of the existing five features of the following rational function (som
et a fraction or decimal then plot as close to the true location as po

Ask by Kelley Turnbull. in the United States

Mar 28,2025

Question

of the existing five features of the following rational function (som et a fraction or decimal then plot as close to the true location as po $$ f(x)=\frac{-2 x-2}{x^{2}-1} $$

UpStudy AI · Accepted Answer

1. Write the function in factored form:
   $$
   f(x)=\frac{-2x-2}{x^2-1}=\frac{-2(x+1)}{(x-1)(x+1)}
   $$
   For $$x\neq -1$$ the factor $$x+1$$ cancels, giving
   $$
   f(x)=-\frac{2}{x-1}.
   $$

2. Domain:  
   The denominator of the original function is zero when
   $$
   x^2-1=(x-1)(x+1)=0 \quad\Longrightarrow\quad x=1 \text{ or } x=-1.
   $$
   Thus, the domain is all real numbers except $$x=-1$$ and $$x=1$$.  
   $$
   \text{Domain: } (-\infty, -1)\cup(-1, 1)\cup(1, \infty).
   $$

3. Hole (Removable Discontinuity):  
   The factor $$x+1$$ cancels, indicating a removable discontinuity (hole) at $$x=-1$$.  
   To find its $$y$$-value, take the limit as $$x\to -1$$ using the simplified form:
   $$
   \lim_{x\to -1} -\frac{2}{x-1}=-\frac{2}{-1-1}=-\frac{2}{-2}=1.
   $$
   So, the hole is at $$(-1, 1)$$.

4. $$x$$-intercept:  
   The $$x$$-intercept is found by setting the numerator of the original function equal to zero:
   $$
   -2x-2=0 \quad\Longrightarrow\quad -2(x+1)=0\quad\Longrightarrow\quad x=-1.
   $$
   However, $$x=-1$$ is excluded from the domain (it is a hole), so there is no $$x$$-intercept.

5. $$y$$-intercept:  
   Substitute $$x=0$$ into the function:
   $$
   f(0)=\frac{-2(0)-2}{0^2-1}=\frac{-2}{-1}=2.
   $$
   So, the $$y$$-intercept is at $$(0, 2)$$.

6. Vertical Asymptote:  
   After canceling, the only factor causing a non-removable discontinuity is from $$x-1=0$$.  
   Thus, there is a vertical asymptote at:
   $$
   x=1.
   $$

7. Horizontal Asymptote:  
   For large values of $$|x|$$, use the simplified function:
   $$
   f(x)=-\frac{2}{x-1}.
   $$
   Taking the limit as $$x\to\pm \infty$$ gives:
   $$
   \lim_{x\to\pm\infty}-\frac{2}{x-1}=0.
   $$
   So, the horizontal asymptote is at:
   $$
   y=0.
   $$

Summary of the five features:
- **Domain:** $$x\neq-1,\,1$$.
- **Hole:** at $$(-1, 1)$$.
- **$$x$$-intercept:** None (the candidate $$x=-1$$ is removed).
- **$$y$$-intercept:** $$(0, 2)$$.
- **Vertical Asymptote:** $$x=1$$.
- **Horizontal Asymptote:** $$y=0$$.
- **Domain:** All real numbers except $$x = -1$$ and $$x = 1$$. - **Hole:** At $$(-1, 1)$$. - **$$y$$-intercept:** $$(0, 2)$$. - **Vertical Asymptote:** $$x = 1$$. - **Horizontal Asymptote:** $$y = 0$$.

of the existing five features of the following rational function (som
et a fraction or decimal then plot as close to the true location as po

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