1. **Problem a: Analyze and graph** $f(x) = \frac{2x - 1}{x^2 - 4}$. 2. **Domain:** The function is undefined where the denominator is zero: solve $x^2 - 4 = 0$ which gives $x = \pm 2$. So, $$\text{Domain} = \{ x \in \mathbb{R} \mid x \neq -2, x \neq 2 \}.$$ 3. **Vertical asymptotes:** At points where the denominator is zero but numerator is not zero. - At $x = -2$, numerator $2(-2) - 1 = -4 - 1 = -5 \neq 0$ vertical asymptote. - At $x = 2$, numerator $2(2) - 1 = 4 - 1 = 3 \neq 0$ vertical asymptote. So vertical asymptotes at $x = -2$ and $x = 2$. 4. **Horizontal asymptote:** Compare degree of numerator (1) and denominator (2). Since degree denominator > numerator, horizontal asymptote is $y = 0$. 5. **Intercepts:** - **x-intercept**: Set numerator zero: $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$. - **y-intercept**: Evaluate at $x=0$: $$f(0) = \frac{2(0) - 1}{0^2 - 4} = \frac{-1}{-4} = \frac{1}{4} = 0.25.$$ (Note: Given approx -0.25 is likely a typo; mathematical evaluation shows $0.25$.) 6. **Range:** Because of vertical asymptotes and shape, $f(x)$ takes all real values except possibly values around horizontal asymptote. More detailed analysis shows range is $\mathbb{R}$ except possibly $y=0$, which is the horizontal asymptote but function can approach zero arbitrarily closely. 7. **Summary for a:** - Domain: $x \neq \pm 2$ - Vertical asymptotes: $x = -2, 2$ - Horizontal asymptote: $y=0$ - x-intercept: $x=\frac{1}{2}$ - y-intercept: $y=0.25$ --- 8. **Problem b: Analyze and graph** $f(x) = \frac{1}{2}x + 20$. 9. **Domain:** All real numbers. 10. **Intercepts:** - y-intercept at $x=0$: $f(0)=20$. - x-intercept where $f(x)=0$: $$0 = \frac{1}{2}x + 20 \Rightarrow x = -40.$$ 11. **Range:** Since linear with slope $\frac{1}{2}$, range is all real numbers. 12. **Asymptotes:** None, linear function. --- 13. **Problem c: Analyze and graph** $f(x) = \frac{2x}{6} - 5$. 14. **Simplify:** $$f(x) = \frac{1}{3}x - 5.$$ 15. **Domain:** All real numbers. 16. **Intercepts:** - y-intercept at $x=0$: $f(0) = -5$. - x-intercept where $f(x) = 0$: $$0 = \frac{1}{3}x - 5 \Rightarrow x = 15.$$ 17. **Range:** All real numbers. 18. **Asymptotes:** None, linear function. **Final summary:** - (a) Rational function with vertical asymptotes at $x=\pm 2$, horizontal asymptote at $y=0$, domain excludes $\pm 2$, intercepts at $x=0.5$ and $y=0.25$. - (b) Linear function with slope $\frac{1}{2}$, y-intercept 20, x-intercept $-40$, domain and range all real numbers. - (c) Linear function simplified to $f(x) = \frac{1}{3}x - 5$, y-intercept $-5$, x-intercept $15$, domain and range all real numbers.