1. **Problem Statement:** Find all vertical asymptotes of the given functions. Vertical asymptotes occur where the denominator is zero and the numerator is non-zero. 2. **Function 9:** $$h(x) = \frac{-6}{9 - x}$$ - Set denominator to zero: $$9 - x = 0$$ - Solve for $x$: $$x = 9$$ - Numerator $-6 \neq 0$, so vertical asymptote at $x = 9$. 3. **Function 10:** $$f(x) = \frac{x + 8}{x^2 (5 - 2x)^3}$$ - Set denominator to zero: $$x^2 (5 - 2x)^3 = 0$$ - This happens if $x^2 = 0$ or $(5 - 2x)^3 = 0$ - Solve $x^2 = 0$: $$x = 0$$ - Solve $(5 - 2x)^3 = 0$: $$5 - 2x = 0 \Rightarrow x = \frac{5}{2}$$ - Check numerator at $x=0$: $0 + 8 = 8 \neq 0$ - Check numerator at $x=\frac{5}{2}$: $\frac{5}{2} + 8 = \frac{5}{2} + \frac{16}{2} = \frac{21}{2} \neq 0$ - Vertical asymptotes at $x = 0$ and $x = \frac{5}{2}$. 4. **Function 11:** $$g(t) = \frac{5t}{t (t + 7) (t - 12)}$$ - Denominator: $$t (t + 7) (t - 12) = 0$$ - Solve each factor: - $t = 0$ - $t + 7 = 0 \Rightarrow t = -7$ - $t - 12 = 0 \Rightarrow t = 12$ - Check numerator at $t=0$: $5 \times 0 = 0$ (numerator zero) - Since numerator and denominator both zero at $t=0$, this is a removable discontinuity, not a vertical asymptote. - Check numerator at $t=-7$: $5 \times (-7) = -35 \neq 0$ - Check numerator at $t=12$: $5 \times 12 = 60 \neq 0$ - Vertical asymptotes at $t = -7$ and $t = 12$. 5. **Function 12:** $$g(z) = \frac{z^2 + 1}{(z^2 - 1)^5 (z + 15)^6}$$ - Denominator zero when: - $(z^2 - 1)^5 = 0 \Rightarrow z^2 - 1 = 0 \Rightarrow z = \pm 1$ - $(z + 15)^6 = 0 \Rightarrow z = -15$ - Check numerator at $z=1$: $1^2 + 1 = 2 \neq 0$ - Check numerator at $z=-1$: $(-1)^2 + 1 = 2 \neq 0$ - Check numerator at $z=-15$: $(-15)^2 + 1 = 225 + 1 = 226 \neq 0$ - Vertical asymptotes at $z = -15, -1, 1$. **Final answers:** - 9: Vertical asymptote at $x=9$ - 10: Vertical asymptotes at $x=0$ and $x=\frac{5}{2}$ - 11: Vertical asymptotes at $t=-7$ and $t=12$ - 12: Vertical asymptotes at $z=-15$, $z=-1$, and $z=1$