Practice 1 Solutions

Describe what the value of a does to the graph of a reciprocal function.

–a reflects the graph. When $| a | > 1$ , the graph will stretch vertically, and when $0 < | a | < 1,$ the graph will compress vertically.

Describe what the value of k does to the graph of a reciprocal function.

+k will shift the graph up k-spaces.
–k will shift the graph down k-spaces.

Name any asymptotes and intercepts from the given graph. Describe the transformation from the parent graph.

$g (x) = \frac{3}{x - 1} + 3$

$a = 3, h = 1, k = 3 \begin{matrix} x ‐ intercept & y ‐ intercept \\ \begin{array}{rcl} 0 & = & \frac{3}{x - 1} + 3 \end{array} & g (0) = \frac{3}{0 - 1} + 3 \\ \begin{array}{rcl} - 3 & = & \frac{3}{x - 1} \end{array} & = \frac{3}{- 1} + 3 \\ \begin{array}{rcl} - 3 (x - 1) & = & 3 \end{array} & = - 3 + 3 \\ \begin{array}{rcl} - 3 x + 3 & = & 3 \end{array} & g (0) = 0 \\ \begin{array}{rcl} - 3 x & = & 0 \end{array} \\ x = 0 \end{matrix}$

Asymptotes: x = 1, y = 3
Intercept(s): (0, 0) 

The function is one space to the right and three spaces up from the parent graph because $h = 1$ and $k = 3 .$ It is stretched vertically by a factor of 3 because $a = 3 .$

$f (x) = \frac{1}{x + 3} - 2$

$a = 1, h = - 3, k = - 2 \begin{matrix} x ‐ intercept & y ‐ intercept \\ 0 = \frac{1}{x + 3} - 2 & f (0) = \frac{1}{0 + 3} - 2 \\ 2 = \frac{1}{x + 3} & = \frac{1}{3} - 2 \\ 2 (x + 3) = 1 & = \frac{1}{3} - \frac{6}{3} \\ 2 x + 6 = 1 & f (0) = - \frac{5}{3} \\ 2 x = - 5 \\ x = - \frac{5}{2} \end{matrix}$

Asymptotes: x = –3, y = –2
Intercept(s): $(- \frac{5}{2}, 0), (0, - \frac{5}{3})$

The function is three spaces left and two spaces down from the parent graph because $h = - 3, k = - 2,$ and $a = 1 .$

$h (x) = \frac{- 2}{x + 5} - 1$

$a = - 2, h = - 5, k = - 1 \begin{matrix} x ‐ intercept & y ‐ intercept \\ \begin{array}{rcl} 0 & = & \frac{- 2}{x + 5} - 1 \\ 1 & = & \frac{- 2}{x + 5} \\ x + 5 & = & - 2 \\ x & = & - 7 \end{array} & \begin{array}{rcl} h (0) & = & \frac{- 2}{0 + 5} - 1 \\ = & - \frac{2}{5} - 1 \\ = & - \frac{2}{5} - \frac{5}{5} \\ h (0) & = & - \frac{7}{5} \end{array} \end{matrix}$

Asymptotes: $x = - 5, y = - 1$

Intercept(s): $(- 7, 0), (0, - \frac{7}{5})$

The graph is reflected and stretched vertically by a factor of 2 because $a = - 2$ . The graph is shifted left 5 and down 1 from the parent graph because $h = - 5$ and $k = - 1 .$

$f (x) = \frac{1}{8 x} + 4$

$a = \frac{1}{8}, h = 0, k = 4 \begin{matrix} x ‐ intercept & y ‐ intercept \\ 0 = \frac{1}{8 x} + 4 & none \\ - 4 = \frac{1}{8 x} \\ - 32 x = 1 \\ x = - \frac{1}{32} \end{matrix}$

Asymptotes: x = 0, y = 4
Intercept(s): $(- \frac{1}{32}, 0)$

The graph is compressed vertically by a factor of $\frac{1}{8}$ because $a = \frac{1}{8}$ . The graph is shifted up 4 from the parent graph because $k = 4 .$

$b (x) = \frac{- 1}{6 (x + 1)}$

$a = - \frac{1}{6}, h = - 1, k = 0 \begin{matrix} x ‐ intercept & y ‐ intercept \\ none & b (0) = \frac{- 1}{6 (0 + 1)} \\ b (0) = \frac{- 1}{6} \end{matrix}$

Asymptotes: x = –1, y =0
Intercept(s): $(0, - \frac{1}{6})$

The graph is compressed vertically and reflected into quadrants II and IV by a factor of $- \frac{1}{6}$ because $a = - \frac{1}{6}$ . The graph is shifted left 1 from the parent graph because $h = - 1 .$

$j (x) = \frac{5}{x}$

$a = 5, h = 0, k = 0$

Asymptotes: x = 0, y = 0
Intercept(s): none

The graph does not shift horizontally or vertically. The graph is stretched vertically by a factor of 5 because $a = 5 .$

Name a, h, and k. Determine the domain and range of the reciprocal function.

$f (x) = \frac{- 1}{3 x} + 6$
$a = - \frac{1}{3}, h = 0, k = 6 Domain : \{x | x \in ℝ, x \neq 0\} Range : \{y | y \in ℝ, y \neq 6\}$

$h (x) = \frac{20}{x + 4} - 5$
$a = 20, h = - 4, k = - 5 Domain : \{x | x \in ℝ, x \neq - 4\} Range : \{y | y \in ℝ, y \neq - 5\}$

$n (x) = \frac{9}{x - 12} + 14$
$a = 9, h = 12, k = 14 Domain : \{x | x \in ℝ, x \neq 12\} Range : \{y | y \in ℝ, y \neq 14\}$

$a (x) = \frac{- 1}{6 (x - 2)} + 4$
$a = - \frac{1}{6}, h = 2, k = 4 Domain : \{x | x \in ℝ, x \neq 2\} Range : \{y | y \in ℝ, y \neq 4\}$

Graph the asymptotes and intercepts. Sketch the hyperbola.

Note

Finding the x- and y-intercepts is not necessary for sketching the hyperbola but may be helpful to consider the scale. The intercepts are included in the Worked Solutions for Practice 1. See More to Explore for information on how to use technology to check the graphs.

$f (x) = \frac{4}{x + 1} - 2$

$a = 4, h = - 1, k = - 2$
Asymptotes: $x = - 1, y = - 2$
The graph is stretched because a = 4.

$\begin{matrix} x ‐ intercept & y ‐ intercept \\ \begin{array}{rcl} 0 & = & \frac{4}{x + 1} - 2 \end{array} & f (0) = \frac{4}{0 + 1} - 2 \\ \begin{array}{rcl} 2 & = & \frac{4}{x + 1} \end{array} & = 4 - 2 \\ \begin{array}{rcl} 2 (x + 1) & = & 4 \end{array} & f (0) = 2 \\ \begin{array}{rcl} 2 x + 2 & = & 4 \end{array} & (0, 2) \\ \begin{array}{rcl} 2 x & = & 2 \\ x & = & 1 \end{array} \\ (1, 0) \end{matrix}$

$g (x) = \frac{- 1}{x - 4} + 1$

$a = - 1, h = 4, k = 1$
Asymptotes: $x = 4, y = 1$
The graph is reflected because a = –1.

$\begin{matrix} x ‐ intercept & y ‐ intercept \\ \begin{array}{rcl} 0 & = & \frac{- 1}{x - 4} + 1 \end{array} & g (0) = \frac{- 1}{0 - 4} + 1 \\ - 1 = \frac{- 1}{x - 4} & = \frac{1}{4} + 1 \\ - 1 (x - 4) = - 1 & g (0) = 1 \frac{1}{4} \\ x - 4 = 1 & (0, 1 \frac{1}{4}) \\ x = 5 \\ (5, 0) \end{matrix}$

$h (x) = \frac{- 1}{5 x} - 3$

$a = - \frac{1}{5}, h = 0, k = - 3$
Asymptotes: $x = 0, y = 3$
The graph is reflected and compressed vertically because $a = - \frac{1}{5}$ .

$\begin{matrix} x ‐ intercept & y ‐ intercept \\ 0 = \frac{- 1}{5 x} - 3 & none \\ 3 = \frac{- 1}{5 x} \\ 15 x = - 1 \\ x = - \frac{1}{15} \\ (- \frac{1}{15}, 0) \end{matrix}$

Note

You may wish to rescale this graph because the y-intercept is so close to the origin.

$f (x) = \frac{3}{x - 1}$

$a = 3, h = 1, k = 0$
Asymptotes: $x = 1, y = 0$
The graph is stretched vertically because a = 3.

$\begin{matrix} x ‐ intercept & y ‐ intercept \\ none & f (0) = \frac{3}{0 - 1} \\ f (0) = - 3 \\ (0, - 3) \end{matrix}$

$v (x) = \frac{1}{x} - 1$

$a = 1, h = 0, k = - 1$
Asymptotes: $x = 0, y = - 1$
The graph is the parent graph shifted down one.

$\begin{matrix} x ‐ intercept & y ‐ intercept \\ 0 = \frac{1}{x} - 1 & none \\ 1 = \frac{1}{x} \\ x = 1 \\ (1, 0) \end{matrix}$

$g (x) = \frac{- 1}{2 (x + 2)} + 5$

$a = - \frac{1}{2}, h = - 2, k = 5$
Asymptotes: $x = - 2, y = 5$

The graph is reflected and compressed vertically because $a = - \frac{1}{2}$ .
$\begin{matrix} x ‐ intercept & y ‐ intercept \\ 0 = \frac{- 1}{2 (x + 2)} + 5 & g (0) = \frac{- 1}{2 (0 + 2) + 5} \\ - 5 = \frac{- 1}{2 (x + 2)} & = - \frac{1}{4} + 5 \\ - 10 (x + 2) = - 1 & g (0) = 4 \frac{3}{4} \\ - 10 x - 20 = - 1 & (0, 4.75) \\ - 10 x = 19 \\ x = - \frac{19}{10} \\ (- 1.9, 0) \end{matrix}$