List the domain and the $$ x $$ - and $$ y $$-intercepts of the following function. Graph the function. Be sure to label all the asymptotes. $$ f(x)=\frac{x+6}{x^{2}-36} $$

Question

UpStudy AI · Accepted Answer

**1. Domain**

The function is  
$$
f(x)=\frac{x+6}{x^2-36}.
$$  
Since the denominator cannot be zero, set  
$$
x^2-36=0 \quad \Rightarrow \quad (x-6)(x+6)=0.
$$  
Thus, $$x=6$$ or $$x=-6$$. So, the domain is  
$$
\{x\in\mathbb{R} : x\neq -6 \text{ and } x\neq 6\}.
$$

**2. Simplification and the Removable Discontinuity**

Factor the denominator and observe the common factor:
$$
f(x)=\frac{x+6}{(x+6)(x-6)}.
$$  
For $$x\neq -6$$ (which is already excluded from the domain), we can cancel the factor $$x+6$$:
$$
f(x)=\frac{1}{x-6} \quad \text{for } x\neq -6.
$$  
The cancellation reveals a removable discontinuity (hole) at $$x=-6$$. To find the $$y$$-value of the hole, substitute $$x=-6$$ into the simplified function:
$$
y=\frac{1}{-6-6}=\frac{1}{-12}=-\frac{1}{12}.
$$
Thus, there is a hole at  
$$
\left(-6, -\frac{1}{12}\right).
$$

**3. Intercepts**

- **$$x$$-intercepts:** Set $$f(x)=0$$.
  
  For the simplified form,  
  $$
  \frac{1}{x-6}=0.
  $$
  This equation has no solution because a nonzero constant divided by an expression is never zero.  
  Although the numerator of the original expression is zero at $$x=-6$$, this point is not in the domain (it is the hole).  
  Hence, there are no $$x$$-intercepts.

- **$$y$$-intercept:** Substitute $$x=0$$:
  
  $$
  f(0)=\frac{0+6}{0^2-36}=\frac{6}{-36}=-\frac{1}{6}.
  $$
  So, the $$y$$-intercept is at  
  $$
  \left(0, -\frac{1}{6}\right).
  $$

**4. Asymptotes**

- **Vertical Asymptote:**  
  In the simplified form, $$f(x)=\frac{1}{x-6}$$, the denominator is zero when $$x-6=0$$, i.e., at $$x=6$$.  
  Therefore, there is a vertical asymptote at  
  $$
  x=6.
  $$

- **Horizontal Asymptote:**  
  As $$x\to\pm\infty$$,  
  $$
  f(x)=\frac{1}{x-6}\to 0.
  $$
  Hence, the horizontal asymptote is at  
  $$
  y=0.
  $$

**5. Graph**

The graph of $$f(x)$$ is essentially that of  
$$
y=\frac{1}{x-6},
$$
with the following features:
  
- A vertical asymptote at $$x=6$$.
- A horizontal asymptote at $$y=0$$.
- A hole at $$\left(-6, -\frac{1}{12}\right)$$.
- No $$x$$-intercepts.
- A $$y$$-intercept at $$\left(0, -\frac{1}{6}\right)$$.

The graph is a hyperbola with two branches:
  
- For $$x>6$$, as $$x$$ increases from just greater than $$6$$, $$y$$ is very large and positive, and as $$x\to\infty$$, $$y\to 0^{+}$$.
- For $$x<6$$ (but $$x\neq -6$$), as $$x$$ approaches $$6$$ from the left, $$y$$ tends to $$-\infty$$. Also, as $$x\to-\infty$$, $$y\to 0^{-}$$.

Remember to mark the hole at $$\left(-6, -\frac{1}{12}\right)$$ on the graph.

A rough sketch would show:

- A dashed vertical line at $$x=6$$ (vertical asymptote).
- A dashed horizontal line at $$y=0$$ (horizontal asymptote).
- A small open circle at $$\left(-6, -\frac{1}{12}\right)$$ indicating the hole.
- The curve approaching these asymptotes as described.
**Domain:** All real numbers except $$x = -6$$ and $$x = 6$$. **Intercepts:** - **$$y$$-intercept:** $$(0, -\frac{1}{6})$$ - **No $$x$$-intercepts** **Asymptotes:** - **Vertical Asymptote:** $$x = 6$$ - **Horizontal Asymptote:** $$y = 0$$ **Graph:** A hyperbola with a hole at $$(-6, -\frac{1}{12})$$, vertical asymptote at $$x = 6$$, and horizontal asymptote at $$y = 0$$.

List the domain and the - and -intercepts of the following function. Graph the function. Be sure to label all the asymptotes.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Pre Calculus Questions