- Decide whether a horizontal translation changes the input or the output of the function. - For a right shift, think about which replacement for \(x\) makes the original graph feature at \(x=0\) occur at \(x=4\).

1. A translation 4 units to the right replaces \(x\) by \(x-4\). 2. Therefore, \(g(x)=\frac{1}{x-4}\).

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

- Compare where the two graphs have their vertical gaps. - Compare the horizontal levels the branches approach. - Use those changes to describe one horizontal and one vertical shift.

1. The dashed parent graph has asymptotes \(x=0\) and \(y=0\). 2. The solid graph has asymptotes \(x=2\) and \(y=3\). 3. The solid graph is shifted \(2\) units right and \(3\) units up.

Right \(2\) units and up \(3\) units.

- What is the value of \(x\) where a graph crosses the \(y\)-axis? - What is the value of \(y\) where a graph crosses the \(x\)-axis? - Substitute those values into the equation one at a time.

1. Find the \(y\)-intercept by substituting \(x = 0\): \(y = \frac{12}{0 + 2} + 3 = 6 + 3 = 9\). Therefore, \(b = 9\). 2. Find the \(x\)-intercept by setting \(y = 0\): \(0 = \frac{12}{x + 2} + 3\). 3. Then \(-3 = \frac{12}{x + 2}\), so \(-3(x + 2) = 12\). Thus, \(x + 2 = -4\) and \(x = -6\). Therefore, \(a = -6\). 4. The correct choice is c).

c) \(a = -6\), \(b = 9\)

- At an \(x\)-intercept, the function value is \(0\). - At a \(y\)-intercept, the input is \(0\). - When solving the equation in part c), first isolate the fraction.

1. To find the \(x\)-intercept, set \(h(x) = 0\): \(0 = \frac{12}{x + 4} - 3\). Then \(3 = \frac{12}{x + 4}\), so \(3(x + 4) = 12\). Thus, \(x + 4 = 4\) and \(x = 0\). The \(x\)-intercept is \((0, 0)\). 2. To find the \(y\)-intercept, substitute \(x = 0\): \(h(0) = \frac{12}{0 + 4} - 3 = 3 - 3 = 0\). The \(y\)-intercept is also \((0, 0)\). 3. To solve \(h(x) = 1\), write \(1 = \frac{12}{x + 4} - 3\). Then \(4 = \frac{12}{x + 4}\), so \(4(x + 4) = 12\). Thus, \(x + 4 = 3\) and \(x = -1\).

a) The \(x\)-intercept is \((0, 0)\). b) The \(y\)-intercept is \((0, 0)\). c) \(x = -1\)

- For the vertical asymptote, look for the x-value that the rational branches approach while their y-values grow in magnitude. - Intersection points lie on both graphs. - For each specified x-value, compare which graph is vertically higher. - An intersection means the two functions have the same output at one input.

1. The branches of \(f\) approach the y-axis without reaching it, so the vertical asymptote is \(x=0\). 2. The two graphs intersect at \((-1,-2)\) and \((2,1)\). 3. At \(x=-2\), the graph of \(f\) is above the graph of \(g\), so \(f(-2)>g(-2)\). At \(x=1\), \(f(1)>g(1)\). At \(x=4\), \(f(4)<g(4)\). 4. At an intersection, both functions have the same y-value at the same x-value. Therefore, that x-coordinate satisfies \(f(x)=g(x)\).

a) \(x=0\) b) \((-1,-2)\) and \((2,1)\) c) \(f(-2)>g(-2)\); \(f(1)>g(1)\); \(f(4)<g(4)\) d) At an intersection, the functions have equal outputs for the same input, so its x-coordinate satisfies \(f(x)=g(x)\).

- When does the denominator equal \(0\)? - What input gives a point on the y-axis? - What equation must be solved to find a zero? - In the form \(\frac{k}{x - x_0} + y_0\), how do \(x_0\) and \(y_0\) affect the graph?

1. The denominator is \(0\) when \(x + 1 = 0\), so the vertical asymptote is \(x = -1\). 2. Set \(x = 0\): \(g(0) = \frac{2}{1} - 4 = -2\). The y-intercept is \((0, -2)\). 3. Solve \(0 = \frac{2}{x + 1} - 4\). Then \(4 = \frac{2}{x + 1}\), so \(4(x + 1) = 2\). Therefore, \(x + 1 = 0.5\) and \(x = -0.5\). 4. Replacing \(x\) with \(x + 1\) translates the graph \(1\) unit left, and subtracting \(4\) translates it \(4\) units down.

a) \(x = -1\) b) \((0, -2)\) c) \(x = -0.5\) d) Translate the graph \(1\) unit left and \(4\) units down.

- How does a horizontal or vertical translation change a function rule? - Which input makes the denominator equal to \(0\)? - Which line does the graph approach as \(|x|\) becomes large? - What input gives the y-intercept?

1. Translating \(3\) units left replaces \(x\) with \(x + 3\), and translating \(2\) units up adds \(2\). Thus, \(g(x) = \frac{1}{x + 3} + 2\). 2. The denominator is \(0\) at \(x = -3\), so the vertical asymptote is \(x = -3\). The vertical translation gives the horizontal asymptote \(y = 2\). 3. Set \(x = 0\): \(g(0) = \frac{1}{3} + 2 = \frac{7}{3}\). The y-intercept is \(\left(0, \frac{7}{3}\right)\).

a) \(g(x) = \frac{1}{x + 3} + 2\) b) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = 2\) c) \(\left(0, \frac{7}{3}\right)\)

- Determine which input makes each denominator zero. - Consider what happens to each output as \(|x|\) becomes large. - Compare the absolute values of the numerators for the same input. - Substitute \(x = 1.5\) directly.

1. Both denominators are zero at \(x = 0\), so each domain is \(\mathbb{R} \setminus \{0\}\). Both graphs have the vertical asymptote \(x = 0\) and the horizontal asymptote \(y = 0\). 2. For the same nonzero \(x\), \(|g(x)| = \frac{0.6}{|x|}\) is less than \(|f(x)| = \frac{3}{|x|}\). The points have the same \(x\)-coordinate, so the point on \(g\) lies closer to the origin. 3. Evaluate: \(f(1.5) = \frac{3}{1.5} = 2\), and \(g(1.5) = \frac{0.6}{1.5} = 0.4\).

a) Both domains are \(\mathbb{R} \setminus \{0\}\), and both graphs have asymptotes \(x = 0\) and \(y = 0\). b) For the same nonzero \(x\), the point on \(g\) lies closer to the origin because \(0.6 < 3\), so its \(y\)-coordinate has smaller absolute value. c) \(f(1.5) = 2\) and \(g(1.5) = 0.4\)

- The constant outside the fraction gives the horizontal asymptote. - Compare the two constant terms to determine the translation. - At an \(x\)-intercept, the function value is \(0\).

1. For a function of the form \(\frac{a}{x} + c\), the horizontal asymptote is \(y = c\). Therefore, the asymptotes are \(y = 4\) for \(h\) and \(y = -1\) for \(k\). 2. The change in the vertical shift is \(-1 - 4 = -5\). Therefore, translating the graph of \(h\) down \(5\) units produces the graph of \(k\). 3. Set \(h(x) = 0\): \(0 = \frac{2}{x} + 4\). Then \(-4 = \frac{2}{x}\), so \(x = -\frac{1}{2}\). The \(x\)-intercept is \(\left(-\frac{1}{2}, 0\right)\).

a) The horizontal asymptotes are \(y = 4\) for \(h\) and \(y = -1\) for \(k\). b) Translate the graph of \(h\) down \(5\) units. c) The \(x\)-intercept of \(h\) is \(\left(-\frac{1}{2}, 0\right)\).

- What relationship must a point satisfy to lie on a function graph? - Evaluate the function at each given \(x\)-coordinate. - Compare the given \(y\)-coordinate with the calculated function value.

1. For \(P_1\), \(g(2) = \frac{3}{2 + 1} + 2 = 3\). This equals the given \(y\)-coordinate, so \(P_1\) lies on the graph. 2. For \(P_2\), \(g(-2) = \frac{3}{-2 + 1} + 2 = -1\). Since the point has \(y = 1\) and \(1 > -1\), \(P_2\) lies above the graph. 3. For \(P_3\), \(g(0) = \frac{3}{0 + 1} + 2 = 5\). This equals the given \(y\)-coordinate, so \(P_3\) lies on the graph.

\(P_1(2, 3)\) and \(P_3(0, 5)\) lie on the graph. Point \(P_2(-2, 1)\) lies above the graph.

- Test one positive and one negative input for each function. - A negative numerator reflects the graph across the \(x\)-axis. - Compare the absolute output values for the same nonzero input.

1. For \(f\), positive inputs give positive outputs and negative inputs give negative outputs. Therefore, its branches lie in Quadrants I and III. 2. For \(g\), positive inputs give negative outputs and negative inputs give positive outputs. Therefore, its branches lie in Quadrants II and IV. 3. For the same nonzero \(x\), \(|f(x)| = \frac{1}{|x|}\) and \(|g(x)| = \frac{3}{|x|}\). The points have the same \(x\)-coordinate, but the point on \(f\) has the smaller absolute \(y\)-coordinate. Therefore, the point on \(f\) is closer to the origin.

a) The branches of \(f\) lie in Quadrants I and III. The branches of \(g\) lie in Quadrants II and IV. b) For the same nonzero \(x\), the point on \(f\) lies closer to the origin because \(\frac{1}{|x|} < \frac{3}{|x|}\).

- Compare the outputs of \(f\) and \(g\) for the same nonzero input. - Determine how a negative sign in front of the entire function changes the graph. - Test one positive and one negative input to identify the quadrants.

1. For the same nonzero input, \(|g(x)| = \frac{5}{|x|}\) is five times \(|f(x)| = \frac{1}{|x|}\). Thus, the graph of \(g\) is a vertical stretch of the graph of \(f\), and corresponding points on \(g\) lie farther from the origin. 2. The graph of \(h(x) = -\frac{1}{x}\) is a reflection of the graph of \(f\) across the \(x\)-axis. It can also be viewed as a reflection across the \(y\)-axis. 3. The branches of \(f\) lie in Quadrants I and III, while the branches of \(h\) lie in Quadrants II and IV.

a) The graph of \(g\) is a vertical stretch of the graph of \(f\) by a factor of \(5\); corresponding points lie farther from the origin. b) The graph of \(h\) is a reflection of \(f\). The branches of \(f\) lie in Quadrants I and III, and the branches of \(h\) lie in Quadrants II and IV.

- What equation must the coordinates of a point satisfy to lie on a graph? - What happens to a figure when it is reflected across one of its own lines of symmetry? - How can substitution verify the reflected point?

1. Evaluate \(f(2) = \frac{4}{2 - 1} + 3 = 7\). Therefore, \(Q\) lies on the graph. 2. A reflection across a line of symmetry maps every point of the graph to another point on the same graph. Since \(Q\) lies on the hyperbola, its reflected point \(R\) must also lie on it. 3. Verify by substitution: \(f(5) = \frac{4}{5 - 1} + 3 = 4\). This matches the \(y\)-coordinate of \(R\).

a) \(f(2) = 7\), so \(Q\) lies on the graph. b) Symmetry implies that the reflection \(R\) also lies on the graph, and \(f(5) = 4\) confirms it.

- To find a \(y\)-intercept, substitute \(x = 0\). - At an \(x\)-intercept, the function value is \(0\). - Isolate the fraction before solving each equation.

1. For \(f\), substitute \(x = 0\): \(f(0) = \frac{8}{0 + 2} - 2 = 2\). Therefore, the \(y\)-intercept is \((0, 2)\). 2. Set \(f(x) = 0\): \(0 = \frac{8}{x + 2} - 2\). Then \(2 = \frac{8}{x + 2}\), so \(x + 2 = 4\) and \(x = 2\). Therefore, the \(x\)-intercept is \((2, 0)\). 3. For \(g\), substitute \(x = 0\): \(g(0) = \frac{-6}{0 - 3} - 1 = 1\). Therefore, the \(y\)-intercept is \((0, 1)\). 4. Set \(g(x) = 0\): \(0 = \frac{-6}{x - 3} - 1\). Then \(1 = \frac{-6}{x - 3}\), so \(x - 3 = -6\) and \(x = -3\). Therefore, the \(x\)-intercept is \((-3, 0)\).

\(f\): \(x\)-intercept \((2, 0)\); \(y\)-intercept \((0, 2)\). \(g\): \(x\)-intercept \((-3, 0)\); \(y\)-intercept \((0, 1)\).

- At a \(y\)-intercept, the input is \(0\). - At an \(x\)-intercept, the function value is \(0\). - Isolate the fraction before clearing the denominator.

1. To find the \(y\)-intercept, substitute \(x = 0\): \(f(0) = \frac{4}{0 - 1} + 2 = -4 + 2 = -2\). Therefore, the \(y\)-intercept is \((0, -2)\). 2. To find the \(x\)-intercept, set \(f(x) = 0\): \(0 = \frac{4}{x - 1} + 2\). 3. Then \(-2 = \frac{4}{x - 1}\), so \(-2(x - 1) = 4\). Thus, \(-2x + 2 = 4\), giving \(x = -1\). 4. Therefore, the \(x\)-intercept is \((-1, 0)\).

The \(x\)-intercept is \((-1, 0)\), and the \(y\)-intercept is \((0, -2)\).

- At an \(x\)-intercept, the function value is \(0\). - At a \(y\)-intercept, the input is \(0\). - Keep track of the decimal value when you isolate the denominator.

1. To find the \(y\)-intercept, substitute \(x = 0\): \(h(0) = \frac{15}{0 + 3} - 2.5 = 5 - 2.5 = 2.5\). Therefore, the \(y\)-intercept is \((0, 2.5)\). 2. To find the \(x\)-intercept, set \(h(x) = 0\): \(0 = \frac{15}{x + 3} - 2.5\). 3. Then \(2.5 = \frac{15}{x + 3}\), so \(2.5(x + 3) = 15\). Thus, \(x + 3 = 6\), giving \(x = 3\). 4. Therefore, the \(x\)-intercept is \((3, 0)\).

The \(x\)-intercept is \((3, 0)\), and the \(y\)-intercept is \((0, 2.5)\).

- Use \(x = 0\) to find the \(y\)-intercept and \(f(x) = 0\) to find the \(x\)-intercept. - A point lies on a graph when its coordinates satisfy the function equation. - Substitute the point's \(x\)-coordinate and compare the result with its \(y\)-coordinate.

1. For the \(y\)-intercept, substitute \(x = 0\): \(f(0) = \frac{8}{0 - 2} - 1 = -4 - 1 = -5\). Therefore, the \(y\)-intercept is \((0, -5)\). 2. For the \(x\)-intercept, set \(f(x) = 0\): \(0 = \frac{8}{x - 2} - 1\). Then \(1 = \frac{8}{x - 2}\), so \(x - 2 = 8\) and \(x = 10\). Therefore, the \(x\)-intercept is \((10, 0)\). 3. To test \(P(6, 1)\), evaluate \(f(6)\): \(f(6) = \frac{8}{6 - 2} - 1 = 2 - 1 = 1\). 4. The calculated value matches the point's \(y\)-coordinate, so \(P(6, 1)\) lies on the graph.

a) The \(x\)-intercept is \((10, 0)\), and the \(y\)-intercept is \((0, -5)\). b) Yes. Since \(f(6) = 1\), the point \(P(6, 1)\) lies on the graph.

- Substitute \(x = 0\) to find a \(y\)-intercept. - Set the function value equal to \(0\) to find an \(x\)-intercept. - Isolate the fraction before solving for \(x\).

1. For \(f\), substitute \(x = 0\): \(f(0) = \frac{6}{2} - 1.5 = 1.5\). Therefore, the \(y\)-intercept is \((0, 1.5)\). 2. Set \(f(x) = 0\): \(0 = \frac{6}{x + 2} - 1.5\). Then \(1.5 = \frac{6}{x + 2}\), so \(x + 2 = 4\) and \(x = 2\). Therefore, the \(x\)-intercept is \((2, 0)\). 3. For \(g\), substitute \(x = 0\): \(g(0) = \frac{-4}{-5} + 0.8 = 1.6\). Therefore, the \(y\)-intercept is \((0, 1.6)\). 4. Set \(g(x) = 0\): \(0 = \frac{-4}{x - 5} + 0.8\). Then \(-0.8 = \frac{-4}{x - 5}\), so \(x - 5 = 5\) and \(x = 10\). Therefore, the \(x\)-intercept is \((10, 0)\).

\(f\): \(x\)-intercept \((2, 0)\); \(y\)-intercept \((0, 1.5)\). \(g\): \(x\)-intercept \((10, 0)\); \(y\)-intercept \((0, 1.6)\).

- How do changes inside and outside a function affect horizontal and vertical position? - What are the coordinates of the origin? - When a function value is given, which variable must be found?

1. Replacing \(x\) with \(x - 3\) translates the graph \(3\) units right, and subtracting \(2\) translates it \(2\) units down. 2. The origin is \((0, 0)\). Evaluate \(g(0) = \frac{6}{-3} - 2 = -4\). Since \(g(0) \ne 0\), the graph does not pass through the origin. 3. Solve \(1 = \frac{6}{x - 3} - 2\). Then \(3 = \frac{6}{x - 3}\), so \(x - 3 = 2\) and \(x = 5\).

a) Translate \(3\) units right and \(2\) units down. b) No, because \(g(0) = -4\). c) \(x = 5\)

- Identify the current asymptotes. - Recall the equations of the coordinate axes. - Determine how far each asymptote must move to coincide with its corresponding axis.

1. The original vertical asymptote is \(x = 2\), and the original horizontal asymptote is \(y = 3\). 2. The y-axis has equation \(x = 0\), so the graph must move \(2\) units left. 3. The x-axis has equation \(y = 0\), so the graph must move \(3\) units down. 4. These translations undo the original shifts, giving the parent function \(g_{\text{new}}(x) = \frac{1}{x}\).

a) Translate \(2\) units left and \(3\) units down. b) \(g_{\text{new}}(x) = \frac{1}{x}\)

- Identify the asymptotes of the original function. - Compare each original asymptote with the corresponding new asymptote. - A horizontal shift changes the input, while a vertical shift changes the output. - Check which input makes the new denominator zero.

1. The graph of \(f\) has vertical asymptote \(x=-3\) and horizontal asymptote \(y=0\). 2. Moving the vertical asymptote from \(x=-3\) to \(x=1\) requires a shift right \(4\) units. 3. Moving the horizontal asymptote from \(y=0\) to \(y=-4\) requires a shift down \(4\) units. 4. Therefore, \(g(x)=f(x-4)-4=\frac{5}{(x-4)+3}-4=\frac{5}{x-1}-4\).

Shift right \(4\) units and down \(4\) units. The function is \(g(x)=\frac{5}{x-1}-4\).

- Substitute the given inputs and compare the outputs. - When does a rational expression equal \(0\)? - What is the sign of \(x - 4\) when \(x > 4\)? - What happens when a fixed positive numerator is divided by an increasing positive denominator?

1. Evaluate the function: \(f(6) = \frac{24}{6 - 4} = 12\), \(f(8) = \frac{24}{8 - 4} = 6\), and \(f(12) = \frac{24}{12 - 4} = 3\). 2. As \(x\) increases, the positive denominator increases, so the function values decrease. 3. A rational expression can equal \(0\) only when its numerator equals \(0\). Since the numerator is always \(24\), \(f(x)\) cannot equal \(0\). 4. For \(x > 4\), both numerator and denominator are positive, so \(f(x) > 0\). Therefore, the function cannot become negative on the stated domain.

a) \(f(6) = 12\), \(f(8) = 6\), and \(f(12) = 3\). The function values decrease as \(x\) increases. b) The function can be neither \(0\) nor negative for \(x > 4\).

- Identify the vertical and horizontal lines the right-hand graph approaches. - Compare the asymptotes of the translated graph with those of the parent function. - How do horizontal and vertical translations change \(\frac{1}{x}\)? - Pay attention to the sign inside the denominator for a left translation.

1. The right graph approaches the vertical line \(x = -3\) and the horizontal line \(y = 2\), so these are its asymptotes. 2. The parent graph has asymptotes \(x = 0\) and \(y = 0\). Moving them to \(x = -3\) and \(y = 2\) translates the graph \(3\) units left and \(2\) units up. 3. Therefore, \(g(x) = \frac{1}{x + 3} + 2\).

a) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = 2\) b) Translate \(3\) units left and \(2\) units up. c) \(g(x) = \frac{1}{x + 3} + 2\)

- Identify the vertical and horizontal lines each graph approaches. - Compare each pair of asymptotes with the asymptotes \(x = 0\) and \(y = 0\) of the parent function. - A horizontal translation changes the expression in the denominator. - A vertical translation adds or subtracts a number outside the fraction.

1. For \(g\), the graph approaches the vertical line \(x = -2\) and the horizontal line \(y = -1\). 2. Translating \(\frac{1}{x}\) left \(2\) units replaces \(x\) with \(x + 2\). Translating it down \(1\) unit subtracts \(1\). Therefore, \(g(x) = \frac{1}{x + 2} - 1\). 3. For \(h\), the graph approaches the vertical line \(x = 1\) and the horizontal line \(y = 2\). 4. Translating \(\frac{1}{x}\) right \(1\) unit replaces \(x\) with \(x - 1\). Translating it up \(2\) units adds \(2\). Therefore, \(h(x) = \frac{1}{x - 1} + 2\).

a) Asymptotes: \(x = -2\) and \(y = -1\); \(g(x) = \frac{1}{x + 2} - 1\) b) Asymptotes: \(x = 1\) and \(y = 2\); \(h(x) = \frac{1}{x - 1} + 2\)

- Which vertical and horizontal lines does the curve approach? - How do you find an x-intercept and a y-intercept from a function rule? - Divide the numerator by the denominator and examine the fractional part for large inputs.

1. The graph approaches the vertical line \(x=2\) and the horizontal line \(y=2\). 2. The numerator is zero at \(x=3\), so the x-intercept is \((3,0)\). Also, \(f(0)=\frac{-6}{-2}=3\), giving y-intercept \((0,3)\). 3. Divide: \(2x-6=2(x-2)-2\), so \(f(x)=2-\frac{2}{x-2}\). 4. For large positive or negative \(x\), the fraction gets close to zero, so \(f(x)\) approaches \(2\) at both ends. The horizontal asymptote is \(y=2\).

a) Vertical asymptote: \(x=2\); horizontal asymptote: \(y=2\). b) X-intercept: \((3,0)\); y-intercept: \((0,3)\). c) \(f(x)=2-\frac{2}{x-2}\); both ends approach \(y=2\).

- Compare the center of the solid hyperbola with the center of the parent graph. - The horizontal asymptote determines the vertical translation. - The vertical asymptote determines the horizontal translation.

1. The solid graph approaches the vertical line \(x = -2\) and the horizontal line \(y = 2\), so these are its asymptotes. 2. Compared with \(f(x) = \frac{1}{x}\), the graph is translated \(2\) units left and \(2\) units up. 3. A translation \(2\) units left replaces \(x\) with \(x + 2\), and a translation \(2\) units up adds \(2\). Therefore, \(g(x) = \frac{1}{x + 2} + 2\). 4. For example, \(g(-1) = \frac{1}{-1 + 2} + 2 = 3\), which agrees with the graph.

The asymptotes are \(x = -2\) and \(y = 2\), and \(g(x) = \frac{1}{x + 2} + 2\).

- Read the vertical asymptote from the denominator shift. - Read the horizontal asymptote from the constant outside the fraction. - Use the branch orientation to determine the sign of the numerator. - Verify a proposed match with one point on the graph.

1. Graph a) has vertical asymptote \(x = 2\) and horizontal asymptote \(y = 1\). It matches rule 1), \(y = \frac{1}{x - 2} + 1\). 2. Graph b) has vertical asymptote \(x = -1\) and horizontal asymptote \(y = 2\). Its branches have the orientation produced by a negative numerator, so it matches rule 2), \(y = -\frac{1}{x + 1} + 2\).

a) 1) \(y = \frac{1}{x - 2} + 1\); b) 2) \(y = -\frac{1}{x + 1} + 2\)

- For \(\frac{1}{x - h} + k\), the vertical asymptote is \(x = h\) and the horizontal asymptote is \(y = k\). - Read both asymptotes from each graph before choosing a function. - Pay attention to the sign inside the denominator.

1. Graph A has vertical asymptote \(x = 3\) and horizontal asymptote \(y = 1\). Therefore, it matches \(f_2(x) = \frac{1}{x - 3} + 1\). 2. Graph B has vertical asymptote \(x = -2\) and horizontal asymptote \(y = -1\). Therefore, it matches \(f_1(x) = \frac{1}{x + 2} - 1\). 3. Graph C has vertical asymptote \(x = 2\) and horizontal asymptote \(y = -2\). Therefore, it matches \(f_4(x) = \frac{1}{x - 2} - 2\). 4. Graph D has vertical asymptote \(x = -1\) and horizontal asymptote \(y = 2\). Therefore, it matches \(f_3(x) = \frac{1}{x + 1} + 2\).

A: 2); B: 1); C: 4); D: 3)

- Identify the lines approached by the translated graph. - Compare the intersection of those asymptotes with the origin. - A right translation produces a minus sign inside the denominator.

1. The graph approaches the vertical line \(x = 1\) and the horizontal line \(y = 2\). 2. Compared with the parent function, the graph is translated \(1\) unit right and \(2\) units up. 3. A translation \(1\) unit right replaces \(x\) with \(x - 1\), and a translation \(2\) units up adds \(2\). 4. Therefore, \(g(x) = \frac{1}{x - 1} + 2\).

The asymptotes are \(x = 1\) and \(y = 2\). The graph is translated \(1\) unit right and \(2\) units up, so \(g(x) = \frac{1}{x - 1} + 2\).

- Use the vertical asymptote to determine the horizontal translation. - Use the horizontal asymptote to determine the vertical translation. - Then update the parent function rule to match both shifts.

1. The vertical asymptote moves from \(x=0\) to \(x=-2\), so the graph shifts left \(2\) units. 2. The horizontal asymptote moves from \(y=0\) to \(y=-1\), so the graph shifts down \(1\) unit. 3. Replacing \(x\) with \(x+2\) shifts the graph left \(2\) units, and subtracting \(1\) shifts it down. 4. Therefore, \(g(x)=\frac{1}{(x+2)^2}-1\).

Shift left \(2\) units and down \(1\) unit; \(g(x)=\frac{1}{(x+2)^2}-1\)

- Identify the vertical and horizontal asymptotes of each translated graph. - Compare the intersection of each pair of asymptotes with the origin. - A horizontal translation changes the expression in the denominator.

1. The graph of \(g\) has vertical asymptote \(x = -2\) and horizontal asymptote \(y = 1\). Therefore, it is translated \(2\) units left and \(1\) unit up. 2. Thus, \(g(x) = \frac{1}{x + 2} + 1\). 3. The graph of \(h\) has vertical asymptote \(x = 3\) and horizontal asymptote \(y = -2\). Therefore, it is translated \(3\) units right and \(2\) units down. 4. Thus, \(h(x) = \frac{1}{x - 3} - 2\).

\(g(x) = \frac{1}{x + 2} + 1\) \(h(x) = \frac{1}{x - 3} - 2\)

- Find the input that makes the denominator zero. - Separate the vertical shift from the reciprocal term. - Check the sign of the reciprocal term on the right side of the excluded input.

1. The denominator is zero at \(x=-1\), so claim 1 is incorrect: the vertical asymptote is \(x=-1\). 2. The function is \(-\frac{2}{x+1}\) shifted up \(3\), so the horizontal asymptote is \(y=3\). Claim 2 is correct. 3. For \(x>-1\), the denominator is positive, so \(-\frac{2}{x+1}<0\). The graph lies below \(y=3\) on the right. Claim 3 is incorrect.

Claims 1 and 3 are incorrect. The vertical asymptote is \(x=-1\), and the right branch lies below \(y=3\). Claim 2 is correct.

- Compare which graphs have a vertical gap at the same x-value. - Among those, compare their heights at an allowed shared input. - For the remaining graph, measure the horizontal change in the vertical asymptote's position.

1. Graphs \(f\) and \(g\) have the same vertical and horizontal asymptotes, but \(g\) is twice as far from the x-axis at each common input. Thus, \(g\) is the vertical stretch. 2. Graph \(h\) has its vertical asymptote at \(x=2\), rather than \(x=0\); it is \(f\) shifted right \(2\) units.

\(g\): vertical stretch by \(2\). \(h\): shift right \(2\) units.

- Average speed equals distance divided by travel time. - Compare the speeds as the allowed time becomes larger. - What would dividing a positive distance by zero mean in this situation?

1. Substituting gives \(v(1.5)=\frac{18}{1.5}=12\,\text{mph}\) and \(v(3)=\frac{18}{3}=6\,\text{mph}\). 2. The graph approaches \(y=0\) as \(t\) grows; the required speed becomes smaller for longer trip times but is never zero for finite positive \(t\). 3. At \(t=0\), the rule would divide by zero; a nonzero-distance trip cannot take zero time at finite speed.

a) \(12\,\text{mph}\) and \(6\,\text{mph}\). b) \(y=0\); longer ride times require a smaller average speed. c) \(t=0\) would make the denominator zero and cannot describe this trip.

- How can substitution determine whether a point lies on a graph? - How does the sign of the numerator affect the orientation of a reciprocal graph relative to its asymptotes? - How can you solve an equation when the variable appears in the denominator?

1. Evaluate the function at \(x = -1\): \(g(-1) = \frac{-2}{-1 + 3} = \frac{-2}{2} = -1\). Therefore, Sofia is correct. 2. Relative to the asymptotes \(x = -3\) and \(y = 0\), a negative numerator places the branches in Quadrants II and IV. Therefore, Lucas is correct. 3. Solve \(\frac{-2}{x + 3} = -0.5\). Multiplying by \(x + 3\) gives \(-2 = -0.5(x + 3)\). Dividing by \(-0.5\) gives \(4 = x + 3\), so \(x = 1\).

a) Yes. \(g(-1) = -1\). b) Lucas is correct. Since the numerator is negative, the branches lie in Quadrants II and IV relative to the asymptotes. c) \(x = 1\)

- The intersection of the asymptotes gives the horizontal and vertical translations. - Set \(g(x) = 0\) to find the x-intercept. - Can a fraction with a nonzero numerator equal \(0\)?

1. The asymptote intersection \((-3, 2)\) means the vertical asymptote is \(x = -3\) and the horizontal asymptote is \(y = 2\). 2. Translating \(\frac{1}{x}\) \(3\) units left and \(2\) units up gives \(g(x) = \frac{1}{x + 3} + 2\). 3. To find the x-intercept, solve \(0 = \frac{1}{x + 3} + 2\). Then \(-2 = \frac{1}{x + 3}\), so \(-2(x + 3) = 1\). 4. Solving \(-2x - 6 = 1\) gives \(x = -3.5\). The x-intercept is \((-3.5, 0)\). 5. The line \(y = 2\) is the horizontal asymptote. Since \(\frac{1}{x + 3}\) can never equal \(0\), \(g(x)\) can never equal \(2\).

a) \(g(x) = \frac{1}{x + 3} + 2\) b) \((-3.5, 0)\) c) The term \(\frac{1}{x + 3}\) is never \(0\), so \(g(x)\) never equals \(2\).

- What effect does each number in the function rule have on the parent graph? - Which input is excluded, and which output cannot occur? - Rearrange the equation one step at a time to isolate \(x\).

1. The negative sign reflects the graph across the x-axis. Replacing \(x\) with \(x - 2\) translates it \(2\) units right, and adding \(5\) translates it \(5\) units up. 2. The denominator cannot equal \(0\), so the domain is \(\mathbb{R} \setminus \{2\}\). The function cannot equal its horizontal asymptote, so the range is \(\mathbb{R} \setminus \{5\}\). 3. Solve \(4 = -\frac{1}{x - 2} + 5\). Then \(-1 = -\frac{1}{x - 2}\), so \(x - 2 = 1\) and \(x = 3\).

a) Reflect across the x-axis, translate \(2\) units right, and translate \(5\) units up. b) Domain: \(\mathbb{R} \setminus \{2\}\); range: \(\mathbb{R} \setminus \{5\}\) c) \(x = 3\)

- Substitute the known input into the appropriate function rule. - What does comparing two \(y\)-values at the same \(x\)-coordinate tell you about vertical position? - How can you solve an equation when the variable is in the denominator? - Check carefully which function is used in each part.

1. Evaluate \(f_1(4) = \frac{4}{4} = 1\). Therefore, \(S\) lies on the graph of \(f_1\). 2. Evaluate \(f_2(4) = \frac{4}{4 - 2} + 1 = 3\). Since the point has \(y = 1\) and \(1 < 3\), \(S\) lies below the graph of \(f_2\). 3. Solve \(5 = \frac{4}{x - 2} + 1\). Then \(4 = \frac{4}{x - 2}\), so \(x - 2 = 1\) and \(x = 3\).

1) \(f_1(4) = 1\), so \(S\) lies on the graph of \(f_1\). 2) \(S\) lies below the graph of \(f_2\), because \(f_2(4) = 3\) and \(1 < 3\). 3) \(x = 3\)

- Compare the denominators and constant terms. - Substitute \(x = 0\) to find the \(y\)-intercepts. - Set the two function rules equal to test for intersections. - A contradiction means there is no common point.

1. Both functions have the denominator \(x - 1\) and the same vertical shift. Therefore, both graphs have the vertical asymptote \(x = 1\) and the horizontal asymptote \(y = 2\). 2. Evaluate \(f(0)\): \(f(0) = \frac{4}{-1} + 2 = -2\). Thus, the \(y\)-intercept of \(f\) is \((0, -2)\). 3. Evaluate \(g(0)\): \(g(0) = \frac{-4}{-1} + 2 = 6\). Thus, the \(y\)-intercept of \(g\) is \((0, 6)\). 4. To find a common point, solve \(\frac{4}{x - 1} + 2 = \frac{-4}{x - 1} + 2\). Subtracting \(2\) and multiplying by \(x - 1\) gives \(4 = -4\), a contradiction. 5. Therefore, the graphs have no common point. They are reflections of each other across \(y = 2\).

a) Both graphs have asymptotes \(x = 1\) and \(y = 2\). b) The \(y\)-intercepts are \((0, -2)\) for \(f\) and \((0, 6)\) for \(g\). c) The graphs have no common point because solving \(f(x) = g(x)\) leads to the contradiction \(4 = -4\).

- Find both intercepts for each function systematically. - If substituting \(x = 0\) gives a function value of \(0\), the graph passes through the origin. - Compare the distances and coordinates of the intercepts relative to the origin.

1. For \(f\), substitute \(x = 0\): \(f(0) = \frac{4}{0 - 2} + 2 = 0\). Therefore, the \(y\)-intercept is \((0, 0)\). 2. Set \(f(x) = 0\): \(0 = \frac{4}{x - 2} + 2\). Then \(-2 = \frac{4}{x - 2}\), so \(-2(x - 2) = 4\), giving \(x = 0\). Thus, the \(x\)-intercept is also \((0, 0)\). 3. For \(g\), substitute \(x = 0\): \(g(0) = \frac{-4}{0 + 2} - 2 = -4\). Therefore, the \(y\)-intercept is \((0, -4)\). 4. Set \(g(x) = 0\): \(0 = \frac{-4}{x + 2} - 2\). Then \(2 = \frac{-4}{x + 2}\), so \(2(x + 2) = -4\), giving \(x = -4\). Thus, the \(x\)-intercept is \((-4, 0)\). 5. The graph of \(f\) passes through the origin. The two intercepts of \(g\) lie on the negative axes, each \(4\) units from the origin; they are reflections of each other across the line \(y = x\).

\(f\): Both intercepts are at \((0, 0)\). \(g\): The \(x\)-intercept is \((-4, 0)\), and the \(y\)-intercept is \((0, -4)\). Comparison: The graph of \(f\) passes through the origin. The intercepts of \(g\) are each \(4\) units from the origin and are reflections across \(y = x\).

- Set each function equal to \(0\) to find its \(x\)-intercept. - The distance from the origin of a point \((x, 0)\) is \(|x|\). - Compare the absolute values of the two \(x\)-coordinates.

1. Set \(f(x) = 0\): \(0 = \frac{10}{x - 2} - 4\). Then \(4 = \frac{10}{x - 2}\), so \(x - 2 = 2.5\) and \(x = 4.5\). The \(x\)-intercept is \((4.5, 0)\). 2. Set \(g(x) = 0\): \(0 = \frac{10}{x - 4} - 2\). Then \(2 = \frac{10}{x - 4}\), so \(x - 4 = 5\) and \(x = 9\). The \(x\)-intercept is \((9, 0)\). 3. The distances from the origin are \(|4.5| = 4.5\) and \(|9| = 9\). Since \(9 > 4.5\), the \(x\)-intercept of \(g\) is farther from the origin.

The \(x\)-intercept of \(f\) is \((4.5, 0)\), and the \(x\)-intercept of \(g\) is \((9, 0)\). Therefore, the intercept of \(g\) is farther from the origin.

- Calculate all four intercepts separately. - The constants outside the fractions produce vertical shifts. - Consider how moving a graph up or down changes where it crosses the \(x\)-axis.

1. For \(f\), substitute \(x = 0\): \(f(0) = \frac{4}{-2} + 1 = -1\). Therefore, the \(y\)-intercept is \((0, -1)\). 2. Set \(f(x) = 0\): \(0 = \frac{4}{x - 2} + 1\). Then \(-1 = \frac{4}{x - 2}\), so \(x - 2 = -4\) and \(x = -2\). Therefore, the \(x\)-intercept is \((-2, 0)\). 3. For \(h\), substitute \(x = 0\): \(h(0) = \frac{4}{-2} - 1 = -3\). Therefore, the \(y\)-intercept is \((0, -3)\). 4. Set \(h(x) = 0\): \(0 = \frac{4}{x - 2} - 1\). Then \(1 = \frac{4}{x - 2}\), so \(x - 2 = 4\) and \(x = 6\). Therefore, the \(x\)-intercept is \((6, 0)\). 5. The constants \(+1\) and \(-1\) shift the same parent graph vertically in opposite directions. Because each graph is at a different vertical position, it crosses the \(x\)-axis at a different input.

a) \(f\): \(x\)-intercept \((-2, 0)\), \(y\)-intercept \((0, -1)\); \(h\): \(x\)-intercept \((6, 0)\), \(y\)-intercept \((0, -3)\). b) The constants shift the graphs vertically in opposite directions, so the graphs cross the \(x\)-axis at different inputs.

- When can a rational expression equal \(0\)? - What happens to the function value as \(|x|\) becomes large? - Which input is excluded because it makes the denominator \(0\)? - How does the term \(-1\) translate the graph?

1. The denominator is \(0\) at \(x = -2\), so the vertical asymptote is \(x = -2\). The horizontal asymptote is \(y = -1\). 2. For the y-intercept, set \(x = 0\): \(f(0) = \frac{4}{2} - 1 = 1\). The y-intercept is \((0, 1)\). 3. For the x-intercept, solve \(0 = \frac{4}{x + 2} - 1\). Then \(1 = \frac{4}{x + 2}\), so \(x + 2 = 4\) and \(x = 2\). The x-intercept is \((2, 0)\). 4. The graph of \(g\) has horizontal asymptote \(y = 0\), and \(\frac{4}{x + 2}\) cannot equal \(0\) because its numerator is nonzero. The graph of \(f\) is translated \(1\) unit down, so it can cross the x-axis, as shown by its zero at \(x = 2\).

a) Vertical asymptote: \(x = -2\); horizontal asymptote: \(y = -1\) b) y-intercept: \((0, 1)\); x-intercept: \((2, 0)\) c) The expression \(\frac{4}{x + 2}\) cannot equal \(0\), but translating the graph down \(1\) unit produces the zero \(x = 2\) for \(f\).

- To test for a \(y\)-intercept, determine whether \(x = 0\) is allowed and, if so, evaluate the function there. - A fraction equals \(0\) only when its numerator is \(0\) and its denominator is not \(0\). - Check the domain of each function.

1. For \(f(x) = \frac{2}{x} + 3\), the domain is \(\mathbb{R} \setminus \{0\}\). Because \(x = 0\) is not allowed, the graph has no \(y\)-intercept. 2. To find the \(x\)-intercept of \(f\), solve \(0 = \frac{2}{x} + 3\). Then \(-3 = \frac{2}{x}\), so \(x = -\frac{2}{3}\). Thus, the \(x\)-intercept is \(\left(-\frac{2}{3}, 0\right)\). 3. For \(g(x) = \frac{2}{x + 3}\), substitute \(x = 0\): \(g(0) = \frac{2}{3}\). Thus, the \(y\)-intercept is \(\left(0, \frac{2}{3}\right)\). 4. The equation \(\frac{2}{x + 3} = 0\) has no solution because a fraction can equal zero only when its numerator is zero, and the numerator is the nonzero constant \(2\). Therefore, \(g\) has no \(x\)-intercept.

\(f\) has no \(y\)-intercept because \(x = 0\) is excluded from its domain. Its \(x\)-intercept is \(\left(-\frac{2}{3}, 0\right)\). \(g\) has the \(y\)-intercept \(\left(0, \frac{2}{3}\right)\) and no \(x\)-intercept because its numerator is never zero.

- At the given \(x\)-intercept, substitute the input and use a function value of \(0\). - A fraction equals \(0\) only when its numerator equals \(0\). - After finding \(q\), substitute \(x = 0\) to find the \(y\)-intercept.

1. An \(x\)-intercept at \(x = 7\) means \(h(7) = 0\). Substitute: \(0 = \frac{12}{7 - 3} + q = 3 + q\), so \(q = -3\). 2. The function is \(h(x) = \frac{12}{x - 3} - 3\). Substitute \(x = 0\): \(h(0) = \frac{12}{-3} - 3 = -7\). Therefore, the \(y\)-intercept is \((0, -7)\). 3. An \(x\)-intercept of \(j\) would require \(\frac{12}{x - 3} = 0\). A fraction can equal zero only when its numerator is zero, but the numerator is the nonzero constant \(12\). Therefore, the equation has no solution and \(j\) has no \(x\)-intercept.

a) \(q = -3\), and the \(y\)-intercept is \((0, -7)\). b) The equation \(\frac{12}{x - 3} = 0\) has no solution because its numerator is the nonzero constant \(12\).

- Compare the vertical and horizontal asymptotes of the two functions. - What translation moves the vertical asymptote from \(x = -1\) to \(x = 2\)? - A given \(y\)-coordinate is a given function value; solve for the corresponding input.

1. The vertical asymptote of \(f\) is \(x = -1\), and that of \(h\) is \(x = 2\). Moving from \(-1\) to \(2\) requires a translation of \(3\) units right. 2. The horizontal asymptote of \(f\) is \(y = 4\), and that of \(h\) is \(y = 1\). Moving from \(4\) to \(1\) requires a translation of \(3\) units down. 3. Solve \(6 = \frac{2}{x + 1} + 4\). Then \(2 = \frac{2}{x + 1}\), so \(x + 1 = 1\) and \(x = 0\). The point is \((0, 6)\).

a) Translate \(3\) units right and \(3\) units down. b) \((0, 6)\)

- Calculate the intercepts of each function separately. - Two intercepts are the same only when both coordinates match. - Track the signs carefully.

1. For \(h_1\), solve \(0 = \frac{2}{x - 1} - 1\). Then \(1 = \frac{2}{x - 1}\), so \(x - 1 = 2\) and \(x = 3\). Its \(x\)-intercept is \((3, 0)\). 2. For \(h_2\), solve \(0 = \frac{-2}{x - 1} + 1\). Then \(-1 = \frac{-2}{x - 1}\), so \(x - 1 = 2\) and \(x = 3\). Its \(x\)-intercept is also \((3, 0)\). 3. Evaluate at \(x = 0\): \(h_1(0) = \frac{2}{-1} - 1 = -3\), so its \(y\)-intercept is \((0, -3)\). 4. Also, \(h_2(0) = \frac{-2}{-1} + 1 = 3\), so its \(y\)-intercept is \((0, 3)\). 5. Therefore, the functions have the same \(x\)-intercept but different \(y\)-intercepts.

The graphs have the same \(x\)-intercept, \((3, 0)\). They have different \(y\)-intercepts: \((0, -3)\) for \(h_1\) and \((0, 3)\) for \(h_2\).

- Find the denominator zero and the limiting output at large \(|x|\). - Use the behavior of a reciprocal function to determine monotonicity on each branch. - Convert the band condition into an absolute-value inequality.

1. The denominator is zero at \(x=8\), so the vertical asymptote is \(x=8\). As \(|x|\to\infty\), the fraction approaches \(0\), so the horizontal asymptote is \(y=-2\). 2. The reciprocal term \(\frac{10}{x-8}\) is strictly decreasing on each interval \((-\infty, 8)\) and \((8, \infty)\). Therefore, \(f\) is strictly decreasing on each interval. 3. The band condition is \(|f(x)+2|<0.2\), or \(\left|\frac{10}{x-8}\right|<0.2\). 4. This is equivalent to \(|x-8|>50\), so \(x>58\) or \(x<-42\). 5. Thus, \(x_0=58\) and \(x_1=-42\).

a) Vertical asymptote: \(x=8\); horizontal asymptote: \(y=-2\). The function is strictly decreasing on \((-\infty, 8)\) and \((8, \infty)\). b) (1) \(x_0=58\) (2) \(x_1=-42\)

- A vertical compression multiplies the entire function value, not just one term. - A vertical shift is added after the compression. - Identify which part of a rational function determines each asymptote. - Consider whether changing only outputs can move a vertical line.

1. Apply the vertical compression to the entire output: \(h(x) = 0.5f(x) = 0.5\left(\frac{6}{x-4}+2\right) = \frac{3}{x-4}+1\). 2. Shift the graph up \(3\) units: \(h(x)+3 = \frac{3}{x-4}+4 = g(x)\). 3. For \(f\), the denominator is zero at \(x=4\), so the vertical asymptote is \(x=4\). Since \(\frac{6}{x-4}\to 0\) as \(x\to\pm\infty\), the horizontal asymptote is \(y=2\). 4. For \(g\), the vertical asymptote is still \(x=4\), and the horizontal asymptote is \(y=4\). 5. The vertical asymptote does not move because the transformations change only output values. The horizontal asymptote changes from \(y=2\) to \(y=0.5\cdot 2+3=4\).

The transformation gives \(0.5f(x)+3 = \frac{3}{x-4}+4 = g(x)\). Asymptotes of \(f\): \(x=4\) and \(y=2\). Asymptotes of \(g\): \(x=4\) and \(y=4\). The vertical asymptote is unchanged. The horizontal asymptote is transformed by \(y\mapsto 0.5y+3\), so \(2\mapsto 4\).

- Begin with \(g(x)=kf(x)+d\). - Compare the coefficient of the rational term and then compare the constant terms. - Identify the input value that makes the denominator zero. - Decide whether transformations applied outside the function change its domain.

1. Write the transformation as \(g(x)=kf(x)+d\): \(g(x)=k\left(-\frac{1}{x+2}+5\right)+d=-\frac{k}{x+2}+5k+d\). 2. Match the rational terms: \(-k=-4\), so \(k=4\). 3. Match the constant terms: \(5k+d=12\). Substituting \(k=4\) gives \(20+d=12\), so \(d=-8\). 4. Both functions are undefined at \(x=-2\). Vertical transformations change outputs but do not change the input value where the denominator is zero, so the vertical asymptote remains \(x=-2\).

The values are \(k=4\) and \(d=-8\). The vertical asymptote remains \(x=-2\) because vertical scaling and shifting do not change the zero of the denominator.

- Factor the coefficient of \(x\) from the denominator. - Match the resulting input to the form \(b(x+c)\). - An input factor changes horizontal scale, while an outside factor changes vertical scale.

1. Factor the denominator: \(2x+4=2(x+2)\). 2. Then \(g(x)=3\left(\frac{1}{2(x+2)}\right)-1=3f(2(x+2))-1\). Thus, one representation has \(a=3\), \(b=2\), \(c=2\), and \(d=-1\). 3. Starting with the graph of \(f\), horizontally compress by a factor of \(\frac{1}{2}\), shift left \(2\) units, vertically stretch by a factor of \(3\), and shift down \(1\) unit.

One valid representation is \(g(x)=3f(2(x+2))-1\). Transformations: horizontally compress by a factor of \(\frac{1}{2}\); shift left \(2\) units; vertically stretch by a factor of \(3\); shift down \(1\) unit.

- In each sequence, apply the second transformation to the result of the first. - A vertical compression multiplies the entire current output. - Check the two final rules at any allowed input, such as \(x=1\).

1. For Sequence A, first form \(0.5f(x)\), then add \(2\): \(g(x)=0.5\left(\frac{1}{x}\right)+2=\frac{0.5}{x}+2\). 2. For Sequence B, first form \(f(x)+2\), then multiply the entire output by \(0.5\): \(h(x)=0.5\left(\frac{1}{x}+2\right)=\frac{0.5}{x}+1\). 3. The constant terms differ, so \(g(x)\ne h(x)\) for every \(x\ne 0\). Therefore, the graphs are not identical.

\(g(x)=\frac{0.5}{x}+2\) and \(h(x)=\frac{0.5}{x}+1\). The functions are not identical because their constant terms differ.

- Factor the quadratic expression in the denominator. - Rewrite \(g\) in the form \(af(x-c)+d\). - Match each parameter with its graphical effect. - Pay attention to the negative sign and the constant outside the fraction.

1. Factor the denominator: \(x^2-4x+4=(x-2)^2\). 2. Rewrite the rule as \(g(x)=-4\left(\frac{1}{(x-2)^2}\right)+1=-4f(x-2)+1\). 3. Shift the graph of \(f\) right \(2\) units. 4. Vertically stretch by a factor of \(4\) and reflect across the x-axis. 5. Shift up \(1\) unit.

Shift right \(2\) units, vertically stretch by a factor of \(4\), reflect across the x-axis, and shift up \(1\) unit.

- A horizontal stretch changes the input of the function. - A vertical shift is added to the entire function value. - Use the zero of the denominator and the constant term to identify the asymptotes. - Decide which transformation affects each asymptote.

1. A horizontal stretch by a factor of \(3\) replaces \(x\) with \(\frac{x}{3}\): \(f\left(\frac{x}{3}\right)=\frac{2}{x/3+4}-3\). 2. Simplify: \(\frac{2}{x/3+4}-3=\frac{2}{(x+12)/3}-3=\frac{6}{x+12}-3\). 3. Shift up \(4\) units: \(f\left(\frac{x}{3}\right)+4=\frac{6}{x+12}+1=g(x)\). 4. The vertical asymptote \(x=-4\) is horizontally stretched to \(x=-12\). The horizontal asymptote \(y=-3\) shifts up to \(y=1\).

\(g(x)=f\left(\frac{x}{3}\right)+4=\frac{6}{x+12}+1\). The vertical asymptote changes from \(x=-4\) to \(x=-12\), and the horizontal asymptote changes from \(y=-3\) to \(y=1\).

- A vertical stretch multiplies the entire function, including its constant term. - A shift left replaces \(x\) with \(x+3\). - Track the horizontal asymptote through the vertical stretch. - Track the vertical asymptote through the horizontal shift.

1. Vertically stretch the entire output: \(4h(x)=4\left(\frac{1}{x-2}+5\right)=\frac{4}{x-2}+20\). 2. Shift left \(3\) units by replacing \(x\) with \(x+3\): \(k(x)=\frac{4}{(x+3)-2}+20=\frac{4}{x+1}+20\). 3. The original vertical asymptote \(x=2\) shifts left \(3\) units to \(x=-1\). 4. The original horizontal asymptote \(y=5\) is vertically stretched by a factor of \(4\), so it becomes \(y=20\).

\(k(x)=\frac{4}{x+1}+20\). Its vertical asymptote is \(x=-1\), and its horizontal asymptote is \(y=20\).

- Rewrite the denominator by factoring out \(2\). - The horizontal shift appears inside the denominator, while the vertical shift is outside the fraction. - Set the function equal to \(0\) to find the x-intercept.

1. The denominator is zero at \(x=4\), so the vertical asymptote is \(x=4\). 2. Since \(\frac{5}{2x-8}\to 0\) as \(x\to\pm\infty\), the horizontal asymptote is \(y=-1.5\). 3. To find the x-intercept, solve \(\frac{5}{2x-8}-1.5=0\). Then \(5=1.5(2x-8)\), so \(17=3x\) and \(x=\frac{17}{3}\). 4. Rewrite \(h(x)=\frac{2.5}{x-4}-1.5\). Starting with \(y=\frac{2.5}{x}\), shift the graph \(4\) units right and \(1.5\) units down.

Vertical asymptote: \(x=4\) Horizontal asymptote: \(y=-1.5\) x-intercept: \(\left(\frac{17}{3}, 0\right)\) Transformation: shift \(y=\frac{2.5}{x}\) right \(4\) units and down \(1.5\) units.

- Rewrite the numerator as a multiple of the denominator plus a remainder. - The constant outside the fraction determines the horizontal asymptote. - The numerator zero in the original form determines the x-intercept.

1. Rewrite the numerator as \(2x+7=2(x+3)+1\). Thus, \(f(x)=2+\frac{1}{x+3}\). 2. The vertical asymptote is \(x=-3\), and the horizontal asymptote is \(y=2\). 3. The x-intercept satisfies \(2x+7=0\), so \(x=-\frac{7}{2}\). 4. The form \(2+\frac{1}{x+3}\) displays the horizontal asymptote directly because the fractional term approaches \(0\). The original numerator-over-denominator form displays the x-intercept directly because it comes from the numerator zero.

a) \(f(x)=2+\frac{1}{x+3}\) b) Horizontal asymptote: \(y=2\); vertical asymptote: \(x=-3\); x-intercept: \(\left(-\frac{7}{2}, 0\right)\) c) The transformed form shows the horizontal asymptote most directly; the original form shows the x-intercept most directly.

- Compare the outputs when the denominator stays the same but the numerator doubles. - Determine the signs of the input and output in each branch. - A nonzero constant numerator prevents the function value from being \(0\). - Substituting \(x = 0\) would create division by zero.

1. Evaluate each function at the six given inputs. The outputs of \(g\) are twice the corresponding outputs of \(f\). 2. For every \(x > 0\), \(g(x) = 2f(x) > f(x)\). The points have the same positive \(x\)-coordinate, but the point on \(g\) has the larger \(y\)-coordinate, so it is farther from the origin. 3. Since both numerator parameters are positive, positive inputs give positive outputs and negative inputs give negative outputs. Thus, the branches lie in Quadrants I and III. 4. The graphs do not cross the \(y\)-axis because \(x = 0\) is excluded from the domain. They do not cross the \(x\)-axis because a fraction with a nonzero constant numerator cannot equal \(0\).

a) <table> <tr><td>\(x\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(6\)</td><td>\(12\)</td></tr> <tr><td>\(f(x)\)</td><td>\(12\)</td><td>\(6\)</td><td>\(4\)</td><td>\(3\)</td><td>\(2\)</td><td>\(1\)</td></tr> <tr><td>\(g(x)\)</td><td>\(24\)</td><td>\(12\)</td><td>\(8\)</td><td>\(6\)</td><td>\(4\)</td><td>\(2\)</td></tr> </table> b) For the same positive input, the point on \(g\) is farther from the origin because \(g(x) = 2f(x)\). c) The branches lie in Quadrants I and III. The graphs cannot cross the \(y\)-axis because \(x = 0\) is excluded, and they cannot cross the \(x\)-axis because their numerators are nonzero.

- Find the value that makes the denominator zero. - Compare the degrees and leading coefficients to find the horizontal asymptote. - Rewrite the numerator using \(x+1\). - In the form \(a\frac{1}{x-h}+k\), identify the reflection, stretch, and translations.

1. The denominator is zero at \(x=-1\), so the domain is \(\mathbb{R}\setminus\{-1\}\). 2. The numerator is zero at \(x=2\), so the x-intercept is \((2, 0)\). 3. Since the denominator is zero and the numerator is nonzero at \(x=-1\), the vertical asymptote is \(x=-1\). The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of the leading coefficients: \(y=1\). 4. Rewrite the numerator as \(x-2=(x+1)-3\): \(f(x)=\frac{x+1-3}{x+1}=1-\frac{3}{x+1}\). Thus, \(a=1\) and \(b=-3\). 5. Starting with \(y=\frac{1}{x}\), stretch vertically by a factor of \(3\), reflect across the x-axis, shift left \(1\) unit, and shift up \(1\) unit.

a) Domain: \(\mathbb{R}\setminus\{-1\}\); x-intercept: \((2, 0)\); vertical asymptote: \(x=-1\); horizontal asymptote: \(y=1\) b) \(f(x)=1-\frac{3}{x+1}\), so \(a=1\) and \(b=-3\). c) Stretch vertically by a factor of \(3\), reflect across the x-axis, shift left \(1\) unit, and shift up \(1\) unit.

- Find the denominator zero and compare leading coefficients. - Rewrite the function as a constant plus a reciprocal term. - Compare the function values at \(1+h\) and \(1-h\).

1. The denominator is zero at \(x=1\). Since \(a\neq 3\), the numerator at \(x=1\) is nonzero, so the vertical asymptote is \(x=1\). 2. The numerator and denominator have the same degree, so the horizontal asymptote is \(y=a\). Their intersection is \(S=(1, a)\). 3. Rewrite the function: \(f_a(x)=a+\frac{a-3}{x-1}\). 4. Let \(x=1+h\), where \(h\neq 0\). Then \(f_a(1+h)-a=\frac{a-3}{h}\), while \(f_a(1-h)-a=-\frac{a-3}{h}\). 5. Thus, points with x-coordinates equally spaced from \(1\) have y-values equally spaced on opposite sides of \(a\). Therefore, the graph is symmetric about \((1, a)\).

a) Vertical asymptote: \(x=1\); horizontal asymptote: \(y=a\) b) The graph is symmetric about \(S=(1, a)\).

- Substitute \(x=0\) to find the y-intercept. - Set the function equal to \(0\) to find the x-intercept. - The triangle's legs lie along the coordinate axes. - Use the area formula for a right triangle.

1. To find \(T\), substitute \(x=0\): \(f(0)=\frac{6}{2}-1=2\). Therefore, \(T=(0,2)\). 2. To find \(N\), set \(f(x)=0\): \(\frac{6}{x+2}-1=0\). Then \(\frac{6}{x+2}=1\), so \(x=4\). Therefore, \(N=(4,0)\). 3. The legs of the right triangle lie on the coordinate axes and have lengths \(4\) and \(2\). 4. Its area is \(A=\frac{1}{2}\cdot4\cdot2=4\) square units.

a) \(N=(4,0)\) and \(T=(0,2)\) b) \(4\) square units

- Read the new asymptotes to determine the translations. - Compare the branch orientation with the parent function. - Use the form \(\frac{a}{x - h} + k\).

1. The solid graph has the vertical asymptote \(x = 3\) and the horizontal asymptote \(y = -1\). 2. These asymptotes indicate a translation \(3\) units to the right and \(1\) unit down. 3. Relative to the new asymptotes, the branch orientation is reversed from the parent function. This indicates a reflection across the \(x\)-axis before the translations. 4. Therefore, the function rule is \(g(x) = -\frac{1}{x - 3} - 1\).

Reflect the graph of \(f\) across the \(x\)-axis, then translate it \(3\) units right and \(1\) unit down. The function rule is \(g(x) = -\frac{1}{x - 3} - 1\).

- Read the lines approached by the graph. - Read the axis crossings. - Check the asymptotes and intercepts of each candidate rule.

1. From the graph, the vertical asymptote is \(x=1\), and the horizontal asymptote is \(y=2\). 2. The graph crosses the y-axis at \((0, -4)\) and the x-axis at \((-2, 0)\). 3. Rule I has vertical asymptote \(x=1\), horizontal asymptote \(y=2\), x-intercept \((-2, 0)\), and y-intercept \((0, -4)\). Therefore, it matches every displayed feature. 4. Rule II has horizontal asymptote \(y=1\), and Rule III has vertical asymptote \(x=-1\), so neither matches.

a) Vertical asymptote: \(x=1\); horizontal asymptote: \(y=2\) b) Y-intercept: \((0, -4)\); x-intercept: \((-2, 0)\) c) I) \(f(x)=\frac{2x+4}{x-1}\)

- The vertical asymptote occurs where the denominator is zero. - The constant added outside the fraction determines the vertical shift. - Analyze the sign, scale factor, and translations one at a time. - Set the function equal to zero to find the x-intercept.

1. The denominator is zero at \(x = -3\), so the vertical asymptote is \(x = -3\). As \(|x|\) increases, \(\frac{-2}{x + 3}\) approaches \(0\), so the horizontal asymptote is \(y = 1\). 2. Starting from \(h(x) = \frac{1}{x}\), reflect across the x-axis, stretch vertically by a factor of \(2\), shift left \(3\) units, and shift up \(1\) unit. 3. Set \(f(x) = 0\): \(0 = \frac{-2}{x + 3} + 1\). Then \(\frac{2}{x + 3} = 1\), so \(x + 3 = 2\) and \(x = -1\).

a) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = 1\) b) Reflect across the x-axis, stretch vertically by \(2\), shift left \(3\), and shift up \(1\). c) The x-intercept is \((-1, 0)\).

- Compare each graph’s vertical and horizontal asymptotes with those of \(\frac{1}{x}\). - Use the quadrants containing the branches to identify a reflection. - Substitute a point with integer coordinates to determine a scale factor.

1. For \(g\), the vertical asymptote moves from \(x=0\) to \(x=-2\), so the graph shifts left \(2\) units. The horizontal asymptote moves from \(y=0\) to \(y=1\), so the graph shifts up \(1\) unit. 2. Therefore, \(g(x)=\frac{1}{x+2}+1\). 3. For \(h\), both asymptotes remain unchanged. The branches move to Quadrants II and IV, so the graph is reflected across the x-axis. 4. The point \((1, -2)\) lies on \(h\), so the reflected graph is also stretched vertically by a factor of \(2\). Therefore, \(h(x)=-\frac{2}{x}\).

\(g\): shift left \(2\) units and up \(1\) unit; \(g(x)=\frac{1}{x+2}+1\) \(h\): reflect across the x-axis and stretch vertically by a factor of \(2\); \(h(x)=-\frac{2}{x}\)

- Identify the horizontal asymptote before reflecting. - A reflection across the horizontal asymptote reverses the sign of the fraction term. - A translation left replaces \(x\) with \(x + d\). - Determine the new vertical asymptote from the final denominator.

1. The horizontal asymptote of \(f\) is \(y = 2\). 2. Reflecting across \(y = 2\) reverses the signed distance from that line. Therefore, the reciprocal term changes sign while the constant remains \(2\): \(r(x) = -\frac{1}{x - 4} + 2\). 3. Translating \(r\) \(3\) units left replaces \(x\) with \(x + 3\): \(g(x) = r(x + 3) = -\frac{1}{(x + 3) - 4} + 2 = -\frac{1}{x - 1} + 2\). 4. The denominator is zero at \(x = 1\), so the vertical asymptote is \(x = 1\). The horizontal asymptote remains \(y = 2\).

\(g(x) = -\frac{1}{x - 1} + 2\); vertical asymptote: \(x = 1\); horizontal asymptote: \(y = 2\)

- First find the intersection of the asymptotes. - What slopes do the lines of symmetry of a rectangular hyperbola have? - Use point-slope form for a line through the center.

1. The vertical asymptote is \(x = -3\), and the horizontal asymptote is \(y = -1\). Their intersection is the center \((-3, -1)\). 2. The two lines of symmetry pass through the center and have slopes \(1\) and \(-1\). 3. For slope \(1\), use point-slope form: \(y + 1 = x + 3\), so \(y = x + 2\). 4. For slope \(-1\), \(y + 1 = -(x + 3)\), so \(y = -x - 4\).

The lines of symmetry are \(y = x + 2\) and \(y = -x - 4\).

- Substitute \(x = 3\) into each function. - In part b), solve both equations for \(x\) before forming the ratio. - Compare outputs for equal inputs and inputs for equal outputs separately.

1. Evaluate at \(x = 3\): \(f(3) = \frac{18}{3} = 6\), and \(g(3) = \frac{36}{3} = 12\). 2. Solve each equation for the input: \(x_f = \frac{18}{y_0}\) and \(x_g = \frac{36}{y_0}\). 3. Therefore, \(\frac{x_g}{x_f} = \frac{36/y_0}{18/y_0} = 2\). 4. For the same positive input, \(g(x) = 2f(x)\), so the point on \(g\) is twice as far above the \(x\)-axis. 5. For the same positive output, \(x_g = 2x_f\), so the point on \(g\) is twice as far to the right of the \(y\)-axis. This is a stretch, not a translation.

a) \(f(3) = 6\) and \(g(3) = 12\) b) \(\frac{x_g}{x_f} = 2\) c) For equal positive inputs, the graph of \(g\) is farther from the \(x\)-axis. For equal positive outputs, it is farther from the \(y\)-axis.

- Can the numerator be written as the denominator plus a constant? - Which input is excluded by the denominator? - What point is formed by the intersection of the asymptotes? - To find intersections with \(y = x\), set \(f(x) = x\).

1. Rewrite the numerator as \(x + 3 = (x - 1) + 4\): \(f(x) = \frac{(x - 1) + 4}{x - 1} = 1 + \frac{4}{x - 1}\). 2. Therefore, the vertical asymptote is \(x = 1\), and the horizontal asymptote is \(y = 1\). 3. Their intersection is the center of symmetry \(Z(1, 1)\). 4. Set \(f(x) = x\): \(\frac{x + 3}{x - 1} = x\). Then \(x + 3 = x(x - 1)\), so \(x^2 - 2x - 3 = 0\). Factoring gives \((x - 3)(x + 1) = 0\), so \(x = 3\) or \(x = -1\). The intersection points are \((3, 3)\) and \((-1, -1)\).

1) \(f(x) = \frac{4}{x - 1} + 1\) 2) Vertical asymptote: \(x = 1\); horizontal asymptote: \(y = 1\) 3) \(Z(1, 1)\) 4) \((3, 3)\) and \((-1, -1)\)