$r(x) = \frac{P(x)}{Q(x)}$, where P(x) and Q(x) are polynomials.

Solve $P(x) = 0$, ensuring that the solutions are not zeros of $Q(x)$.

$(-\infty, a)$, $(a, b)$, $(b, \infty)$ where a and b are critical points.

$Q(a) = 0$ and $P(a) \ne 0$ for $r(x) = \frac{P(x)}{Q(x)}$.

Analyze the sign of $r(x)$ in intervals defined by zeros and vertical asymptotes.

$ax^2 + bx + c = a(x - r_1)(x - r_2)$, where $r_1$ and $r_2$ are the roots.

Domain = {$x \in \mathbb{R} | Q(x) \ne 0$}

$\frac{P(x)}{Q(x)} = \frac{(x-a)R(x)}{(x-a)S(x)} = \frac{R(x)}{S(x)}$, if $x \ne a$.

$y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

$r(x) = \frac{(x-a)P(x)}{(x-a)}$, where the discontinuity is at $x=a$.

1. Factor: $f(x) = \frac{(x-1)(x+1)}{x+1}$. 2. Domain: $x \ne -1$. 3. Simplify: $f(x) = x-1, x \ne -1$. 4. Numerator zero: $x = 1$.

1. Critical points: $x = 2, x = -3$. 2. Intervals: $(-\infty, -3), (-3, 2), (2, \infty)$. 3. Test values. 4. Solution: $(-\infty, -3) \cup (2, \infty)$.

1. Denominator zero: $x - 5 = 0$. 2. Solve: $x = 5$. 3. Check numerator: $f(5)$ is undefined. 4. Vertical asymptote: $x = 5$.

1. Critical points: $x = -1, x = 2$. 2. Intervals: $(-\infty, -1), (-1, 2), (2, \infty)$. 3. Test values. 4. Solution: $(-1, 2)$.

1. Denominator: $x - 2$. 2. Set $x - 2 \ne 0$. 3. Solve: $x \ne 2$. 4. Domain: All real numbers except 2.

1. Factor the numerator: $x^2 - 5x + 6 = (x-2)(x-3)$. 2. Set numerator to zero: $(x-2)(x-3) = 0$. 3. Solve for x: $x = 2, 3$. 4. Check domain: $x \ne 1$. 5. Zeros: $x = 2, 3$.

1. Find critical points: $x = -4, x = 1$. 2. Create intervals: $(-\infty, -4], [-4, 1), (1, \infty)$. 3. Test values in each interval. 4. Include endpoints where function equals zero. 5. Solution: $[-4, 1)$.

1. Factor the denominator: $x^2 - 9 = (x - 3)(x + 3)$. 2. Set denominator equal to zero: $(x - 3)(x + 3) = 0$. 3. Solve for x: $x = 3, -3$. 4. Check numerator at these points. 5. Vertical asymptotes: $x = 3, x = -3$.

1. Factor the numerator: $x^2 - 16 = (x - 4)(x + 4)$. 2. Find critical points: $x = -4, 4, -2$. 3. Create intervals: $(-\infty, -4), (-4, -2), (-2, 4), (4, \infty)$. 4. Test values in each interval. 5. Solution: $(-4, -2) \cup (4, \infty)$.

1. Factor the denominator: $x^2 - 4x + 3 = (x - 1)(x - 3)$. 2. Set denominator not equal to zero: $(x - 1)(x - 3) \ne 0$. 3. Solve for x: $x \ne 1, 3$. 4. Domain: All real numbers except 1 and 3.

A function that can be expressed as the quotient of two polynomials.

The real zeros of the numerator that are also in the domain of the function.

A vertical line x = a where the function approaches infinity or negative infinity as x approaches a.

All real numbers except for the values that make the denominator equal to zero.

The process of determining the sign of a rational function on different intervals defined by its zeros and asymptotes.

The zeros of the numerator and the zeros of the denominator.

The x-values where the function equals zero (numerator zeros).

The x-values where the function is undefined (denominator zeros).

Values used to check the sign of the function in each interval created by zeros and asymptotes.

To cancel out common factors in the numerator and denominator.