1. We are asked to determine the function rule (voorschrift) of a homographic function given certain conditions. 2. A homographic function has the general form: $$f(x) = \frac{ax + b}{cx + d}$$ where $a$, $b$, $c$, and $d$ are constants and $cx + d \neq 0$. 3. Important properties: - The zero (nulwaarde) is the value of $x$ for which $f(x) = 0$, so $ax + b = 0$. - The vertical asymptote (pool) is where the denominator is zero, so $cx + d = 0$. - The horizontal asymptote is the limit of $f(x)$ as $x \to \pm \infty$, which is $\frac{a}{c}$ if $c \neq 0$. 4. For the first problem (c): - Zero at $x = -\frac{1}{3}$ means: $$a\left(-\frac{1}{3}\right) + b = 0 \implies -\frac{a}{3} + b = 0 \implies b = \frac{a}{3}$$ - Horizontal asymptote at $y = 2$ means: $$\lim_{x \to \infty} f(x) = \frac{a}{c} = 2 \implies a = 2c$$ - The graph passes through $P(-1, -8)$, so: $$f(-1) = \frac{a(-1) + b}{c(-1) + d} = -8$$ Substitute $b = \frac{a}{3}$ and $a = 2c$: $$\frac{-a + \frac{a}{3}}{-c + d} = -8 \implies \frac{-\frac{2a}{3}}{-c + d} = -8$$ Replace $a = 2c$: $$\frac{-\frac{4c}{3}}{-c + d} = -8 \implies \frac{-\frac{4c}{3}}{-c + d} = -8$$ Multiply both sides by $-c + d$: $$-\frac{4c}{3} = -8(-c + d) = 8c - 8d$$ Rearranged: $$8c - 8d + \frac{4c}{3} = 0 \implies 8c + \frac{4c}{3} = 8d$$ Calculate left side: $$8c = \frac{24c}{3}$$ So: $$\frac{24c}{3} + \frac{4c}{3} = \frac{28c}{3} = 8d \implies d = \frac{28c}{24} = \frac{7c}{6}$$ 5. We can choose $c = 6$ to clear denominators: Then $a = 2c = 12$, $b = \frac{a}{3} = 4$, and $d = \frac{7c}{6} = 7$. 6. The function is: $$f(x) = \frac{12x + 4}{6x + 7}$$ 7. For the second problem (d): - Vertical asymptote (pool) at $x = -\frac{1}{2}$ means: $$c\left(-\frac{1}{2}\right) + d = 0 \implies -\frac{c}{2} + d = 0 \implies d = \frac{c}{2}$$ - Horizontal asymptote at $y = -5$ means: $$\lim_{x \to \infty} f(x) = \frac{a}{c} = -5 \implies a = -5c$$ - The graph passes through $P(1, 2)$, so: $$f(1) = \frac{a(1) + b}{c(1) + d} = 2$$ Substitute $a = -5c$ and $d = \frac{c}{2}$: $$\frac{-5c + b}{c + \frac{c}{2}} = 2 \implies \frac{-5c + b}{\frac{3c}{2}} = 2$$ Multiply both sides by $\frac{3c}{2}$: $$-5c + b = 2 \times \frac{3c}{2} = 3c$$ Solve for $b$: $$b = 3c + 5c = 8c$$ 8. Choose $c = 1$ for simplicity: Then $a = -5$, $b = 8$, $d = \frac{1}{2}$. 9. The function is: $$f(x) = \frac{-5x + 8}{x + \frac{1}{2}} = \frac{-5x + 8}{x + 0.5}$$ Final answers: - (c) $$f(x) = \frac{12x + 4}{6x + 7}$$ - (d) $$f(x) = \frac{-5x + 8}{x + 0.5}$$