Question

1. $x - \frac{7}{3+x}$
2. $\frac{x^2 - 5}{x^2 + 5x - 14} - \frac{x + 3}{x + 7}$
3. $\frac{10n}{n^2 - 25} - \frac{5}{n - 5}$
4. $\frac{6x - 3x^2}{x^2 - 5x + 6} \cdot \frac{x^2 - 5x - 6}{3x + 3}$

find the vertical asymptotes.

1. $f(x) = \frac{5}{(x + 4)(3x - 1)}$
2. $f(x) = \frac{x - 1}{x^2 + 5x - 6}$

determine the vertical and horizontal asymptotes and holes if there are any.

1. $g(x) = \frac{2x + 8}{3x - 12}$
2. $f(t) = \frac{x^2 + x - 6}{x^2 - x - 12}$

1. write the equation of the graph in the format $f(x) = a\left(\frac{1}{x - h}\

ight) + k$.

determine the vertical and horizontal asymptotes and holes if there are any.

1. solve. $\frac{10}{x^2 - 2x} + \frac{4}{x} = \frac{5}{x - 2}$
2. $\frac{3}{4x} + \frac{1}{8} = \frac{7}{4x}$

(image for 77): a graph with vertical asymptote $x = -3$, horizontal asymptote $y = -2$, and a point $(2, 0)$)

Explanation:

Answer:

$x = - 4,x=\frac{1}{3}$
$x = 1$
Vertical asymptote: $x = 4$, horizontal asymptote: $y=\frac{2}{3}$
Vertical asymptote: $x = 4$, horizontal asymptote: $y = 1$, hole at $x=-3$
$f(x)=2(\frac{1}{x + 3})-2$
$x = 2$ (extraneous), no solution
$x = 4$