1. **Problem Statement:** We are asked to provide 3 examples of rational functions: a monomial, a binomial, and a trinomial. Then, for each function, find the x-intercept, y-intercept, domain, range, vertical asymptote, and horizontal asymptote. --- 2. **Definition and Formula:** A rational function is a function of the form $$f(x) = \frac{P(x)}{Q(x)}$$ where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$. - **Monomial:** A polynomial with one term. - **Binomial:** A polynomial with two terms. - **Trinomial:** A polynomial with three terms. --- 3. **Examples:** **(a) Monomial Rational Function:** $$f(x) = \frac{x}{x-1}$$ **(b) Binomial Rational Function:** $$g(x) = \frac{x^2 + 1}{x - 2}$$ **(c) Trinomial Rational Function:** $$h(x) = \frac{x^2 + 3x + 2}{x^2 - 4}$$ --- 4. **Analysis for each function:** ### (a) $$f(x) = \frac{x}{x-1}$$ - **x-intercept:** Set numerator to zero: $$x=0$$ so x-intercept is $(0,0)$. - **y-intercept:** Set $x=0$: $$f(0) = \frac{0}{0-1} = 0$$ so y-intercept is $(0,0)$. - **Domain:** All real numbers except where denominator is zero: $$x-1=0 \Rightarrow x=1$$ so domain is $$(-\infty,1) \cup (1,\infty)$$. - **Range:** All real numbers except $y=1$ (horizontal asymptote). - **Vertical asymptote:** At $x=1$. - **Horizontal asymptote:** Since degrees of numerator and denominator are equal (both 1), horizontal asymptote is ratio of leading coefficients: $$y=\frac{1}{1}=1$$. --- ### (b) $$g(x) = \frac{x^2 + 1}{x - 2}$$ - **x-intercept:** Set numerator to zero: $$x^2 + 1=0$$ has no real roots, so no x-intercept. - **y-intercept:** Set $x=0$: $$g(0) = \frac{0 + 1}{0 - 2} = \frac{1}{-2} = -\frac{1}{2}$$ so y-intercept is $(0,-\frac{1}{2})$. - **Domain:** Denominator zero at $x=2$, so domain is $$(-\infty,2) \cup (2,\infty)$$. - **Vertical asymptote:** At $x=2$. - **Horizontal asymptote:** Degree numerator (2) > degree denominator (1), so no horizontal asymptote (there is an oblique asymptote). --- ### (c) $$h(x) = \frac{x^2 + 3x + 2}{x^2 - 4}$$ - Factor numerator: $$x^2 + 3x + 2 = (x+1)(x+2)$$ - Factor denominator: $$x^2 - 4 = (x-2)(x+2)$$ - **x-intercept:** Set numerator zero: $$x=-1, -2$$ so x-intercepts at $(-1,0)$ and $(-2,0)$. - **y-intercept:** Set $x=0$: $$h(0) = \frac{0 + 0 + 2}{0 - 4} = \frac{2}{-4} = -\frac{1}{2}$$ so y-intercept is $(0,-\frac{1}{2})$. - **Domain:** Denominator zero at $x=2$ and $x=-2$, so domain is $$(-\infty,-2) \cup (-2,2) \cup (2,\infty)$$. - **Vertical asymptotes:** At $x=2$ and $x=-2$. - **Horizontal asymptote:** Degrees numerator and denominator both 2, so horizontal asymptote is ratio of leading coefficients: $$y=\frac{1}{1}=1$$. - Note: The factor $(x+2)$ cancels in numerator and denominator, so at $x=-2$ there is a hole, not a vertical asymptote. --- 5. **Summary Table:** | Function | x-intercept(s) | y-intercept | Domain | Vertical Asymptote(s) | Horizontal Asymptote | |---|---|---|---|---|---| | $\frac{x}{x-1}$ | $0$ | $0$ | $x \neq 1$ | $x=1$ | $y=1$ | | $\frac{x^2+1}{x-2}$ | None | $-\frac{1}{2}$ | $x \neq 2$ | $x=2$ | None | | $\frac{x^2+3x+2}{x^2-4}$ | $-1, -2$ | $-\frac{1}{2}$ | $x \neq \pm 2$ | $x=2$ (vertical asymptote), $x=-2$ (hole) | $y=1$ | --- This completes the requested examples and their properties.