Test 5 (Lessons 9–10): Rational Equations and Functions Solutions

Solve. Check for extraneous solutions.

$\frac{1}{x} + \frac{3 x}{x + 3} = \frac{8 x^{2} - x + 6}{3 x^{2} + 9 x}$

$LCD : 3 x (x + 3); x \neq - 3, 0$

$\begin{array}{rcl} \frac{1 (3) (x + 3)}{3 x (x + 3)} + \frac{3 x (3 x)}{3 x (x + 3)} & = & \frac{8 x^{2} - x + 6}{3 x (x + 3)} \\ 3 (x + 3) + 3 x (3 x) & = & 8 x^{2} - x + 6 \\ 3 x + 9 + 9 x^{2} & = & 8 x^{2} - x + 6 \\ x^{2} + 4 x + 3 & = & 0 \\ (x + 3) (x + 1) & = & 0 \\ x & = & - 3, - 1 \end{array}$

–3 is extraneous

x = –1

$\frac{x - 4}{3 x} = \frac{x - 4}{x + 3}$

$\begin{array}{rcl} x & \neq & - 3, 0 \\ (x - 4) (x + 3) & = & 3 x (x - 4) \\ x^{2} - x - 12 & = & 3 x^{2} - 12 x \\ 0 & = & 2 x^{2} - 11 x + 12 \\ (2 x - 3) (x - 4) & = & 0 \end{array}$

$x = \frac{3}{2}, 4$

Name the asymptotes and intercepts for: $h (x) = \frac{- 1}{x + 2} + 1$

Asymptotes: x = –2, y = 1

x-int $(- 1, 0)$

x-intercept:
$\begin{array}{rcl} 0 & = & \frac{- 1}{x + 2} + 1 \\ - 1 & = & \frac{- 1}{x + 2} \\ - 1 (x + 2) & = & - 1 \\ x + 2 & = & 1 \\ x & = & - 1 \\ (- 1, 0) \end{array}$

y-int $(0, \frac{1}{2})$

y-intercept:
$h (0) = \frac{- 1}{0 + 2} + 1 h (0) = - \frac{1}{2} + 1 h (0) = \frac{1}{2} (0, \frac{1}{2})$

Name the domain and range in set notation.

domain: $\{x | x \in ℝ, x \neq - 2\}$
range: $\{h (x) | h (x) \in ℝ, h (x) \neq 1\}$

Sandy can mow the lawn in 45 minutes. When Mike helps, the job can be completed in 18 minutes. How long would it take for Mike to mow the lawn alone?

	Sandy		Mike		Together
time	45		t		18
rate	$\frac{1}{45}$	+	$\frac{1}{t}$	=	$\frac{1}{18}$

$\frac{1}{45} + \frac{1}{t} = \frac{1}{18} LCD : 90 t 90 t (\frac{1}{45} + \frac{1}{t} = \frac{1}{18}) 2 t + 90 = 5 t 90 = 3 t$

t = 30 Mike can mow the lawn in 30 minutes.

Name the domain and range for the function: $y = \frac{1}{2 (x - 1)} + 8$

$a = \frac{1}{2}, h = 1, k = 8 2 (x - 1) = 0 x \neq 1$

domain: $\{x | x \in ℝ, x \neq 1\}$
range: $\{y | y \in ℝ, y \neq 8\}$

Solve. Check for extraneous solutions.

$\frac{4}{x + 2} + \frac{4}{x - 4} = \frac{8}{x - 4}$

$L C D : (x + 2) (x - 4) : x \neq - 2, 4$

$\begin{array}{rcl} \frac{4 (x - 4)}{(x + 2) (x - 4)} + \frac{4 (x + 2)}{(x + 2) (x - 4)} & = & \frac{8 (x + 2)}{(x + 2) (x - 4)} \\ 4 x - 16 + 4 x + 8 & = & 8 x + 16 \\ 8 x - 8 & = & 8 x + 16 \\ - 8 & = & 16 \end{array}$

no solution

$\frac{2 x + 4}{2 x - 1} - 2 = \frac{17 - x}{2 x^{2} + 5 x - 3}$

$LCD : (2 x - 1) (x + 3); x \neq - 3, \frac{1}{2}$

$\begin{array}{rcl} \frac{(2 x + 4) (x + 3)}{(2 x - 1) (x + 3)} + \frac{- 2 (2 x^{2} + 5 x - 3)}{(2 x - 1) (x + 3)} & = & \frac{17 - x}{(2 x^{2} + 5 x - 3)} \\ 2 x^{2} + 10 x + 12 - 4 x^{2} - 10 x + 6 & = & 17 - x \\ - 2 x^{2} + 18 & = & 17 - x \\ 0 & = & 2 x^{2} - x - 1 \\ (2 x + 1) (x - 1) & = & 0 \end{array}$

$x = - \frac{1}{2}, 1$

For problems 9–10, use the function: $f (x) = \frac{3}{x - 2} - 1$

Name a, h, and k. Then find the asymptotes and intercepts algebraically for the function.

a = 3, h = 2, k = –1

Asymptotes:
x = 2

y = –1

x-intercept:

$\begin{array}{rcl} 0 & = & \frac{3}{x - 2} - 1 \\ 1 & = & \frac{3}{x - 2} \\ x - 2 & = & 3 \\ x & = & 5 \end{array}$

(5, 0)

y-intercept:

$f (0) = \frac{3}{0 - 2} - 1 f (0) = - \frac{3}{2} - 1 f (0) = - 2.5$

(0, –2.5)

Describe the transformation from the parent and sketch a graph. Only label the asymptotes and intercepts.

The graph will stretch vertically by a factor of 3. It will also shift 2 spaces right and 1 space down (as compared to the parent graph).