Graphs of Rational Functions: Additional Practice (Set 2)

Factor everything. The numerator is $x(x-1)(x+1)$ and the denominator is $(x-1)(x+1)$, so the domain excludes $x=1$ and $x=-1$.

$f(x) = \frac{x(x-1)(x+1)}{(x-1)(x+1)} = x \qquad (x \neq 1,\; x \neq -1)$

Both denominator factors cancel completely, so neither excluded input gives a vertical asymptote. The graph is the line $y=x$ with two points punched out, at the heights the line reaches there: $(1,1)$ and $(-1,-1)$.

Factor and cancel first, because a zero of the original numerator counts only if it is still in the domain.

$f(x) = \frac{(2x-1)(x+2)}{x(x+2)} = \frac{2x-1}{x} \qquad (x \neq 0,\; x \neq -2)$

The value $x=-2$ is a zero of the original numerator, but it is excluded from the domain, so it is a hole rather than an intercept. The surviving numerator gives the intercept.

$2x - 1 = 0 \;\Longrightarrow\; x = \frac12$

So the $x$-intercept is $\left(\frac12, 0\right)$.

Divide. Dividing $x^3+8$ by $x^2+4$ gives quotient $x$ with remainder $8-4x$, since $x(x^2+4) = x^3+4x$ and $(x^3+8) - (x^3+4x) = 8-4x$.

$f(x) = x + \frac{8-4x}{x^2+4}$

So the oblique asymptote is $y=x$, and the curve meets it exactly where the correction term is zero, which happens where the remainder is zero.

$8 - 4x = 0 \;\Longrightarrow\; x = 2$

Check it: $f(2) = \frac{8+8}{4+4} = 2$, and the line $y=x$ is also at height $2$ there, so the crossing point is $(2,2)$. A constant remainder could never vanish, but this remainder is linear, so it can.

A horizontal asymptote $y=0$ needs the denominator's degree to beat the numerator's, and a vertical asymptote at $x=5$ needs the factor $(x-5)$ downstairs, so the function looks like $\frac{c}{x-5}$ for some nonzero constant $c$. The point $(0,1)$ pins down $c$.

$\frac{c}{0-5} = 1 \;\Longrightarrow\; c = -5$

So $f(x) = \frac{-5}{x-5}$, and indeed $f(0) = \frac{-5}{-5} = 1$. The second choice passes through $(0,-1)$, the third has the horizontal asymptote $y=1$ (its degrees tie), and the fourth has its vertical asymptote at $x=-5$.

Factor both parts: the numerator is $(2x-1)(x+3)$ and the denominator is $(x+3)(x-2)$, so the domain excludes $-3$ and $2$.

$f(x) = \frac{(2x-1)(x+3)}{(x+3)(x-2)} = \frac{2x-1}{x-2} \qquad (x \neq -3,\; x \neq 2)$

The factor $(x+3)$ cancels completely, so $x=-3$ is a hole, at height $\frac{2(-3)-1}{-3-2} = \frac{-7}{-5} = \frac 75$. The factor $(x-2)$ survives, so $x=2$ is a vertical asymptote. The degrees tie in the original ($2$ and $2$), so the horizontal asymptote is the ratio of leading coefficients, $\frac 21 = 2$.

From the division, $f(x) = 2x + 1 + \frac{-2x-2}{x^2+1}$, so the curve sits exactly one correction term away from the line $y = 2x+1$. It meets the line where that correction is zero, which is where the remainder is zero.

$-2x - 2 = 0 \;\Longrightarrow\; x = -1$

Check it: $f(-1) = \frac{-2+1-1}{1+1} = \frac{-2}{2} = -1$, and the line $y=2x+1$ is also at $-1$ there, so the crossing point is $(-1,-1)$. The point $(0,1)$ lies on the line but not on the curve, since $f(0) = -1$.

The degrees tie and both leading coefficients are $1$, so the horizontal asymptote is $y=1$. Set the function equal to $1$ and clear the denominator, which is never zero.

$x^2 - 3x = x^2 + 1 \;\Longrightarrow\; -3x = 1 \;\Longrightarrow\; x = -\frac{1}{3}$

The quadratic terms cancel, which is what always happens when the degrees tie, so what is left is linear and has exactly one solution. Check it: at $x=-\frac13$ the numerator is $\frac19 + 1 = \frac{10}{9}$ and the denominator is $\frac19 + 1 = \frac{10}{9}$, so $f\left(-\frac13\right) = 1$.

Factor both parts. The numerator is $3x(x+2)$ and the denominator is $(x-2)(x+2)$, so the domain excludes $2$ and $-2$.

$f(x) = \frac{3x(x+2)}{(x-2)(x+2)} = \frac{3x}{x-2} \qquad (x \neq 2,\; x \neq -2)$

The factor $(x+2)$ cancels completely, so $x=-2$ is a hole (at height $\frac{3(-2)}{-2-2} = \frac32$), not an asymptote. Only $x=2$ is a vertical asymptote. The degrees tie in the original, so the horizontal asymptote is the ratio of the leading coefficients, $\frac 31 = 3$.

Compare the degrees: $4$ on top against $2$ on the bottom. The top outgrows the bottom, so there is no horizontal asymptote, and the gap is two rather than one, so the quotient of the division is not linear and there is no oblique asymptote either.

$\frac{x^4+1}{x^2-1} = x^2 + 1 + \frac{2}{x^2-1}$

(Check by multiplying back: $(x^2+1)(x^2-1) + 2 = x^4 - 1 + 2 = x^4+1$.) The curve does follow something far out, but it is the parabola $y = x^2+1$, not a straight line.

The degrees tie, so the horizontal asymptote is the ratio of the leading coefficients, $y = 4$. Set the function equal to $4$ and clear the denominator (its zeros, $x = \pm 1$, are not in the domain, so watch for them at the end).

$4x^2 + x = 4(x^2-1) \;\Longrightarrow\; 4x^2 + x = 4x^2 - 4 \;\Longrightarrow\; x = -4$

The quadratic terms cancel, leaving a linear equation, and $x=-4$ is safely in the domain. Check it: $f(-4) = \frac{64-4}{16-1} = \frac{60}{15} = 4$, so the crossing point is $(-4,4)$. The origin is on the graph, but it is not on the line $y=4$.

A hole needs the denominator's factor $(x-3)$ to cancel, so $(x-3)$ must divide the numerator. By the factor theorem, that means the numerator must be zero at $x=3$.

$3^2 + c = 0 \;\Longrightarrow\; c = -9$

Check it: with $c=-9$ the function is $\frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3$ for $x \neq 3$, the line $y=x+3$ with a hole at $(3,6)$. For any other $c$ the numerator is nonzero at $x=3$, so the factor stays in the denominator and $x=3$ is a vertical asymptote.

Factor both parts. The numerator is $(x-2)(x+2)$ and the denominator is $(x-2)(x-3)$, so the domain excludes $2$ and $3$.

$f(x) = \frac{(x-2)(x+2)}{(x-2)(x-3)} = \frac{x+2}{x-3} \qquad (x \neq 2,\; x \neq 3)$

The factor $(x-2)$ cancels completely, so $x=2$ is a hole, at height $\frac{2+2}{2-3} = -4$. The factor $(x-3)$ survives, so $x=3$ is a vertical asymptote. The degrees tie, so the horizontal asymptote is $y=1$, and the surviving numerator is zero at $x=-2$, which is in the domain, so the $x$-intercept is $(-2,0)$.