, so the quadratic $x^2-2x+1={\left(x-1\right)}^2=0$x² − 2x + 1 = (x − 1)² = 0 has one repeated root, x = 1. Only one point of the curve has height 2: the turning point (1, 2). In the same way y = −2 gives x = −1 and the turning point (−1, −2).

This is why the method also finds turning points: the curve touches the edge of its band at one point and turns back.

A second example

Take $y=\frac{x^2+3}{x-1}$y = (x² + 3)/(x − 1). Multiply by x − 1: yx $-y=x^2+3$− y = x² + 3, so $x^2-$x² − yx + (y + 3) = 0.

The discriminant is $y^2-4\left(y+3\right)=y^2-4y-12=\left(y-6\right)\left(y+2\right)$y² − 4(y + 3) = y² − 4y − 12 = (y − 6)(y + 2). It is at least 0 when $y\le -2$y ≤ −2 or $y\ge 6$y ≥ 6, so the curve cannot take any value with −2 < y < 6.

At y = 6 the quadratic is $x^2-6x+9={\left(x-3\right)}^2=0$x² − 6x + 9 = (x − 3)² = 0, so x = 3, and at y = −2 it is $x^2+2x+1={\left(x+1\right)}^2=0$x² + 2x + 1 = (x + 1)² = 0, so x = −1. The turning points are (3, 6) and (−1, −2): $f\left(3\right)=\frac{12}{2}=6$f(3) = 12/2 = 6 and $f\left(-1\right)=\frac{4}{-2}=-2$f(−1) = 4/(−2) = −2.

When the quadratic is only linear

For $y=\frac{x+1}{x-2}$y = (x + 1)/(x − 2), the same rearrangement gives yx − 2y = x + 1, so x(y − 1) = 2y + 1 and $x=\frac{2y+1}{y-1}$x = (2y + 1)/(y − 1). This works for every y except y = 1, where it divides by 0.

So this curve takes every value except 1, and y = 1 is its horizontal asymptote.

The usual slips

Dropping the 4. The discriminant is $b^2-4ac$b² − 4ac, so $x^2-$x² − yx + k = 0 gives $y^2-4k$y² − 4k, not $y^2-k$y² − k.

Answering with the values y can take. $b^2-4ac\ge 0$b² − 4ac ≥ 0 describes the heights that are reached; the values y cannot take are where $b^2-4ac<0$b² − 4ac < 0.

Using > instead of $\ge$≥. When the discriminant is 0 there is still one real x, so the edge values are reached, at the turning points.