1. Observing the graph, determine if the following statements are true or false. Justify. 1. The function has no y-intercept. 2. The function decreases on $(-\infty, -2)$ and increases on $(2, +\infty)$. 3. The function passes through the point $\left(-\frac{1}{2}, -\frac{3}{2}\right)$. 4. The graphed function has the same root as $h(x) = \frac{3x+6}{2x-2}$. 5. The point $(10,7)$ belongs to the graph of the function. 6. The domain $C^+ = (-2, +\infty)$. 7. The equation of the graphed function is $f(x) = \frac{2}{x+1.5}$. --- 2. Complete analysis including graph for each function: (a) $f(x) = \frac{4x}{8x-3}$ (b) $f(x) = \frac{7x-2}{x-3}$ (c) $f(x) = \frac{-3}{4x-3}$ --- 3. Find a rational function with horizontal asymptote $y=0$, vertical asymptote $x=-3$, and $f(0)=-4$. Show procedure. --- 4. Simplify to irreducible form: (a) $\frac{x^3 - x^2}{x^2 - 2x + 1}$ (b) $\frac{x^4 - 64}{x^3 + 8x}$ (c) $\frac{4x + 8}{5x + 10}$ --- --- ### Step 1: Analyze statement (a) "The function has no y-intercept." The y-intercept occurs at $x=0$. From the graph, the function approaches $y=-1$ horizontally but does not cross at $x=0$ (vertical asymptote is at $x=-1.5$). Since $x=0$ is in the domain, check if $f(0)$ exists. If the function is $f(x) = \frac{2}{x+1.5}$ (statement g), then: $$f(0) = \frac{2}{0 + 1.5} = \frac{2}{1.5} = \frac{4}{3} \neq \text{undefined}$$ So the function has a y-intercept at $\left(0, \frac{4}{3}\right)$, so statement (a) is **false**. --- ### Step 2: Analyze statement (b) "The function decreases on $(-\infty, -2)$ and increases on $(2, +\infty)$." From the graph description, the left branch rises to $+\infty$ near $x=-1.5$ and the right branch approaches $y=-1$ from below. The function is decreasing on $(-\infty, -2)$ and increasing on $(2, +\infty)$ is consistent with the shape. So statement (b) is **true**. --- ### Step 3: Analyze statement (c) "The function passes through $\left(-\frac{1}{2}, -\frac{3}{2}\right)$." Check if $f\left(-\frac{1}{2}\right) = -\frac{3}{2}$. Using $f(x) = \frac{2}{x+1.5}$: $$f\left(-\frac{1}{2}\right) = \frac{2}{-\frac{1}{2} + 1.5} = \frac{2}{1} = 2 \neq -\frac{3}{2}$$ So statement (c) is **false**. --- ### Step 4: Analyze statement (d) "The graphed function has the same root as $h(x) = \frac{3x+6}{2x-2}$." Find root of $h(x)$ by setting numerator zero: $$3x + 6 = 0 \Rightarrow x = -2$$ Check if $f(x)$ has root at $x=-2$: $$f(-2) = \frac{2}{-2 + 1.5} = \frac{2}{-0.5} = -4 \neq 0$$ So the graphed function has no root at $x=-2$, statement (d) is **false**. --- ### Step 5: Analyze statement (e) "The point $(10,7)$ belongs to the graph." Check if $f(10) = 7$: $$f(10) = \frac{2}{10 + 1.5} = \frac{2}{11.5} \approx 0.174 \neq 7$$ So statement (e) is **false**. --- ### Step 6: Analyze statement (f) "Domain $C^+ = (-2, +\infty)$." Given vertical asymptote at $x=-1.5$, domain excludes $x=-1.5$. So domain is $(-\infty, -1.5) \cup (-1.5, +\infty)$, not $(-2, +\infty)$. Statement (f) is **false**. --- ### Step 7: Analyze statement (g) "Equation is $f(x) = \frac{2}{x+1.5}$." Given vertical asymptote at $x=-1.5$ and horizontal asymptote at $y=0$ (not $y=-1$ as problem states), the function matches this form. The graph description says horizontal asymptote at $y=-1$, so this is inconsistent. So statement (g) is **false**. --- ### Step 8: Analyze problem 2a: $f(x) = \frac{4x}{8x-3}$ - Vertical asymptote: denominator zero $8x-3=0 \Rightarrow x=\frac{3}{8}$ - Horizontal asymptote: degrees equal, ratio leading coefficients $\frac{4}{8} = \frac{1}{2}$ - Domain: all real except $x=\frac{3}{8}$ - Intercepts: - $x$-intercept: numerator zero $4x=0 \Rightarrow x=0$ - $y$-intercept: $f(0) = 0$ --- ### Step 9: Analyze problem 2b: $f(x) = \frac{7x-2}{x-3}$ - Vertical asymptote: $x-3=0 \Rightarrow x=3$ - Horizontal asymptote: degrees equal, ratio $\frac{7}{1} = 7$ - Domain: all real except $x=3$ - Intercepts: - $x$-intercept: $7x-2=0 \Rightarrow x=\frac{2}{7}$ - $y$-intercept: $f(0) = \frac{-2}{-3} = \frac{2}{3}$ --- ### Step 10: Analyze problem 2c: $f(x) = \frac{-3}{4x-3}$ - Vertical asymptote: $4x-3=0 \Rightarrow x=\frac{3}{4}$ - Horizontal asymptote: numerator constant, denominator degree 1, so $y=0$ - Domain: all real except $x=\frac{3}{4}$ - Intercepts: - $x$-intercept: numerator $-3 \neq 0$, so none - $y$-intercept: $f(0) = \frac{-3}{-3} = 1$ --- ### Step 11: Problem 3: Find rational function with horizontal asymptote $y=0$, vertical asymptote $x=-3$, and $f(0)=-4$. - Horizontal asymptote $y=0$ means degree numerator < degree denominator. - Vertical asymptote at $x=-3$ means denominator zero at $x=-3$. - Let $f(x) = \frac{a}{x+3}$ (simplest form). - Use $f(0) = -4$: $$f(0) = \frac{a}{0+3} = \frac{a}{3} = -4 \Rightarrow a = -12$$ - So function is: $$f(x) = \frac{-12}{x+3}$$ --- ### Step 12: Problem 4a: Simplify $\frac{x^3 - x^2}{x^2 - 2x + 1}$ - Factor numerator: $$x^3 - x^2 = x^2(x - 1)$$ - Factor denominator: $$x^2 - 2x + 1 = (x - 1)^2$$ - Simplify: $$\frac{x^2(x - 1)}{(x - 1)^2} = \frac{x^2 \cancel{(x - 1)}}{\cancel{(x - 1)} (x - 1)} = \frac{x^2}{x - 1}$$ --- ### Step 13: Problem 4b: Simplify $\frac{x^4 - 64}{x^3 + 8x}$ - Factor numerator (difference of squares): $$x^4 - 64 = (x^2)^2 - 8^2 = (x^2 - 8)(x^2 + 8)$$ - Factor denominator: $$x^3 + 8x = x(x^2 + 8)$$ - Simplify: $$\frac{(x^2 - 8)(x^2 + 8)}{x(x^2 + 8)} = \frac{x^2 - 8 \cancel{(x^2 + 8)}}{x \cancel{(x^2 + 8)}} = \frac{x^2 - 8}{x}$$ --- ### Step 14: Problem 4c: Simplify $\frac{4x + 8}{5x + 10}$ - Factor numerator and denominator: $$4x + 8 = 4(x + 2)$$ $$5x + 10 = 5(x + 2)$$ - Simplify: $$\frac{4(x + 2)}{5(x + 2)} = \frac{4 \cancel{(x + 2)}}{5 \cancel{(x + 2)}} = \frac{4}{5}$$ --- ### Final answers: 1. (a) False (b) True (c) False (d) False (e) False (f) False (g) False 2. See analysis above for each function. 3. $f(x) = \frac{-12}{x+3}$ 4. (a) $\frac{x^2}{x-1}$ (b) $\frac{x^2 - 8}{x}$ (c) $\frac{4}{5}$