Values a Rational Function Cannot Take

A band the curve skips

$y = \frac{{x^{2} + 1}}{x}$ is the same as $y = x + \frac{1}{x}$. 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 $y = \frac{{x^{2} + 1}}{x}$ can be solved for x.

Multiply both sides by x: yx $= x^{2} + 1$. Collect everything on one side: $x^{2} -$ 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 $b^{2} - 4ac \ge 0$. Here that is $\left(-y\right)^{2} - 4 \times 1 \times 1 \ge 0$, so $y^{2} \ge 4$.

$y^{2} \ge 4$ means $y \le -2$ or $y \ge 2$. 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 $x^{2} - 2x + 1 = \left(x - 1\right)^{2} = 0$ 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 $y = \frac{{x^{2} + 3}}{{x - 1}}$. Multiply by x − 1: yx $- y = x^{2} + 3$, so $x^{2} -$ yx + (y + 3) = 0.

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

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

When the quadratic is only linear

For $y = \frac{{x + 1}}{{x - 2}}$, the same rearrangement gives yx − 2y = x + 1, so x(y − 1) = 2y + 1 and $x = \frac{{2y + 1}}{{y - 1}}$. 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 $b^{2} - 4ac$, so $x^{2} -$ yx + k = 0 gives $y^{2} - 4k$, not $y^{2} - k$.

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

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.