Rational Functions and Holes - AP Precalculus

Approaches $x + 1$, but undefined at $x = 1$. $\frac{(x-1)(x+1)}{x-1}$ approaches $x+1 = 2$ as $x \to 1$.

Function approaches a limit but is not defined at the hole. Limiting value exists despite the discontinuity.

$f(x) = x - 2$ for $x \neq 2$. $\frac{(x-2)^2}{x-2}$ simplifies to $(x-2)$ with hole.

They determine the horizontal asymptote when degrees are equal. When degrees are equal, their ratio gives horizontal asymptote.

Horizontal asymptote at $y = 2$. Equal degrees: $\frac{4}{2} = 2$ gives horizontal asymptote.

A point where the function is not defined due to a hole. Gap in graph that can be 'filled' by canceling factors.

Horizontal asymptote at $y = \frac{3}{2}$. Equal degrees: ratio of leading coefficients is $\frac{3}{2}$.

Hole at $x = 1$, as $(x-1)$ is a common factor. $\frac{(x-1)(x+1)}{(x-1)^2}$ has $(x-1)$ in common.

$f(0) = \frac{p(0)}{q(0)}$, if defined. Evaluate function at $x = 0$ if denominator nonzero.

Hole at $x = 5$. $\frac{(x-5)(x+5)}{x(x-5)}$ has $(x-5)$ common factor.

Occurs as $x \to \infty$, based on the degrees of $p(x)$ and $q(x)$. Determined by comparing degrees of numerator and denominator.

Describes $f(x)$ as $x \to \pm\infty$, often related to asymptotes. Behavior of function values as $x$ approaches infinity.

As $x \to \pm\infty$, $f(x) \to \pm\infty$. Numerator degree exceeds denominator, so no horizontal limit.

Removable discontinuity at $x = 3$. $\frac{(x-3)(x+3)}{x-3}$ cancels at $x = 3$.

Occurs when degree of $p(x)$ is one more than $q(x)$. Creates diagonal asymptote from polynomial long division.

Quotient is $x + 2$, remainder is 0. $x^2 + 3x + 2 = (x+1)(x+2)$ divides evenly.

Creates a hole in the graph at the factor's root. Removable discontinuity where factors cancel out.

Common factor is $x$. Factor $x$ appears in both numerator and denominator.

Dominates $q(x)$, leading to polynomial-like behavior. Function grows without bound as $x$ increases.

A vertical asymptote in a rational function. Infinite discontinuity that cannot be removed by cancellation.

All real numbers except where $q(x) = 0$. Excludes values that make the denominator zero.

Quotient is $x + 2$, remainder is 0. $x^2 + 3x + 2 = (x+1)(x+2)$ divides evenly.

$f(0) = \frac{p(0)}{q(0)}$, if defined. Evaluate function at $x = 0$ if denominator nonzero.

$x$-intercept at $x = 3$. $\frac{(x-3)(x+3)}{x+3}$ has zero at $x = 3$.

Determine where the graph crosses axes. Show where function crosses or touches coordinate axes.

Common factor is $x$. Factor $x$ appears in both numerator and denominator.

Occurs when degree of $p(x)$ is one more than $q(x)$. Creates diagonal asymptote from polynomial long division.

Slant asymptote is $y = x + 1$. Divide $x^2 + 1$ by $x - 1$ using long division.

Quotient determines slant asymptote if degrees differ by 1. Provides slant asymptote when numerator degree exceeds by one.