Rational Functions and Zeros - AP Precalculus
Card 1 of 30
Find the domain of $f(x) = \frac{4x + 1}{x^2 - 9}$.
Find the domain of $f(x) = \frac{4x + 1}{x^2 - 9}$.
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Domain: $x \neq 3, x \neq -3$. Set $x^2 - 9 = 0$ to find where the function is undefined.
Domain: $x \neq 3, x \neq -3$. Set $x^2 - 9 = 0$ to find where the function is undefined.
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How do you determine a hole in a rational function?
How do you determine a hole in a rational function?
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If $p(x)$ and $q(x)$ share a common factor. Common factors can be cancelled, creating a removable discontinuity.
If $p(x)$ and $q(x)$ share a common factor. Common factors can be cancelled, creating a removable discontinuity.
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What is the horizontal asymptote for $f(x) = \frac{x + 1}{x + 2}$?
What is the horizontal asymptote for $f(x) = \frac{x + 1}{x + 2}$?
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The horizontal asymptote is $y = 1$. When degrees are equal, the horizontal asymptote is $\frac{1}{1} = 1$.
The horizontal asymptote is $y = 1$. When degrees are equal, the horizontal asymptote is $\frac{1}{1} = 1$.
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What is the slant asymptote of $f(x) = \frac{x^2 + 1}{x}$?
What is the slant asymptote of $f(x) = \frac{x^2 + 1}{x}$?
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The slant asymptote is $y = x$. Divide $x^2 + 1$ by $x$ using polynomial long division.
The slant asymptote is $y = x$. Divide $x^2 + 1$ by $x$ using polynomial long division.
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Describe when a rational function has no horizontal asymptote.
Describe when a rational function has no horizontal asymptote.
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When the degree of $p(x)$ is greater than $q(x)$. When numerator degree exceeds denominator degree, no horizontal asymptote exists.
When the degree of $p(x)$ is greater than $q(x)$. When numerator degree exceeds denominator degree, no horizontal asymptote exists.
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Find the zeros of $f(x) = \frac{3x^2 - 12}{x + 2}$.
Find the zeros of $f(x) = \frac{3x^2 - 12}{x + 2}$.
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The zeros are $x = 2$ and $x = -2$. Factor the numerator: $3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)$.
The zeros are $x = 2$ and $x = -2$. Factor the numerator: $3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)$.
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Identify a removable discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$.
Identify a removable discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$.
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Removable discontinuity at $x = 1$. The factor $(x-1)$ cancels from numerator and denominator.
Removable discontinuity at $x = 1$. The factor $(x-1)$ cancels from numerator and denominator.
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State the horizontal asymptote of $f(x) = \frac{2x^2 + 3}{x^2 + 1}$.
State the horizontal asymptote of $f(x) = \frac{2x^2 + 3}{x^2 + 1}$.
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The horizontal asymptote is $y = 2$. Equal degrees give horizontal asymptote $y = \frac{2}{1} = 2$.
The horizontal asymptote is $y = 2$. Equal degrees give horizontal asymptote $y = \frac{2}{1} = 2$.
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What is the degree of the polynomial $p(x) = 3x^4 - 2x^2 + 1$?
What is the degree of the polynomial $p(x) = 3x^4 - 2x^2 + 1$?
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The degree is 4. The highest power term determines the degree of a polynomial.
The degree is 4. The highest power term determines the degree of a polynomial.
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What is the horizontal asymptote for $f(x) = \frac{2x^2}{x^2 + 1}$?
What is the horizontal asymptote for $f(x) = \frac{2x^2}{x^2 + 1}$?
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The horizontal asymptote is $y = 2$. Equal degrees give horizontal asymptote $y = \frac{2}{1} = 2$.
The horizontal asymptote is $y = 2$. Equal degrees give horizontal asymptote $y = \frac{2}{1} = 2$.
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Identify a removable discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$.
Identify a removable discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$.
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Removable discontinuity at $x = 1$. The factor $(x-1)$ cancels from numerator and denominator.
Removable discontinuity at $x = 1$. The factor $(x-1)$ cancels from numerator and denominator.
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Find the domain of $f(x) = \frac{4x + 1}{x^2 - 9}$.
Find the domain of $f(x) = \frac{4x + 1}{x^2 - 9}$.
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Domain: $x \neq 3, x \neq -3$. Set $x^2 - 9 = 0$ to find where the function is undefined.
Domain: $x \neq 3, x \neq -3$. Set $x^2 - 9 = 0$ to find where the function is undefined.
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Find the zero of $f(x) = \frac{2x - 4}{x + 3}$.
Find the zero of $f(x) = \frac{2x - 4}{x + 3}$.
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The zero is $x = 2$. Set $2x - 4 = 0$ to find where the numerator equals zero.
The zero is $x = 2$. Set $2x - 4 = 0$ to find where the numerator equals zero.
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Find the zeros of $f(x) = \frac{3x^2 - 12}{x + 2}$.
Find the zeros of $f(x) = \frac{3x^2 - 12}{x + 2}$.
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The zeros are $x = 2$ and $x = -2$. Factor the numerator: $3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)$.
The zeros are $x = 2$ and $x = -2$. Factor the numerator: $3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)$.
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State the condition for a vertical asymptote in a rational function.
State the condition for a vertical asymptote in a rational function.
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Occurs where $q(x) = 0$ and $p(x) \neq 0$. The denominator is zero while the numerator is non-zero.
Occurs where $q(x) = 0$ and $p(x) \neq 0$. The denominator is zero while the numerator is non-zero.
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What is the behavior of $f(x) = \frac{1}{x}$ as $x \to 0$?
What is the behavior of $f(x) = \frac{1}{x}$ as $x \to 0$?
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As $x \to 0$, $f(x) \to \pm \infty$. The function has a vertical asymptote at $x = 0$.
As $x \to 0$, $f(x) \to \pm \infty$. The function has a vertical asymptote at $x = 0$.
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Identify the horizontal asymptote of $f(x) = \frac{x^2 + 3}{x^2 - 1}$.
Identify the horizontal asymptote of $f(x) = \frac{x^2 + 3}{x^2 - 1}$.
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The horizontal asymptote is $y = 1$. When degrees are equal, the horizontal asymptote equals the ratio of leading coefficients.
The horizontal asymptote is $y = 1$. When degrees are equal, the horizontal asymptote equals the ratio of leading coefficients.
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What is the behavior of $f(x) = \frac{1}{x}$ as $x \to 0$?
What is the behavior of $f(x) = \frac{1}{x}$ as $x \to 0$?
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As $x \to 0$, $f(x) \to \pm \infty$. The function has a vertical asymptote at $x = 0$.
As $x \to 0$, $f(x) \to \pm \infty$. The function has a vertical asymptote at $x = 0$.
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For $f(x) = \frac{x^2 - 4}{x - 2}$, what is the discontinuity?
For $f(x) = \frac{x^2 - 4}{x - 2}$, what is the discontinuity?
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Discontinuity (hole) at $x = 2$. Factor and cancel: $(x-2)$ appears in both numerator and denominator.
Discontinuity (hole) at $x = 2$. Factor and cancel: $(x-2)$ appears in both numerator and denominator.
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What happens to $f(x) = \frac{1}{x-3}$ as $x \to 3$?
What happens to $f(x) = \frac{1}{x-3}$ as $x \to 3$?
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$f(x) \to \pm \infty$, vertical asymptote at $x = 3$. As $x$ approaches 3, the denominator approaches 0 while numerator stays 1.
$f(x) \to \pm \infty$, vertical asymptote at $x = 3$. As $x$ approaches 3, the denominator approaches 0 while numerator stays 1.
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Identify the horizontal asymptote of $f(x) = \frac{5x}{x^2 + 4}$.
Identify the horizontal asymptote of $f(x) = \frac{5x}{x^2 + 4}$.
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The horizontal asymptote is $y = 0$. Numerator degree is less than denominator degree, so $y = 0$.
The horizontal asymptote is $y = 0$. Numerator degree is less than denominator degree, so $y = 0$.
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What is the intercept of $f(x) = \frac{3}{x-2}$?
What is the intercept of $f(x) = \frac{3}{x-2}$?
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Vertical intercept at $(0, -\frac{3}{2})$. Substitute $x = 0$ into the function to find the y-intercept.
Vertical intercept at $(0, -\frac{3}{2})$. Substitute $x = 0$ into the function to find the y-intercept.
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What is the degree of $q(x) = x^3 - 7x$?
What is the degree of $q(x) = x^3 - 7x$?
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The degree is 3. The highest power term $x^3$ has degree 3.
The degree is 3. The highest power term $x^3$ has degree 3.
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What is the general form of a rational function?
What is the general form of a rational function?
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$f(x) = \frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials. This defines a rational function as a ratio of two polynomial functions.
$f(x) = \frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials. This defines a rational function as a ratio of two polynomial functions.
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How do you determine a hole in a rational function?
How do you determine a hole in a rational function?
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If $p(x)$ and $q(x)$ share a common factor. Common factors can be cancelled, creating a removable discontinuity.
If $p(x)$ and $q(x)$ share a common factor. Common factors can be cancelled, creating a removable discontinuity.
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Identify the horizontal asymptote of $f(x) = \frac{x^2 + 3}{x^2 - 1}$.
Identify the horizontal asymptote of $f(x) = \frac{x^2 + 3}{x^2 - 1}$.
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The horizontal asymptote is $y = 1$. When degrees are equal, the horizontal asymptote equals the ratio of leading coefficients.
The horizontal asymptote is $y = 1$. When degrees are equal, the horizontal asymptote equals the ratio of leading coefficients.
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What is a zero of a rational function?
What is a zero of a rational function?
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A value of $x$ for which $f(x) = 0$. A zero occurs when the numerator equals zero but the denominator doesn't.
A value of $x$ for which $f(x) = 0$. A zero occurs when the numerator equals zero but the denominator doesn't.
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State the condition for a vertical asymptote in a rational function.
State the condition for a vertical asymptote in a rational function.
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Occurs where $q(x) = 0$ and $p(x) \neq 0$. The denominator is zero while the numerator is non-zero.
Occurs where $q(x) = 0$ and $p(x) \neq 0$. The denominator is zero while the numerator is non-zero.
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Find the zero of $f(x) = \frac{2x - 4}{x + 3}$.
Find the zero of $f(x) = \frac{2x - 4}{x + 3}$.
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The zero is $x = 2$. Set $2x - 4 = 0$ to find where the numerator equals zero.
The zero is $x = 2$. Set $2x - 4 = 0$ to find where the numerator equals zero.
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Which term denotes values not in the domain of a rational function?
Which term denotes values not in the domain of a rational function?
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These are the poles or vertical asymptotes. Points where the denominator equals zero make the function undefined.
These are the poles or vertical asymptotes. Points where the denominator equals zero make the function undefined.
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