$r(x) = \frac{P(x)}{Q(x)}$, 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 $x=a$.

They are the x-intercepts of the graph, where the function's value is zero, provided they are in the domain.

They indicate where the function is undefined, leading to vertical asymptotes or holes in the graph.

Zeros and vertical asymptotes are the points where a rational function can change its sign (positive to negative or vice versa).

It involves testing values within each interval created by zeros and asymptotes to determine the function's sign in that interval.

To ensure that the zeros found are actually part of the function and not points where the function is undefined.

The function approaches infinity (or negative infinity) as x gets closer to the asymptote.

Simplification helps to identify and remove any removable discontinuities (holes) and avoids incorrect zero identification.

By using interval analysis, testing values in each interval defined by zeros and asymptotes.

Factoring helps to identify zeros and simplify the function, making it easier to analyze its behavior.

Rational functions are built from polynomial functions, and understanding polynomial zeros and behavior is essential for analyzing rational functions.

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.