1. **Problem a:** Find the domain, intercepts, range, and asymptotes of $$f(x)=\frac{2x-1}{x^2-4}$$ and sketch its graph. 2. **Domain:** The denominator cannot be zero. $$x^2-4=0 \implies x^2=4 \implies x=\pm 2$$. Thus, domain is all real numbers except $$x=2$$ and $$x=-2$$. 3. **Intercepts:** - **x-intercept:** Set numerator equal to zero. $$2x-1=0 \implies x=\frac{1}{2}$$. So, x-intercept: $$(\frac{1}{2}, 0)$$. - **y-intercept:** Set $$x=0$$. $$f(0)=\frac{2\cdot0 -1}{0-4}=\frac{-1}{-4}=\frac{1}{4}$$. So, y-intercept is $$(0, \frac{1}{4})$$. 4. **Asymptotes:** - **Vertical asymptotes:** Values where denominator is zero and numerator is nonzero. At $$x=2$$ and $$x=-2$$, vertical asymptotes occur. - **Horizontal asymptote:** Degree numerator is 1, denominator degree 2. Horizontal asymptote is $$y=0$$. 5. **Range:** Since horizontal asymptote is $$y=0$$ and vertical asymptotes at $$x=\pm 2$$, the function takes all real values except possibly $$y=0$$. Exact analysis shows range: all real $$y$$ except possibly $$0$$. ------- 6. **Problem b:** For $$f(x)=\frac{1}{2}x + 20$$, a linear function. - Domain: all real numbers. - Intercepts: Set $$f(x)=0$$ for x-intercept: $$0=\frac{1}{2}x+20 \implies x=-40$$ x-intercept: $$(-40,0)$$. y-intercept: $$f(0)=20$$. y-intercept: $$(0,20)$$. - No asymptotes. - Range: all real numbers. ------- 7. **Problem c:** $$f(x)=\frac{2x}{6} - 5 = \frac{x}{3} - 5$$. - Domain: all real numbers. - Intercepts: x-intercept: $$0=\frac{x}{3} - 5 \implies x=15$$ So, $$(15,0)$$. y-intercept: $$f(0) = -5$$ So, $$(0,-5)$$. - No asymptotes. - Range: all real numbers. ---