Punto 1 (Valor 2,0). Encuentre el dominio y rango de la siguiente función, con su respectiva grafica en 2 D $$ \begin{array}{c}f(x)=\frac{x+1}{x^{2}-4 x-12} \ f(x)=\sqrt{\frac{3-x}{x+2}}\end{array} $$

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Alright, I need to find the domain and range of the given functions and provide their respective 2D graphs. The functions are: 1. $$ f(x) = \frac{x + 1}{x^2 - 4x - 12} $$ 2. $$ f(x) = \sqrt{\frac{3 - x}{x + 2}} $$ **Function 1: $$ f(x) = \frac{x + 1}{x^2 - 4x - 12} $$** First, I'll analyze the first function. The domain of a rational function is all real numbers except those that make the denominator zero. So, I need to find the values of $$ x $$ that make $$ x^2 - 4x - 12 = 0 $$. Let's solve $$ x^2 - 4x - 12 = 0 $$: Using the quadratic formula: $$ x = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} $$ So, $$ x = 6 $$ or $$ x = -2 $$. Therefore, the domain of $$ f(x) $$ is all real numbers except $$ x = 6 $$ and $$ x = -2 $$. Next, to find the range, I'll consider the behavior of the function. As $$ x $$ approaches $$ 6 $$ from the left and right, the denominator approaches zero, making $$ f(x) $$ approach positive or negative infinity. Similarly, as $$ x $$ approaches $$ -2 $$, $$ f(x) $$ also approaches infinity or negative infinity. To find the horizontal asymptote, I'll look at the degrees of the numerator and denominator. Both are degree 1, so the horizontal asymptote is the ratio of the leading coefficients, which is $$ \frac{1}{1} = 1 $$. For the range, since the function can take on all real values except possibly the horizontal asymptote, and considering the behavior around the vertical asymptotes, the range is all real numbers except $$ y = 1 $$. **Function 2: $$ f(x) = \sqrt{\frac{3 - x}{x + 2}} $$** For the second function, the domain is determined by the expression inside the square root and the denominator of the fraction. First, the denominator $$ x + 2 $$ cannot be zero, so $$ x \neq -2 $$. Next, the expression inside the square root must be non-negative: $$ \frac{3 - x}{x + 2} \geq 0 $$ To solve this inequality, I'll find the critical points where the numerator or denominator is zero: - Numerator: $$ 3 - x = 0 $$ ⇒ $$ x = 3 $$ - Denominator: $$ x + 2 = 0 $$ ⇒ $$ x = -2 $$ Now, I'll test intervals around these critical points: 1. $$ x < -2 $$: Both $$ 3 - x > 0 $$ and $$ x + 2 < 0 $$, so $$ \frac{3 - x}{x + 2} < 0 $$ ⇒ Not in domain. 2. $$ -2 < x < 3 $$: $$ 3 - x > 0 $$ and $$ x + 2 > 0 $$, so $$ \frac{3 - x}{x + 2} > 0 $$ ⇒ In domain. 3. $$ x > 3 $$: $$ 3 - x < 0 $$ and $$ x + 2 > 0 $$, so $$ \frac{3 - x}{x + 2} < 0 $$ ⇒ Not in domain. Therefore, the domain is $$ -2 < x \leq 3 $$. For the range, since the square root function outputs non-negative values, $$ f(x) \geq 0 $$. The maximum value occurs when $$ x = 3 $$, $$ f(3) = \sqrt{0} = 0 $$. As $$ x $$ approaches $$ -2 $$ from the right, $$ f(x) $$ approaches infinity. Therefore, the range is $$ [0, \infty) $$. **Graphical Representation:** For both functions, I'll sketch their graphs in 2D. 1. **First Function ($$ f(x) = \frac{x + 1}{x^2 - 4x - 12} $$):** - Vertical asymptotes at $$ x = 6 $$ and $$ x = -2 $$. - Horizontal asymptote at $$ y = 1 $$. - The graph will have two branches approaching the asymptotes and intersecting the horizontal asymptote. 2. **Second Function ($$ f(x) = \sqrt{\frac{3 - x}{x + 2}} $$):** - Domain: $$ -2 < x \leq 3 $$. - The graph will start at $$ x = -2 $$ (undefined) and increase to $$ x = 3 $$ where $$ f(x) = 0 $$, then continue upwards as $$ x $$ approaches infinity. **Función 1: $$ f(x) = \frac{x + 1}{x^2 - 4x - 12} $$** - **Dominio:** Todos los números reales excepto $$ x = 6 $$ y $$ x = -2 $$. - **Rango:** Todos los números reales excepto $$ y = 1 $$. **Función 2: $$ f(x) = \sqrt{\frac{3 - x}{x + 2}} $$** - **Dominio:** $$ -2 < x \leq 3 $$. - **Rango:** $$ [0, \infty) $$. **Gráficas en 2D:** 1. **Función 1:** - **Asíntotas Verticales:** $$ x = 6 $$ y $$ x = -2 $$. - **Asíntota Horizontal:** $$ y = 1 $$. - La gráfica tendrá dos ramas que se acercan a las asíntotas y cruzan la asíntota horizontal. 2. **Función 2:** - **Dominio:** $$ -2 < x \leq 3 $$. - **Rango:** $$ [0, \infty) $$. - La gráfica comenzará en $$ x = -2 $$ (no definida) y aumentará hasta $$ x = 3 $$ donde $$ f(x) = 0 $$, luego continuará hacia arriba a medida que $$ x $$ se acerca a infinito.

Punto 1 (Valor 2,0). Encuentre el dominio y rango de la siguiente función, con su respectiva grafica en 2 D \[ \begin{array}{c}f(x)=\frac{x+1}{x^{2}-4 x-12} \\ f(x)=\sqrt{\frac{3-x}{x+2}}\end{array} \]

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