1. **Problem Statement:** Find all vertical asymptotes of the given functions by identifying where the denominator is zero and verifying by plugging values near those points. 2. **Recall:** Vertical asymptotes occur where the denominator is zero and the numerator is non-zero. --- **9.** $h(x) = \frac{-6}{9 - x}$ - Denominator zero when $9 - x = 0 \Rightarrow x = 9$ - Plug in values near 9: - At $x=8.9$, denominator $= 9 - 8.9 = 0.1$ (small positive), $h(8.9) = -6/0.1 = -60$ (large negative) - At $x=9.1$, denominator $= 9 - 9.1 = -0.1$ (small negative), $h(9.1) = -6/-0.1 = 60$ (large positive) - Since function tends to $\pm \infty$ near $x=9$, vertical asymptote at $x=9$. --- **10.** $f(x) = \frac{x + 8}{x^2 (5 - 2x)^3}$ - Denominator zero when $x^2 = 0$ or $(5 - 2x)^3 = 0$ - Solve: - $x^2 = 0 \Rightarrow x=0$ - $5 - 2x = 0 \Rightarrow x = \frac{5}{2} = 2.5$ - Plug near $x=0$: - $f(0.1) = \frac{0.1 + 8}{(0.1)^2 (5 - 0.2)^3} = \frac{8.1}{0.01 \times 4.8^3} > 0$ large - $f(-0.1) = \frac{-0.1 + 8}{0.01 \times (5 + 0.2)^3} = \frac{7.9}{0.01 \times 5.2^3} > 0$ large - Plug near $x=2.5$: - $f(2.49) = \frac{2.49 + 8}{(2.49)^2 (5 - 4.98)^3} = \frac{10.49}{(2.49)^2 (0.02)^3}$ very large positive - $f(2.51) = \frac{2.51 + 8}{(2.51)^2 (5 - 5.02)^3} = \frac{10.51}{(2.51)^2 (-0.02)^3}$ very large negative - Vertical asymptotes at $x=0$ and $x=2.5$. --- **11.** $g(t) = \frac{5t}{t (t + 7) (t - 12)}$ - Denominator zero when $t=0$, $t=-7$, or $t=12$ - Simplify numerator and denominator: - $g(t) = \frac{5t}{t (t + 7) (t - 12)} = \frac{\cancel{5t}}{\cancel{t} (t + 7) (t - 12)} = \frac{5}{(t + 7)(t - 12)}$, $t \neq 0$ - So $t=0$ is a removable discontinuity, not a vertical asymptote. - Check near $t=-7$: - $g(-7.1) = \frac{5}{(-7.1 + 7)(-7.1 - 12)} = \frac{5}{(-0.1)(-19.1)} = \frac{5}{1.91} > 0$ - $g(-6.9) = \frac{5}{(0.1)(-18.9)} = \frac{5}{-1.89} < 0$ - Check near $t=12$: - $g(11.9) = \frac{5}{(11.9 + 7)(11.9 - 12)} = \frac{5}{18.9 \times (-0.1)} < 0$ - $g(12.1) = \frac{5}{(12.1 + 7)(0.1)} > 0$ - Vertical asymptotes at $t = -7$ and $t = 12$. --- **12.** $g(z) = \frac{z^2 + 1}{(z^2 - 1)^5 (z + 15)^6}$ - Denominator zero when $z^2 - 1 = 0$ or $z + 15 = 0$ - Solve: - $z^2 - 1 = 0 \Rightarrow z = \pm 1$ - $z + 15 = 0 \Rightarrow z = -15$ - Plug near $z=1$: - $g(0.9) = \frac{0.9^2 + 1}{(0.9^2 - 1)^5 (0.9 + 15)^6} = \frac{1.81}{(-0.19)^5 (15.9)^6}$ very large negative (since $(-0.19)^5$ is negative) - $g(1.1) = \frac{1.1^2 + 1}{(1.1^2 - 1)^5 (1.1 + 15)^6} = \frac{2.21}{(0.21)^5 (16.1)^6}$ very large positive - Plug near $z=-1$: - $g(-1.1) = \frac{(-1.1)^2 + 1}{((-1.1)^2 - 1)^5 (-1.1 + 15)^6} = \frac{2.21}{(0.21)^5 (13.9)^6}$ very large positive - $g(-0.9) = \frac{(-0.9)^2 + 1}{((-0.9)^2 - 1)^5 (-0.9 + 15)^6} = \frac{1.81}{(-0.19)^5 (14.1)^6}$ very large negative - Plug near $z=-15$: - $g(-15.1) = \frac{(-15.1)^2 + 1}{((-15.1)^2 - 1)^5 (-15.1 + 15)^6} = \frac{228.01}{(227.01)^5 (-0.1)^6}$ very large positive - $g(-14.9) = \frac{(-14.9)^2 + 1}{((-14.9)^2 - 1)^5 (0.1)^6}$ very large positive - Vertical asymptotes at $z = -15, -1, 1$. --- **Final answers:** - 9: Vertical asymptote at $x=9$ - 10: Vertical asymptotes at $x=0$ and $x=2.5$ - 11: Vertical asymptotes at $t=-7$ and $t=12$ (removable discontinuity at $t=0$) - 12: Vertical asymptotes at $z=-15$, $z=-1$, and $z=1$