- Read the coordinates where the curve crosses each coordinate axis. - At the y-intercept, the input is \(0\); use that graph-read point in the given function form. - After finding \(k\), rewrite the function as one fraction before checking the x-intercept. - An x-intercept must be a point in the function's domain.

1. From the graph, the y-intercept is \((0,-6)\) and the x-intercept is \((3,0)\). 2. At the y-intercept, \(x=0\). Substitute \((0,-6)\) into the given form: \(-6=2+\frac{k}{1}\), so \(k=-8\). 3. Thus, \(g(x)=2-\frac{8}{x+1}=\frac{2x-6}{x+1}\). 4. For the x-intercept, set the numerator equal to zero: \(2x-6=0\), so \(x=3\). Since \(3+1\ne0\), the x-intercept is \((3,0)\), agreeing with the graph. 5. At \(x=-1\), the denominator is zero, so \(g\) is undefined there. Therefore, \(x=-1\) cannot be an x-intercept.

a) x-intercept: \((3,0)\); y-intercept: \((0,-6)\) b) \(k=-8\); the exact x-intercept is \((3,0)\) c) \(x=-1\) is excluded from the domain because the denominator is zero.

- The asymptotes determine \(h\) and \(k\) directly. - Substitute the given point after writing the function with an unknown \(a\). - Solve the resulting equation for \(a\).

1. The vertical asymptote \(x = 3\) gives \(h = 3\), and the horizontal asymptote \(y = 2\) gives \(k = 2\). 2. The function has the form \(f(x) = \frac{a}{x - 3} + 2\). 3. Substitute \(P(5, 3)\): \(3 = \frac{a}{5 - 3} + 2\). 4. This gives \(1 = \frac{a}{2}\), so \(a = 2\). 5. Therefore, \(f(x) = \frac{2}{x - 3} + 2\).

\(a = 2\), \(h = 3\), and \(k = 2\); \(f(x) = \frac{2}{x - 3} + 2\)

- Use the vertical asymptote to determine the denominator shift. - Use the horizontal asymptote to determine the vertical shift. - Substitute the given \(y\)-intercept to solve for \(a\).

1. The vertical asymptote \(x = -1\) means \(-d = -1\), so \(d = 1\). 2. The horizontal asymptote \(y = 4\) gives \(c = 4\). Therefore, \(g(x) = \frac{a}{x + 1} + 4\). 3. Substitute the \(y\)-intercept \((0, 1)\): \(1 = \frac{a}{0 + 1} + 4\). 4. Thus, \(1 = a + 4\), so \(a = -3\). 5. The complete function rule is \(g(x) = \frac{-3}{x + 1} + 4\).

\(a = -3\), and \(g(x) = \frac{-3}{x + 1} + 4\)

- The coordinates of the asymptote intersection give the horizontal and vertical shifts. - An \(x\)-intercept has a function value of \(0\). - Substitute all known values into the general form to solve for the scale factor.

1. The asymptote intersection \((2, -1)\) gives \(h_0 = 2\) and \(k = -1\). Therefore, \(h(x) = \frac{a}{x - 2} - 1\). 2. The \(x\)-intercept at \(x = 4\) means that \((4, 0)\) lies on the graph. Substitute: \(0 = \frac{a}{4 - 2} - 1\). 3. Then \(1 = \frac{a}{2}\), so \(a = 2\). 4. Therefore, the function rule is \(h(x) = \frac{2}{x - 2} - 1\).

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

- A point on the graph must satisfy the function equation. - Substitute the point's \(x\)-coordinate for \(x\) and its \(y\)-coordinate for \(f(x)\). - Isolate the parameter step by step.

1. Substitute the coordinates of \(P(1, 1)\): \(1 = \frac{k}{1 + 3} - 2\). 2. Then \(1 = \frac{k}{4} - 2\), so \(3 = \frac{k}{4}\). 3. Multiply by \(4\): \(k = 12\).

\(k = 12\)

- The horizontal translation and vertical translation determine the asymptotes. - Set the denominator equal to zero to locate the vertical asymptote. - Substitute the coordinates of the given point to find \(a\).

1. The horizontal asymptote \(y=4\) gives \(y_0=4\). 2. The vertical asymptote \(x=-1\) gives \(x_0=-1\). 3. Thus, \(f(x)=\frac{a}{x+1}+4\). Substitute \(P=(1, 5)\): \(5=\frac{a}{2}+4\). 4. Therefore, \(1=\frac{a}{2}\), so \(a=2\). The function is \(f(x)=\frac{2}{x+1}+4\).

\(a=2\), \(x_0=-1\), \(y_0=4\), and \(f(x)=\frac{2}{x+1}+4\)

- An x-intercept comes from a numerator zero that is in the domain. - A vertical asymptote comes from a noncanceling denominator zero. - An even multiplicity produces the same sign on both sides of the asymptote. - Compare the behavior of \(\frac{1}{x}\) and \(\frac{1}{x^2}\).

1. An x-intercept at \(x=5\) requires the numerator factor \((x-5)\). 2. A vertical asymptote at \(x=-3\) without a sign change can be created by the even-power denominator factor \((x+3)^2\). 3. Therefore, one possible function is \(f(x)=\frac{x-5}{(x+3)^2}\).

One possible formula is \(f(x)=\frac{x-5}{(x+3)^2}\).

- Use a denominator factor to exclude \(x=4\). - Use an odd multiplicity to create a sign change. - Match numerator and denominator degrees and leading coefficients to obtain \(y=-2\).

1. To exclude only \(x=4\) and create a sign-changing vertical asymptote, use the denominator factor \((x-4)\) to an odd power. Choose \(x-4\). 2. For horizontal asymptote \(y=-2\), use numerator and denominator polynomials of the same degree with leading-coefficient ratio \(-2\). 3. One example is \(f(x)=\frac{-2x}{x-4}\). At \(x=4\), the numerator equals \(-8\), so the factor does not cancel, and the vertical asymptote remains.

One possible formula is \(f(x)=\frac{-2x}{x-4}\).

- First identify the vertical and horizontal asymptotes. - Use the asymptotes to determine the horizontal and vertical translations. - Test a point with integer coordinates to determine the sign of the numerator. - Notice which two regions formed by the asymptotes contain the branches.

1. The graph approaches the vertical line \(x = 1\) and the horizontal line \(y = 2\). Therefore, the function has the form \(f(x) = \frac{a}{x - 1} + 2\). 2. The point \((0, 3)\) lies on the graph. Substitute it into the function rule: \(3 = \frac{a}{0 - 1} + 2\). 3. Solving gives \(3 = -a + 2\), so \(a = -1\). 4. Thus, \(f(x) = -\frac{1}{x - 1} + 2\), which is choice b).

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

- The vertical asymptote gives \(h\). - The horizontal asymptote gives \(k\). - Substitute the marked point after finding \(h\) and \(k\).

1. The vertical asymptote is \(x = 2\), so \(h = 2\). 2. The horizontal asymptote is \(y = 1\), so \(k = 1\). The function has the form \(f(x) = \frac{a}{x - 2} + 1\). 3. The marked point is \((5, 2)\). Substitute it: \(2 = \frac{a}{5 - 2} + 1\). 4. This gives \(1 = \frac{a}{3}\), so \(a = 3\). 5. Therefore, \(f(x) = \frac{3}{x - 2} + 1\).

\(h = 2\), \(k = 1\), and \(a = 3\); \(f(x) = \frac{3}{x - 2} + 1\)

- Use the vertical asymptote to find \(h\). - Use the horizontal asymptote to find \(k\). - Substitute the marked point to solve for \(a\).

1. The vertical asymptote is \(x = -3\), so \(h = -3\). 2. The horizontal asymptote is \(y = -1\), so \(k = -1\). The function has the form \(f(x) = \frac{a}{x + 3} - 1\). 3. The marked point is \((-1, -2)\). Substitute it: \(-2 = \frac{a}{-1 + 3} - 1\). 4. This gives \(-1 = \frac{a}{2}\), so \(a = -2\). 5. Therefore, \(f(x) = -\frac{2}{x + 3} - 1\).

\(a = -2\), \(h = -3\), and \(k = -1\); \(f(x) = -\frac{2}{x + 3} - 1\)

- The vertical asymptote gives the value of \(h\). - The horizontal asymptote gives the value of \(k\). - Substitute the marked point to determine \(a\).

1. The vertical asymptote is \(x = 3\), so \(h = 3\). 2. The horizontal asymptote is \(y = -4\), so \(k = -4\). The function has the form \(f(x) = \frac{a}{x - 3} - 4\). 3. The marked point is \((6, -2)\). Substitute it: \(-2 = \frac{a}{6 - 3} - 4\). 4. This gives \(2 = \frac{a}{3}\), so \(a = 6\). 5. Therefore, \(f(x) = \frac{6}{x - 3} - 4\).

\(a = 6\), \(h = 3\), and \(k = -4\); \(f(x) = \frac{6}{x - 3} - 4\)

- What horizontal value do the branches get closer to far from the origin? - How does that value appear in the function form? - Read the marked point and substitute its coordinates to find the missing coefficient.

1. The graph approaches the horizontal line \(y=2\), so \(c=2\) and \(f(x)=\frac{a}{x}+2\). The vertical asymptote is the y-axis, \(x=0\), consistent with the stated form. 2. Read the point \(P=(1,-1)\) from the graph. Substitution gives \(-1=\frac{a}{1}+2\), so \(a=-3\). 3. Therefore, \(f(x)=-\frac{3}{x}+2\).

\(a=-3\), \(c=2\); \(f(x)=-\frac{3}{x}+2\).

- Read the asymptotes carefully; each grid interval represents \(0.5\) unit. - Use the vertical asymptote to write the denominator. - Substitute the marked point to determine the numerator.

1. The vertical asymptote is \(x = -2.5\), and the horizontal asymptote is \(y = -0.5\). 2. Therefore, the function has the form \(f(x) = \frac{a}{x + 2.5} - 0.5\). 3. The marked point is \((-1.5, 0.5)\). Substitute it: \(0.5 = \frac{a}{-1.5 + 2.5} - 0.5\). 4. This gives \(1 = \frac{a}{1}\), so \(a = 1\). 5. Thus, \(f(x) = \frac{1}{x + 2.5} - 0.5\).

\(f(x) = \frac{1}{x + 2.5} - 0.5\)

- Evaluate the function at \(x=0\). - Read the y-intercept of each graph. - For these functions, the y-intercept is also the extreme value.

1. At \(x=0\), \(j_a(0)=a\). Therefore, the y-intercept directly gives the parameter. 2. Graph p has y-intercept \(2\), so \(a=2\). Graph q has y-intercept \(4\), so \(a=4\). Graph r has y-intercept \(-1\), so \(a=-1\).

p) \(a=2\) q) \(a=4\) r) \(a=-1\)

- Identify the horizontal line approached by the graph. - Determine where the graph has a vertical break. - Substitute a point with integer coordinates to find the numerator. - A vertical translation adds a constant outside the fraction.

1. The graph approaches the horizontal line \(y = 3\), so the function has the form \(f(x) = \frac{a}{x} + 3\). 2. The vertical asymptote is the y-axis, \(x = 0\), so no horizontal translation is needed. 3. The point \((1, 4)\) lies on the graph. Substitute it: \(4 = \frac{a}{1} + 3\). 4. Therefore, \(a = 1\), and \(f(x) = \frac{1}{x} + 3\).

One possible function rule is \(f(x) = \frac{1}{x} + 3\).

- Use the vertical asymptote to find \(h\). - Use the horizontal asymptote to find \(k\). - Substitute a point on the graph to solve for \(a\).

1. The vertical asymptote is \(x = -1\), so \(h = -1\). 2. The horizontal asymptote is \(y = -2\), so \(k = -2\). Thus, \(f(x) = \frac{a}{x + 1} - 2\). 3. The graph crosses the y-axis at \((0, -1)\). Substitute this point: \(-1 = \frac{a}{0 + 1} - 2\). 4. Solving gives \(a = 1\). 5. Therefore, \(f(x) = \frac{1}{x + 1} - 2\).

\(h = -1\), \(k = -2\), and \(a = 1\); \(f(x) = \frac{1}{x + 1} - 2\)

- Read the two lines that the graph approaches. - Use these to fill in the unknown shifts in a translated reciprocal-function rule. - Choose a point with readable coordinates from the grid to determine the remaining coefficient.

1. The graph has vertical asymptote \(x=3\) and horizontal asymptote \(y=1\). 2. Use \(f(x)=\frac{a}{x-3}+1\). 3. The curve passes through \((4,3)\). Substituting gives \(3=\frac{a}{4-3}+1\), so \(a=2\). 4. Therefore, \(f(x)=\frac{2}{x-3}+1\).

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

- What does the horizontal asymptote tell you about the constant term? - Substitute the known graph point to find the other coefficient. - Compare the signs of the two fractional terms for positive and negative inputs. - In the final equation, isolate the fraction before solving for the input.

1. The horizontal asymptote gives \(c=-2\). Substituting \(A(4,-1)\) gives \(-1=\frac{a}{4}-2\), so \(a=4\). 2. Hence \(p(x)=\frac{4}{x}-2\), while \(q(x)=-\frac{4}{x}-2\). Their common asymptotes are \(x=0\) and \(y=-2\). 3. Relative to the center \((0,-2)\), the branches of \(p\) lie above and right, and below and left. The branches of \(q\) lie above and left, and below and right. They are reflections across \(y=-2\). 4. Solve \(-1.5=\frac{4}{x}-2\): \(\frac{4}{x}=0.5\), so \(x=8\).

a) \(a=4\), \(c=-2\) b) The branches switch from upper-right/lower-left to upper-left/lower-right relative to \((0,-2)\); the graphs are reflected across \(y=-2\). c) \(x=8\)

- If a point lies on a graph, its coordinates satisfy the function rule. - Rearrange the resulting equation to isolate \(k\). - Write the complete function rule before checking the second point. - Compare the point's \(y\)-coordinate with the function value at the same input.

1. Substitute \(Q(6, 1)\): \(1 = \frac{k}{6 - 4} - 3\). Then \(4 = \frac{k}{2}\), so \(k = 8\). 2. The function is \(h(x) = \frac{8}{x - 4} - 3\). 3. Evaluate at \(x = 2\): \(h(2) = \frac{8}{2 - 4} - 3 = -7\). 4. Since the point has \(y = -5\) and \(-5 > -7\), point \(R\) lies above the graph.

1) \(k = 8\) 2) Point \(R(2, -5)\) lies above the graph.

- Use the vertical asymptote to determine the horizontal shift \(h\). - Substitute each intercept into the function to create two equations. - Combine the equations to eliminate one variable.

1. The vertical asymptote \(x = 2\) gives \(h = 2\), so \(f(x) = \frac{a}{x - 2} + k\). 2. Substitute the \(x\)-intercept \((4, 0)\): \(0 = \frac{a}{4 - 2} + k\), so \(\frac{a}{2} + k = 0\). 3. Substitute the \(y\)-intercept \((0, -1)\): \(-1 = \frac{a}{0 - 2} + k\), so \(-\frac{a}{2} + k = -1\). 4. Add the two equations: \(2k = -1\), so \(k = -\frac{1}{2}\). 5. Substitute \(k = -\frac{1}{2}\) into \(\frac{a}{2} + k = 0\): \(\frac{a}{2} - \frac{1}{2} = 0\), so \(a = 1\).

\(a = 1\) and \(k = -\frac{1}{2}\)

- How are the horizontal and vertical shifts related to the asymptotes? - Substitute the known graph point into the partially filled function rule. - Compare the output's signed distance from the horizontal asymptote before and after the sign change.

1. The asymptotes give \(h=-2\) and \(k=1\), so \(f(x)=\frac{a}{x+2}+1\). 2. Substitute \(P(0,2)\): \(2=\frac{a}{2}+1\), giving \(a=2\). 3. Thus \(f(x)=\frac{2}{x+2}+1\). Replacing \(a=2\) with \(a=-2\) reverses the sign of the fractional term, reflecting the graph across its horizontal asymptote \(y=1\).

a) \(h=-2\), \(k=1\), \(a=2\) b) \(f(x)=\frac{2}{x+2}+1\) c) The graph is reflected across \(y=1\).

- Substitute the coordinates of each intercept into the general function rule. - The two substitutions create a system of two equations. - Eliminate one variable or express one variable in terms of the other.

1. Substitute the \(y\)-intercept \((0, 3)\): \(3 = \frac{a}{0 + 1} + c\), so \(a + c = 3\). 2. Substitute the \(x\)-intercept \((2, 0)\): \(0 = \frac{a}{2 + 1} + c\), so \(\frac{a}{3} + c = 0\). 3. Subtract the second equation from the first: \(a - \frac{a}{3} = 3\). Thus, \(\frac{2a}{3} = 3\), so \(a = 4.5\). 4. Substitute \(a = 4.5\) into \(\frac{a}{3} + c = 0\): \(1.5 + c = 0\), so \(c = -1.5\). 5. Therefore, the function rule is \(f(x) = \frac{4.5}{x + 1} - 1.5\).

\(a = 4.5\), \(c = -1.5\), and \(f(x) = \frac{4.5}{x + 1} - 1.5\)

- A point on the graph must satisfy the function equation. - Substitute the coordinates of \(P\) first to determine \(k\). - After finding \(k\), use \(x = 0\) for the \(y\)-intercept and \(g(x) = 0\) for the \(x\)-intercept.

1. Because \(P(2, -1)\) lies on the graph, substitute \(x = 2\) and \(g(x) = -1\): \(-1 = \frac{k}{2 + 2} - 4\). 2. Then \(3 = \frac{k}{4}\), so \(k = 12\). 3. The function is \(g(x) = \frac{12}{x + 2} - 4\). For the \(y\)-intercept, substitute \(x = 0\): \(g(0) = \frac{12}{2} - 4 = 2\). Therefore, the \(y\)-intercept is \((0, 2)\). 4. For the \(x\)-intercept, set \(g(x) = 0\): \(0 = \frac{12}{x + 2} - 4\). Then \(4 = \frac{12}{x + 2}\), so \(4(x + 2) = 12\). Thus, \(x = 1\). 5. Therefore, the \(x\)-intercept is \((1, 0)\).

a) \(k = 12\) b) The \(x\)-intercept is \((1, 0)\), and the \(y\)-intercept is \((0, 2)\).

- Passing through the origin means that \(f(0) = 0\). - Use the known point to solve for the parameter. - After finding \(a\), solve for all \(x\)-intercepts and remember that every \(y\)-intercept has \(x = 0\).

1. Because the graph passes through \((0, 0)\), substitute \(x = 0\) and \(f(x) = 0\): \(0 = \frac{a}{0 - 4} + 2\). 2. Then \(0 = -\frac{a}{4} + 2\), so \(\frac{a}{4} = 2\) and \(a = 8\). 3. With \(a = 8\), the function is \(f(x) = \frac{8}{x - 4} + 2\). Its \(y\)-intercept is determined by the single input \(x = 0\), which already gives \((0, 0)\). 4. To check for another \(x\)-intercept, solve \(0 = \frac{8}{x - 4} + 2\). Then \(-2 = \frac{8}{x - 4}\), so \(x - 4 = -4\) and \(x = 0\). 5. Therefore, the origin is the only intercept of the graph.

a) \(a = 8\) b) No. The only \(x\)-intercept is \((0, 0)\), and \(x = 0\) can produce only one \(y\)-intercept, so there are no additional intercepts.

- At an \(x\)-intercept, the function value is \(0\). - Substitute the known input and output to solve for \(k\). - To find the \(y\)-intercept, substitute \(x = 0\).

1. An \(x\)-intercept at \(x = 5\) means that \((5, 0)\) lies on the graph. Substitute \(x = 5\) and \(g_k(x) = 0\): \(0 = \frac{k}{5 - 2} + 3\). 2. Then \(0 = \frac{k}{3} + 3\), so \(-3 = \frac{k}{3}\) and \(k = -9\). 3. The function is \(g(x) = \frac{-9}{x - 2} + 3\). For the \(y\)-intercept, substitute \(x = 0\): \(g(0) = \frac{-9}{0 - 2} + 3 = 4.5 + 3 = 7.5\). 4. Therefore, the \(y\)-intercept is \((0, 7.5)\).

a) \(k = -9\) b) The \(y\)-intercept is \((0, 7.5)\).

- At every point on the \(y\)-axis, the input is \(0\). - Use the known intercept to solve for the parameter before finding the other intercept. - At an \(x\)-intercept, the function value is \(0\).

1. Substitute the \(y\)-intercept \((0, -2)\): \(-2 = \frac{8}{0 - d} + 2\). 2. Then \(-4 = \frac{8}{-d}\). Multiplying by \(-d\) gives \(4d = 8\), so \(d = 2\). 3. The function is \(g(x) = \frac{8}{x - 2} + 2\). To find the \(x\)-intercept, set \(g(x) = 0\): \(0 = \frac{8}{x - 2} + 2\). 4. Then \(-2 = \frac{8}{x - 2}\), so \(-2(x - 2) = 8\). Thus, \(-2x + 4 = 8\), giving \(x = -2\). 5. Therefore, the \(x\)-intercept is \((-2, 0)\).

a) \(d = 2\) b) The \(x\)-intercept is \((-2, 0)\).

- Substitute the coordinates of the known intercept to find the parameter. - At an \(x\)-intercept, the function value is \(0\). - Determine \(k\) before beginning part b).

1. Substitute the \(y\)-intercept \((0, 1)\): \(1 = \frac{k}{0 + 4} - 2\). 2. Then \(3 = \frac{k}{4}\), so \(k = 12\). 3. The function is \(g(x) = \frac{12}{x + 4} - 2\). To find the \(x\)-intercept, set \(g(x) = 0\): \(0 = \frac{12}{x + 4} - 2\). 4. Then \(2 = \frac{12}{x + 4}\), so \(2(x + 4) = 12\). Thus, \(x + 4 = 6\) and \(x = 2\). 5. Therefore, the \(x\)-intercept is \((2, 0)\).

a) \(k = 12\) b) The \(x\)-intercept is \((2, 0)\).

- First find the input where \(f(x) = 0\). - Having the same \(x\)-intercept means that \(g\) must also equal \(0\) at that input. - Substitute the intercept into \(g\) and solve for \(a\).

1. Set \(f(x) = 0\): \(0 = \frac{4}{x - 2} + 2\). 2. Then \(-2 = \frac{4}{x - 2}\), so \(-2(x - 2) = 4\). Thus, \(x - 2 = -2\) and \(x = 0\). The \(x\)-intercept of \(f\) is \((0, 0)\). 3. For \(g\) to have the same \(x\)-intercept, \(g(0) = 0\). Substitute: \(0 = \frac{a}{0 - 2} + 4\). 4. Then \(0 = -\frac{a}{2} + 4\), so \(\frac{a}{2} = 4\) and \(a = 8\).

a) The \(x\)-intercept of \(f\) is \((0, 0)\). b) \(a = 8\)

- Substitute the coordinates of the given \(y\)-intercept to find \(a\). - Set the function equal to \(0\) to find the \(x\)-intercept. - In part c), consider when a fraction can equal \(0\).

1. Substitute the \(y\)-intercept \((0, 2)\): \(2 = \frac{24}{0 + a} - 4\). 2. Then \(6 = \frac{24}{a}\), so \(a = 4\). 3. The function is \(h(x) = \frac{24}{x + 4} - 4\). To find the \(x\)-intercept, set \(h(x) = 0\): \(0 = \frac{24}{x + 4} - 4\). 4. Then \(4 = \frac{24}{x + 4}\), so \(4(x + 4) = 24\). Thus, \(x = 2\), and the \(x\)-intercept is \((2, 0)\). 5. To test whether \(h(x) = -4\), write \(-4 = \frac{24}{x + 4} - 4\). This simplifies to \(0 = \frac{24}{x + 4}\). 6. A fraction with a nonzero numerator cannot equal \(0\), so the equation has no solution. Therefore, the graph never reaches \(y = -4\).

a) \(a = 4\) b) The \(x\)-intercept is \((2, 0)\). c) Setting \(h(x) = -4\) gives \(\frac{24}{x + 4} = 0\), which has no solution because the numerator is nonzero.

- First evaluate the first function at \(x = 0\). - Equal \(y\)-intercepts mean equal function values at \(x = 0\). - A rational expression can equal \(0\) only when its numerator is \(0\).

1. Substitute \(x = 0\) into \(h_1\): \(h_1(0) = \frac{4}{0 - 2} + 1 = -1\). Therefore, the \(y\)-intercept is \((0, -1)\). 2. For \(h_2\) to have the same \(y\)-intercept, require \(h_2(0) = -1\). Since \(h_2(0) = \frac{k}{1} = k\), it follows that \(k = -1\). 3. For \(h_1\), solve \(0 = \frac{4}{x - 2} + 1\). Then \(-1 = \frac{4}{x - 2}\), so \(x - 2 = -4\) and \(x = -2\). Thus, \(h_1\) has the \(x\)-intercept \((-2, 0)\). 4. For \(h_2(x) = \frac{-1}{x + 1}\), the numerator is the nonzero constant \(-1\). Therefore, the function can never equal \(0\), so \(h_2\) has no \(x\)-intercept.

a) The \(y\)-intercept of \(h_1\) is \((0, -1)\). b) \(k = -1\) c) \(h_2\) has no \(x\)-intercept because \(\frac{-1}{x + 1}\) cannot equal \(0\).

- At an \(x\)-intercept, the output is \(0\). - Use the given intercept to solve for the parameter. - After finding \(a\), substitute \(x = 0\).

1. The \(x\)-intercept at \(x = -4\) means \(g(-4) = 0\). Substitute: \(0 = \frac{a}{-4 + 3} + 4\). 2. Then \(0 = -a + 4\), so \(a = 4\). 3. The function is \(g(x) = \frac{4}{x + 3} + 4\). Substitute \(x = 0\): \(g(0) = \frac{4}{3} + 4 = \frac{16}{3}\). 4. Therefore, the \(y\)-intercept is \(\left(0, \frac{16}{3}\right)\).

a) \(a = 4\) b) The \(y\)-intercept is \(\left(0, \frac{16}{3}\right)\).

- Compare the two formulas relative to their common horizontal asymptote. - For the same allowed \(x\), how do the vertical displacements from \(y=3\) compare? - Substitute the coordinates of \(Q\) into \(g\), then test the same input in \(h\).

1. Both graphs have vertical asymptote \(x=2\) and horizontal asymptote \(y=3\). Changing \(a\) to \(-a\) places corresponding points the same distance on opposite sides of \(y=3\), so the graphs are reflections across \(y=3\). 2. Substitute \(Q(1,1)\) into \(g\): \(1=\frac{a}{1-2}+3\). Then \(1=-a+3\), so \(a=2\). 3. With \(a=2\), \(h(x)=\frac{-2}{x-2}+3\). Evaluate \(h(1)=\frac{-2}{-1}+3=5\). 4. Since \(h(1)=5\ne1\), \(Q(1,1)\) does not lie on the graph of \(h\).

a) The graphs are reflections of each other across \(y=3\). b) \(a=2\) c) No. Since \(h(1)=5\), \(Q(1,1)\) is not on the graph of \(h\).

- Substitute the given \(y\)-intercept to find \(a\). - Write the complete function rule before finding the next intercept. - At an \(x\)-intercept, the function value is \(0\).

1. Substitute \((0, 3)\): \(3 = \frac{a}{0 - 2} + 4\). 2. Then \(-1 = -\frac{a}{2}\), so \(a = 2\). The function is \(g(x) = \frac{2}{x - 2} + 4\). 3. Set \(g(x) = 0\): \(0 = \frac{2}{x - 2} + 4\). Then \(-4 = \frac{2}{x - 2}\), so \(-4(x - 2) = 2\). 4. Thus, \(-4x + 8 = 2\), so \(-4x = -6\) and \(x = 1.5\). 5. Therefore, the \(x\)-intercept is \((1.5, 0)\).

\(a = 2\), and the \(x\)-intercept is \((1.5, 0)\).

- Two unknown parameters require two independent equations. - Substitute each point into the general function rule. - Subtract the equations to eliminate one parameter.

1. Substitute \(A(2, 4)\): \(4 = \frac{a}{2} + c\). 2. Substitute \(B(4, 3)\): \(3 = \frac{a}{4} + c\). 3. Subtract the second equation from the first: \(1 = \frac{a}{2} - \frac{a}{4} = \frac{a}{4}\). Therefore, \(a = 4\). 4. Substitute \(a = 4\) into \(4 = \frac{a}{2} + c\): \(4 = 2 + c\), so \(c = 2\). 5. The function rule is \(h(x) = \frac{4}{x} + 2\).

\(a = 4\), \(c = 2\), and \(h(x) = \frac{4}{x} + 2\)

- A point on the graph must satisfy the function equation. - Use the given point to determine the unknown parameter. - After finding the complete function rule, set the function equal to \(0\).

1. Substitute \(P(0, 1.5)\): \(1.5 = \frac{a}{0 + 4} - 2\). 2. Then \(3.5 = \frac{a}{4}\), so \(a = 14\). 3. The function is \(f(x) = \frac{14}{x + 4} - 2\). Set \(f(x) = 0\): \(0 = \frac{14}{x + 4} - 2\). 4. Then \(2 = \frac{14}{x + 4}\), so \(2(x + 4) = 14\). Thus, \(x + 4 = 7\) and \(x = 3\). 5. Therefore, the \(x\)-intercept is \((3, 0)\).

a) \(a = 14\) b) The \(x\)-intercept is \((3, 0)\).

- At an \(x\)-intercept, the output is \(0\). - Substitute both coordinates of the given point. - Compare what happens when \(q = 0\) with what happens when \(q \neq 0\). - A rational expression with a nonzero constant numerator cannot equal \(0\).

1. When \(q = -2\), solve \(0 = \frac{6}{x} - 2\). Then \(2 = \frac{6}{x}\), so \(x = 3\). The \(x\)-intercept is \((3, 0)\). 2. For the graph to pass through \(P(3, 4)\), substitute \(x = 3\) and \(f(x) = 4\): \(4 = \frac{6}{3} + q\). Thus, \(4 = 2 + q\), so \(q = 2\). 3. An \(x\)-intercept requires \(0 = \frac{6}{x} + q\). If \(q \neq 0\), then \(x = -\frac{6}{q}\), so there is an \(x\)-intercept. 4. If \(q = 0\), the equation becomes \(0 = \frac{6}{x}\), which has no solution because the numerator is nonzero. Therefore, \(q = 0\) is the only value for which the graph has no \(x\)-intercept.

a) The \(x\)-intercept is \((3, 0)\). b) \(q = 2\) c) \(q = 0\), because \(\frac{6}{x}\) can never equal \(0\).

- A rational expression with a nonzero constant numerator cannot equal \(0\). - Compare the cases \(k=0\) and \(k\ne0\) separately. - A linear function cannot contain the variable in a denominator.

1. Substitute \(A(4,1)\): \(1=k-\frac{8}{4}=k-2\). Therefore, \(k=3\). 2. An x-intercept satisfies \(0=k-\frac{8}{x}\). 3. If \(k\ne0\), then \(\frac{8}{x}=k\), so \(x=\frac{8}{k}\). Therefore, there is exactly one x-intercept. 4. If \(k=0\), then the equation is \(0=-\frac{8}{x}\), which has no solution because the numerator is nonzero. Therefore, there is no x-intercept. 5. Iva's claim is false. Changing \(k\) only changes the vertical shift. The term \(-\frac{8}{x}\) remains for every \(k\), so the function is never linear.

a) \(k=3\) b) If \(k\ne0\), there is exactly one x-intercept at \(x=\frac{8}{k}\). If \(k=0\), there is no x-intercept. c) The claim is false. The nonzero term \(-\frac{8}{x}\) remains for every value of \(k\), so the function is never linear.

- Use denominator factors to create the excluded domain values. - Use a numerator factor to create the required zero. - Make sure no numerator factor cancels a denominator factor. - More than one correct function may be possible.

1. To exclude exactly \(x=3\) and \(x=5\), choose the denominator \((x-3)(x-5)\). 2. To create exactly one zero at \(x=-2\), choose the numerator \(x+2\). 3. No factor cancels, and \(-2\) is in the domain. Therefore, one possible function is \(g(x)=\frac{x+2}{(x-3)(x-5)}\).

One possible formula is \(g(x)=\frac{x+2}{(x-3)(x-5)}\). The denominator excludes \(3\) and \(5\), and the numerator has its only zero at \(-2\).

- A hole can be created with a common numerator and denominator factor. - A noncanceling denominator factor creates a vertical asymptote. - Use the degrees and leading coefficients to control the horizontal asymptote. - Use an odd multiplicity at the vertical asymptote to produce a sign change.

1. A hole at \(x=1\) requires a common factor \((x-1)\) in the numerator and denominator. 2. A vertical asymptote with a sign change at \(x=-2\) can be created by including the factor \((x+2)\) to the first power in the denominator and not in the numerator. 3. To obtain the horizontal asymptote \(y=2\), use numerator and denominator polynomials of the same degree with leading-coefficient ratio \(2\). 4. One example is \(f(x)=\frac{2x(x-1)}{(x-1)(x+2)}\). It has the required hole, vertical asymptote, and horizontal asymptote.

One possible formula is \(f(x)=\frac{2x(x-1)}{(x-1)(x+2)}\).

- Represent a double zero with a squared numerator factor. - Choose a quadratic denominator that is always positive. - Use the y-intercept to determine the scale factor. - More than one correct function may be possible.

1. A double zero at \(x=-1\) requires \(p(x)=a(x+1)^2\). 2. For the domain to be all real numbers, choose a quadratic denominator with no real zeros, such as \(q(x)=x^2+1\). 3. The y-intercept condition gives \(2=f(0)=\frac{a(0+1)^2}{0^2+1}=a\), so \(a=2\). 4. Therefore, one possible function is \(f(x)=\frac{2(x+1)^2}{x^2+1}\).

One possible formula is \(f(x)=\frac{2(x+1)^2}{x^2+1}\).

- Numerator zeros produce x-intercepts when the denominator is nonzero. - Denominator zeros determine excluded domain values. - An even power in the denominator produces the same sign on both sides of a vertical asymptote. - A nonzero constant numerator has no zeros.

1. For part a, place \((x-4)\) in the numerator and \((x+1)\) in the denominator. One example is \(f(x)=\frac{x-4}{x+1}\). 2. For part b, use denominator factors \((x-2)(x+2)\) and numerator factor \(x\). One example is \(f(x)=\frac{x}{(x-2)(x+2)}=\frac{x}{x^2-4}\). 3. For part c, use a nonzero constant numerator so there are no zeros. A squared denominator factor creates a vertical asymptote without a sign change. One example is \(f(x)=\frac{1}{(x-5)^2}\).

a) \(f(x)=\frac{x-4}{x+1}\) b) \(f(x)=\frac{x}{x^2-4}\) c) \(f(x)=\frac{1}{(x-5)^2}\)

- Use numerator factors to create the required zeros and their multiplicities. - Use denominator factors to create excluded values and vertical asymptotes. - Use the ratio of leading coefficients to control a horizontal asymptote. - Substitute a given point to determine an unknown scale factor.

1. For part a, use numerator factors \((x-3)(x+3)\). To obtain horizontal asymptote \(y=2\), choose a denominator of the same degree with no common factors and make the leading-coefficient ratio \(2\). One example is \(g(x)=\frac{2(x^2-9)}{x^2+1}\). 2. For part b, a nonzero constant numerator gives no zeros, and the factor \((x-1)\) creates the excluded value. Let \(g(x)=\frac{c}{x-1}\). Since \(g(0)=4\), \(4=\frac{c}{-1}\), so \(c=-4\). Thus, \(g(x)=\frac{-4}{x-1}\). 3. For part c, tangency to the x-axis at \(x=2\) requires a double zero, so use \((x-2)^2\) in the numerator. The y-axis is \(x=0\), so place \(x\) in the denominator. One example is \(g(x)=\frac{(x-2)^2}{x}\).

a) \(g(x)=\frac{2(x^2-9)}{x^2+1}\) b) \(g(x)=\frac{-4}{x-1}\) c) \(g(x)=\frac{(x-2)^2}{x}\)

- Choose the degrees and leading coefficients so the function decreases without bound. - Use a denominator factor to create the vertical asymptote. - It may help to begin with a polynomial plus a proper rational expression. - Check that the numerator is not zero at the vertical asymptote.

1. To make \(f(x)\to-\infty\), choose a rational function whose polynomial part has a negative leading term. Start with \(f(x)=-x+\frac{1}{x-5}\). 2. The denominator factor \((x-5)\) creates a vertical asymptote at \(x=5\), and the numerator of the fraction is nonzero there. 3. Write the expression as one rational function: \(f(x)=\frac{-x(x-5)+1}{x-5}=\frac{-x^2+5x+1}{x-5}\). 4. As \(x\to\infty\), the term \(-x\) dominates, so \(f(x)\to-\infty\). At \(x=5\), the numerator equals \(1\), so the vertical asymptote is not canceled.

One possible function is \(f(x)=\frac{-x^2+5x+1}{x-5}\).

- Use the vertical asymptotes to construct the denominator. - Use equal numerator and denominator degrees for a nonzero horizontal asymptote. - Passing through the origin requires \(f(0)=0\). - Check that no denominator factor cancels.

1. To create vertical asymptotes at \(x=-1\) and \(x=2\), choose \(q(x)=(x+1)(x-2)=x^2-x-2\). 2. For horizontal asymptote \(y=1.5\), use a quadratic numerator with leading coefficient \(1.5\). 3. Passing through the origin requires the numerator to equal zero at \(x=0\). Choose \(p(x)=1.5x^2\). 4. Therefore, one possible function is \(f(x)=\frac{1.5x^2}{x^2-x-2}\). The numerator is nonzero at \(x=-1\) and \(x=2\), so both vertical asymptotes remain.

One possible formula is \(f(x)=\frac{1.5x^2}{x^2-x-2}\).

- Use a numerator factor to create the zero. - Use an even-power denominator factor to avoid a sign change at the asymptote. - Check the signs of the numerator and denominator near \(x=-1\).

1. A zero at \(x=3\) requires the numerator factor \((x-3)\). 2. A vertical asymptote at \(x=-1\) without a sign change can be produced by the squared denominator factor \((x+1)^2\). 3. For \(f(x)=\frac{x-3}{(x+1)^2}\), the denominator approaches \(0\) through positive values as \(x\) approaches \(-1\) from either side, while the numerator approaches \(-4\). Therefore, the function falls without bound on both sides, so all conditions are satisfied.

One possible formula is \(f(x)=\frac{x-3}{(x+1)^2}\).

- Use an even power in the denominator to avoid a sign change at the vertical asymptote. - Write the function as the horizontal-asymptote value plus a rational term that approaches zero. - Substitute the given point to determine the remaining parameter. - Check that the numerator is nonzero at \(x=0\).

1. A vertical asymptote at \(x=0\) without a sign change can be created with the denominator \(x^2\). 2. To build in the horizontal asymptote \(y=1\), use \(h(x)=1+\frac{a}{x^2}\). 3. Substitute \(P=(2, 2)\): \(2=1+\frac{a}{2^2}\). Thus, \(1=\frac{a}{4}\), so \(a=4\). 4. Therefore, \(h(x)=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\).

One possible formula is \(h(x)=\frac{x^2+4}{x^2}\).

- Use the vertical asymptote to choose a denominator factor. - Use the ratio of leading coefficients to obtain the horizontal asymptote. - Substitute \(x=0\) to use the y-intercept. - Check that no factor cancels at \(x=3\).

1. Use the form \(f(x)=\frac{ax+b}{x-3}\), which has a possible vertical asymptote at \(x=3\). 2. The horizontal asymptote is the ratio of leading coefficients, so \(a=-1\). 3. The y-intercept condition gives \(2=f(0)=\frac{b}{-3}\), so \(b=-6\). 4. Therefore, one possible function is \(f(x)=\frac{-x-6}{x-3}\). The numerator is nonzero at \(x=3\), so the vertical asymptote remains.

One possible formula is \(f(x)=\frac{-x-6}{x-3}\).

- A vertical asymptote comes from a noncanceling denominator factor. - A hole comes from a common numerator and denominator factor. - Use the leading-coefficient ratio for the horizontal asymptote. - Substitute the given point to determine the remaining factor.

1. Use the denominator factors \((x-4)(x-1)\). The factor \((x-4)\) creates the vertical asymptote, while \((x-1)\) will create a hole. 2. Include \((x-1)\) in the numerator so it cancels algebraically but remains excluded from the original domain. 3. To obtain horizontal asymptote \(y=2\), use a quadratic numerator with leading coefficient twice that of the denominator. Let \(h(x)=\frac{2(x-r)(x-1)}{(x-4)(x-1)}\). 4. Substitute \(P=(0, 1)\): \(1=\frac{2(0-r)(-1)}{(-4)(-1)}=\frac{r}{2}\), so \(r=2\). 5. Therefore, one possible function is \(h(x)=\frac{2(x-2)(x-1)}{(x-4)(x-1)}\).

One possible formula is \(h(x)=\frac{2(x-2)(x-1)}{(x-4)(x-1)}\).

- An even function is symmetric about the y-axis. - Use denominator factors for the two vertical asymptotes. - Tangency at an x-intercept requires an even-multiplicity zero. - Use the leading-coefficient ratio for the horizontal asymptote.

1. Symmetry about the y-axis suggests an even function containing only even powers of \(x\). 2. The vertical asymptotes suggest the denominator \((x-3)(x+3)=x^2-9\). 3. Tangency to the x-axis at the origin requires a zero of even multiplicity, so use the numerator factor \(x^2\). 4. To obtain horizontal asymptote \(y=-1\), the ratio of leading coefficients must be \(-1\). Therefore, choose numerator \(-x^2\). 5. One possible function is \(h(x)=\frac{-x^2}{x^2-9}\).

One possible formula is \(h(x)=\frac{-x^2}{x^2-9}\).

- Use \(V=\pi r^2h\) for the cylinder volume. - The open tank consists of one circular base and one lateral surface. - Identify which term of the cost function dominates at each end of the domain.

1. The volume equation is \(\pi r^2h=2\). Therefore, \(h(r)=\frac{2}{\pi r^2}\) for \(r>0\). 2. The base costs \(30\pi r^2\). The lateral area is \(2\pi rh\), so the side costs \(20(2\pi rh)=40\pi rh\). Substitute \(h(r)\): \(K(r)=30\pi r^2+40\pi r\left(\frac{2}{\pi r^2}\right)=30\pi r^2+\frac{80}{r}\). 3. As \(r\to0^+\), \(\frac{80}{r}\to\infty\), so \(K(r)\to\infty\). A very narrow tank must be extremely tall, making the side area very large. 4. As \(r\to\infty\), \(30\pi r^2\to\infty\), so \(K(r)\to\infty\). A very wide tank requires an extremely large base.

a) \(h(r)=\frac{2}{\pi r^2}\), \(r>0\) b) \(K(r)=30\pi r^2+\frac{80}{r}\), \(r>0\) c) \(K(r)\to\infty\) as \(r\to0^+\) and as \(r\to\infty\). The side cost dominates for very small \(r\), and the base cost dominates for very large \(r\).

- The center coordinates determine the vertical and horizontal asymptotes. - Substitute the given point to find \(k\). - Combine the terms using a common denominator. - Set \(x=0\) for the y-intercept and set the numerator equal to zero for the x-intercept.

1. The center \(Z=(4, 3)\) gives the asymptotes \(x=4\) and \(y=3\). Therefore, \(h(x)=\frac{k}{x-4}+3\). 2. Substitute \(P=(2, 5.5)\): \(5.5=\frac{k}{2-4}+3\). Thus, \(2.5=\frac{k}{-2}\), so \(k=-5\). 3. Therefore, \(h(x)=\frac{-5}{x-4}+3=\frac{-5+3(x-4)}{x-4}=\frac{3x-17}{x-4}\). 4. For the y-intercept, \(h(0)=\frac{-17}{-4}=\frac{17}{4}=4.25\), giving \((0, 4.25)\). 5. For the x-intercept, solve \(3x-17=0\). Thus, \(x=\frac{17}{3}\), giving \(\left(\frac{17}{3}, 0\right)\).

a) \(h(x)=\frac{-5}{x-4}+3\) b) \(h(x)=\frac{3x-17}{x-4}\) c) x-intercept: \(\left(\frac{17}{3}, 0\right)\); y-intercept: \((0, 4.25)\)

- Add the two fractions using a common denominator. - Convert a decimal part of an hour to minutes by multiplying by \(60\). - Identify where a denominator approaches \(0\), and compare degrees for the end behavior. - Interpret the sign of the ground speed \(x-40\).

1. Use the common denominator \((x+40)(x-40)=x^2-1600\): \(T(x)=\frac{600(x-40)+600(x+40)}{x^2-1600}=\frac{1200x}{x^2-1600}\). 2. \(T(160)=\frac{1200\cdot160}{160^2-1600}=8\), so the trip takes \(8\) hours. 3. \(T(200)=\frac{1200\cdot200}{200^2-1600}=6.25\), so the trip takes \(6\) hours \(15\) minutes. 4. In the relevant domain, the vertical asymptote is \(x=40\). Since the denominator has greater degree than the numerator, the horizontal asymptote is \(y=0\). 5. If \(0<x<40\), then the plane's ground speed into the wind, \(x-40\), is negative. The plane cannot reach the destination against the wind, so the travel-time expression is not physically meaningful.

a) \(T(x)=\frac{1200x}{x^2-1600}\) b) At \(160\,\text{mph}\): \(8\) hours; at \(200\,\text{mph}\): \(6\) hours \(15\) minutes c) Vertical asymptote: \(x=40\); horizontal asymptote: \(y=0\) d) The headwind exceeds the plane's airspeed, so the plane cannot make forward progress on one leg.

- Use a numerator factor for the zero. - Use an odd denominator multiplicity for a sign-changing vertical asymptote. - Use an even denominator multiplicity for no sign change. - Make the numerator degree smaller than the denominator degree to obtain horizontal asymptote \(y=0\).

1. Use the numerator factor \((x+3)\) to create the zero at \(x=-3\). 2. Use the first-power denominator factor \((x-2)\) to create a sign-changing vertical asymptote at \(x=2\). 3. Use the squared denominator factor \((x+1)^2\) to create a vertical asymptote without a sign change at \(x=-1\). 4. The numerator has degree \(1\), and the denominator has degree \(3\), so the horizontal asymptote is \(y=0\). 5. One possible function is \(f(x)=\frac{x+3}{(x-2)(x+1)^2}\).

One possible formula is \(f(x)=\frac{x+3}{(x-2)(x+1)^2}\).

- A point on the graph must satisfy the function equation. - Compare \(g(x)\) with \(f(x)\) after finding \(a\). - Determine the sign of the quotient when the numerator and denominator have opposite signs.

1. Substitute \(P(0.5, 8)\): \(8 = \frac{a}{0.5}\). Multiplying by \(0.5\) gives \(a = 4\). 2. Therefore, \(g(x) = \frac{4}{x} = 4f(x)\). The graph of \(g\) is a vertical stretch of the graph of \(f\) by a factor of \(4\). 3. If \(a < 0\), then \(x\) and \(g(x)\) have opposite signs. Positive inputs give negative outputs, placing one branch in Quadrant IV; negative inputs give positive outputs, placing the other branch in Quadrant II.

a) \(a = 4\) b) The graph of \(g\) is a vertical stretch of the graph of \(f\) by a factor of \(4\). c) For \(a < 0\), the branches lie in Quadrants II and IV because the input and output have opposite signs.

- Which lines does the graph approach without reaching? - Which input makes the denominator \(0\), and how does that determine \(d\)? - What remains of the function as \(|x|\) becomes very large? - Substitute a clearly readable point to find \(a\).

1. The graph approaches the vertical line \(x = -2\) and the horizontal line \(y = 1\), so these are the asymptotes. 2. Since the vertical asymptote is \(x = d\), \(d = -2\). Since the horizontal asymptote is \(y = c\), \(c = 1\). 3. The function has the form \(f(x) = \frac{a}{x + 2} + 1\). 4. Substitute the graph point \(P(1, 0)\): \(0 = \frac{a}{1 + 2} + 1\). Then \(-1 = \frac{a}{3}\), so \(a = -3\). 5. Therefore, \(f(x) = -\frac{3}{x + 2} + 1\).

a) Vertical asymptote: \(x = -2\); horizontal asymptote: \(y = 1\) b) \(d = -2\) and \(c = 1\) c) \(a = -3\); \(f(x) = -\frac{3}{x + 2} + 1\)

- Identify the lines that the graph approaches. - How do those asymptotes determine \(c\) and \(d\)? - Substitute a clearly readable point to find \(a\). - Which input gives the y-intercept? - What equation gives the x-intercept?

1. The graph approaches the vertical line \(x = 2\) and the horizontal line \(y = 1\), so these are the asymptotes. 2. Therefore, \(d = 2\) and \(c = 1\), giving \(f(x) = \frac{a}{x - 2} + 1\). 3. Substitute the graph point \((4, 3)\): \(3 = \frac{a}{4 - 2} + 1\). Then \(2 = \frac{a}{2}\), so \(a = 4\). Thus, \(f(x) = \frac{4}{x - 2} + 1\). 4. For the y-intercept, set \(x = 0\): \(f(0) = \frac{4}{-2} + 1 = -1\), so the y-intercept is \((0, -1)\). 5. For the x-intercept, solve \(0 = \frac{4}{x - 2} + 1\). Then \(\frac{4}{x - 2} = -1\), so \(x - 2 = -4\) and \(x = -2\). The x-intercept is \((-2, 0)\).

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

- Identify the vertical and horizontal values approached by the graph. - Exclude the x-value of the vertical asymptote from the domain. - Use the asymptotes to write the function with an unknown \(a\). - Substitute an integer-coordinate point to solve for \(a\).

1. The graph approaches the vertical line \(x = 2\) and the horizontal line \(y = -1\). 2. Therefore, the domain is \(\mathbb{R} \setminus \{2\}\), and the function has the form \(f(x) = \frac{a}{x - 2} - 1\). 3. The point \((3, 1)\) lies on the graph. Substitution gives \(1 = \frac{a}{3 - 2} - 1\). 4. Solving gives \(a = 2\). 5. Thus, \(h = 2\), \(k = -1\), and \(f(x) = \frac{2}{x - 2} - 1\).

a) Vertical asymptote: \(x = 2\); horizontal asymptote: \(y = -1\) b) \(\mathbb{R} \setminus \{2\}\) c) \(a = 2\), \(h = 2\), and \(k = -1\); \(f(x) = \frac{2}{x - 2} - 1\)

- Which x-value separates the branches, and what does it tell you about the denominator \(x-d\)? - What height do the branches approach far from that gap? Compare it with the term \(c\). - Once you know the rule, compare the sign of \(f(x)-c\) on either side of the excluded input.

1. The graph approaches the vertical line \(x = 2\), so the excluded input is \(x = 2\) and the vertical asymptote is \(x = 2\). 2. Since the denominator is \(x - d\), the vertical asymptote gives \(d = 2\). 3. The graph approaches the horizontal line \(y = 1\), so the horizontal asymptote is \(y = 1\) and \(c = 1\). 4. The function rule is \(f(x) = \frac{3}{x - 2} + 1\). 5. Subtract the horizontal asymptote: \(f(x)-1=\frac{3}{x-2}\). This is negative for \(x<2\) and positive for \(x>2\). The left branch lies below \(y=1\).

a) Excluded input: \(x = 2\); vertical asymptote: \(x = 2\); \(d = 2\) b) Horizontal asymptote: \(y = 1\); \(c = 1\) c) To the left of \(x=2\), since \(f(x)-1=\frac{3}{x-2}<0\) there.

- Identify the vertical and horizontal lines the graph approaches. - Use the asymptotes to write the function with an unknown numerator. - Substitute a point on the graph to determine \(a\). - A y-intercept has x-coordinate \(0\).

1. The graph approaches the vertical line \(x = 3\) and the horizontal line \(y = 2\). 2. Therefore, \(h = 3\), \(k = 2\), and \(f(x) = \frac{a}{x - 3} + 2\). 3. The point \((4, 0)\) lies on the graph. Substitution gives \(0 = \frac{a}{4 - 3} + 2\), so \(a = -2\). 4. Thus, \(f(x) = -\frac{2}{x - 3} + 2\). 5. For the y-intercept, set \(x = 0\): \(f(0) = -\frac{2}{-3} + 2 = \frac{2}{3} + 2 = \frac{8}{3}\). 6. The y-intercept is \(\left(0, \frac{8}{3}\right)\), or approximately \((0, 2.67)\).

a) Vertical asymptote: \(x = 3\); horizontal asymptote: \(y = 2\) b) \(f(x) = -\frac{2}{x - 3} + 2\) c) \(\left(0, \frac{8}{3}\right)\), approximately \((0, 2.67)\)

- Which lines does the graph approach without crossing? - How do the asymptotes determine the horizontal and vertical translations? - Use a point with integer coordinates to find the scale factor. - Check the intercepts algebraically against the graph.

1. The graph approaches the vertical line \(x = -2\) and the horizontal line \(y = 1\), so these are the asymptotes. 2. The asymptotes give \(d = -2\) and \(e = 1\). Thus, \(f(x) = \frac{a}{x + 2} + 1\). 3. Use the exact point \(P(-1, 3)\): \(3 = \frac{a}{-1 + 2} + 1\), so \(a = 2\). Therefore, \(f(x) = \frac{2}{x + 2} + 1\). 4. For the y-intercept, set \(x = 0\): \(f(0) = \frac{2}{2} + 1 = 2\). The y-intercept is \((0, 2)\). 5. For the x-intercept, solve \(0 = \frac{2}{x + 2} + 1\). Then \(\frac{2}{x + 2} = -1\), so \(x + 2 = -2\) and \(x = -4\). The x-intercept is \((-4, 0)\).

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

- Identify the vertical and horizontal lines the graph approaches. - How do the asymptotes determine \(d\) and \(e\)? - Use a clearly readable point to find \(a\). - Read where the graph crosses each coordinate axis. - Verify an x-intercept by setting the function equal to \(0\), and verify a y-intercept by setting \(x = 0\).

1. The graph shows the vertical asymptote \(x = 2\) and the horizontal asymptote \(y = 2\), so \(d = 2\) and \(e = 2\). 2. Write \(f(x) = \frac{a}{x - 2} + 2\). Using the point \(P(3, -2)\), \(-2 = \frac{a}{3 - 2} + 2\), so \(a = -4\). Therefore, \(f(x) = -\frac{4}{x - 2} + 2\). 3. From the graph, the x-intercept is \((4, 0)\), and the y-intercept is \((0, 4)\). 4. For the x-intercept, solve \(0 = -\frac{4}{x - 2} + 2\). Then \(2 = \frac{4}{x - 2}\), so \(x - 2 = 2\) and \(x = 4\). 5. For the y-intercept, \(f(0) = -\frac{4}{-2} + 2 = 4\). Both calculations confirm the graph.

a) Asymptotes: \(x = 2\) and \(y = 2\); \(f(x) = -\frac{4}{x - 2} + 2\) b) x-intercept: \((4, 0)\); y-intercept: \((0, 4)\) c) Solving \(f(x) = 0\) gives \(x = 4\), and \(f(0) = 4\).

- Identify the vertical and horizontal lines the graph approaches. - How do horizontal and vertical translations determine \(d\) and \(e\)? - Read one marked point from the graph and substitute its coordinates into the partially completed formula.

1. The graph approaches the vertical line \(x=2\) and the horizontal line \(y=1\). 2. Therefore, \(d=2\) and \(e=1\), so \(f(x)=\frac{a}{x-2}+1\). 3. The graph marks \(P=(4,0)\). Substitute it: \(0=\frac{a}{4-2}+1\). Then \(-1=\frac{a}{2}\), so \(a=-2\). 4. The complete function rule is \(f(x)=-\frac{2}{x-2}+1\). The marked point \(Q=(0,2)\) gives the same value of \(a\).

a) Vertical asymptote: \(x=2\); horizontal asymptote: \(y=1\) b) \(d=2\) and \(e=1\) c) \(a=-2\); \(f(x)=-\frac{2}{x-2}+1\)

- Identify the vertical and horizontal lines each graph approaches. - Use the asymptotes to fill in the translations in the reciprocal-function form. - Substitute a clear point on each graph to find the remaining numerator constant. - Check the resulting rule against the displayed intercept.

1. For \(f\), the graph shows vertical asymptote \(x=-1\) and horizontal asymptote \(y=1\). Thus, \(f(x)=1+\frac{a}{x+1}\). 2. The graph passes through \((2, 0)\). Substituting gives \(0=1+\frac{a}{3}\), so \(a=-3\). Therefore, \(f(x)=1-\frac{3}{x+1}\). 3. For \(g\), the graph shows vertical asymptote \(x=2\) and horizontal asymptote \(y=-2\). Thus, \(g(x)=-2+\frac{b}{x-2}\). 4. The graph passes through \((4, 0)\). Substituting gives \(0=-2+\frac{b}{2}\), so \(b=4\). Therefore, \(g(x)=-2+\frac{4}{x-2}\).

a) For \(f\): vertical asymptote \(x=-1\), horizontal asymptote \(y=1\). For \(g\): vertical asymptote \(x=2\), horizontal asymptote \(y=-2\). b) \(f(x)=1-\frac{3}{x+1}\) and \(g(x)=-2+\frac{4}{x-2}\)

- Identify the horizontal and vertical lines the graph approaches, and locate the x-intercept. - The asymptotes determine \(a\) and \(c\) in the given form. - Substitute the zero to find \(b\). - Set the function equal to \(1\) and solve.

1. From the graph, the horizontal asymptote is \(y=3\), the vertical asymptote is \(x=2\), and the zero is \(x=-1\). 2. Therefore, \(f(x)=3+\frac{b}{x-2}\). Use \(f(-1)=0\): \(0=3+\frac{b}{-3}\), so \(b=9\). Thus, \(f(x)=3+\frac{9}{x-2}\). 3. Set \(f(x)=1\): \(3+\frac{9}{x-2}=1\). Then \(\frac{9}{x-2}=-2\), so \(x-2=-4.5\) and \(x=-2.5\).

a) Horizontal asymptote: \(y=3\); vertical asymptote: \(x=2\); zero: \(x=-1\) b) \(f(x)=3+\frac{9}{x-2}\) c) \(x=-2.5\)

- What information do the asymptotes give about the unknown parameters? - Read \(P\)'s coordinates from the graph to find the remaining parameter. - Equate the two functions, keeping track of excluded inputs. - Can the numerator of the simplified expression for \(h\) ever be zero?

1. The vertical asymptote is \(x=2\) and the horizontal asymptote is \(y=1\), so \(f(x)=\frac{a}{x-2}+1\). 2. The plotted point is \(P(1,3)\). Substitution gives \(3=\frac{a}{1-2}+1\), so \(a=-2\) and \(f(x)=-\frac{2}{x-2}+1\). 3. Equate the functions: \(-\frac{2}{x-2}+1=-x+2\), where \(x\ne2\). Multiplying by \(x-2\) gives \(-2=(-x+1)(x-2)=-x^2+3x-2\). 4. Thus, \(x(x-3)=0\), giving \(x=0\) or \(x=3\). The intersection points are \((0,2)\) and \((3,-1)\). 5. Because \(h(x)=-\frac{2}{x-2}\), its numerator is never zero. Hence \(h\) has no zeros.

a) Vertical asymptote: \(x=2\); horizontal asymptote: \(y=1\). b) \(f(x)=-\frac{2}{x-2}+1\). c) \((0,2)\) and \((3,-1)\). d) \(h(x)=-\frac{2}{x-2}\) cannot equal zero for any allowed \(x\).

- Identify the vertical and horizontal asymptotes first. - Connect the asymptotes to the parameters \(d\) and \(c\). - Substitute a clearly readable point to determine \(a\). - Relate each parameter to a transformation of \(\frac{1}{x^2}\). - Set the function equal to \(0\) to find its zeros.

1. The vertical asymptote is \(x=2\), so \(d=2\). The horizontal asymptote is \(y=-1\), so \(c=-1\). 2. The point \((0, 0)\) lies on the graph. Substituting it gives \(0=\frac{a}{(0-2)^2}-1=\frac{a}{4}-1\), so \(a=4\). 3. Therefore, \(f(x)=\frac{4}{(x-2)^2}-1\). 4. Starting with \(p(x)=\frac{1}{x^2}\), shift right \(2\) units, stretch vertically by a factor of \(4\), and shift down \(1\) unit. 5. To find the zeros, solve \(\frac{4}{(x-2)^2}-1=0\). Then \((x-2)^2=4\), so \(x-2=\pm2\). Thus, \(x=0\) or \(x=4\).

a) \(a=4\), \(c=-1\), and \(d=2\), so \(f(x)=\frac{4}{(x-2)^2}-1\) b) Shift right \(2\) units, stretch vertically by a factor of \(4\), and shift down \(1\) unit. c) \(x=0\) and \(x=4\)

- Which missing points are marked by open circles, and where does a curve run sharply up or down? - Find a simpler rational rule that fits each curve away from its hole. - How could you introduce one excluded input without changing the other function values?

1. Graph \(f\) approaches the vertical line \(x=1\) and has a hole at \((3,2.5)\). 2. The curve follows \(\frac{x+2}{x-1}\) except for \(x=3\). Include a common factor: \(f(x)=\frac{(x+2)(x-3)}{(x-1)(x-3)}=\frac{x^2-x-6}{x^2-4x+3}\). 3. Graph \(g\) approaches the vertical line \(x=-2\) and has a hole at \((1,0)\). 4. The curve follows \(\frac{x-1}{x+2}\) except for \(x=1\). Include a common factor: \(g(x)=\frac{(x-1)^2}{(x+2)(x-1)}=\frac{x^2-2x+1}{x^2+x-2}\). 5. The canceled factors leave holes at the original excluded inputs; the uncanceled denominator factors create the vertical asymptotes.

a) Hole: \((3,2.5)\); vertical asymptote: \(x=1\). b) Hole: \((1,0)\); vertical asymptote: \(x=-2\). c) \(f(x)=\frac{(x+2)(x-3)}{(x-1)(x-3)}\) and \(g(x)=\frac{(x-1)^2}{(x+2)(x-1)}\).

- What does the pair of vertical asymptotes tell you about the denominator? - What do the highest-degree terms tell you about the horizontal asymptote? - Read one marked point to find the remaining parameter, then use the other to check.

1. The vertical asymptotes are \(x=-1\) and \(x=1\), so \(c=1\), and the domain excludes \(-1\) and \(1\). 2. The horizontal asymptote is \(y=2\). Because the numerator and denominator have the same degree, \(a=2\). 3. The marked points are \(A=(0,8)\) and \(B=(2,0)\). Using \(A\), \(8=\frac{b}{-1}\), so \(b=-8\). 4. Therefore, \(f(x)=\frac{2x^2-8}{x^2-1}\), and point \(B\) verifies \(f(2)=0\). 5. The y-intercept is \((0,8)\). Solving \(2x^2-8=0\) gives x-intercepts \((-2,0)\) and \((2,0)\).

a) Domain: \(\mathbb{R}\setminus\{-1,1\}\); vertical asymptotes: \(x=-1\), \(x=1\); horizontal asymptote: \(y=2\). b) \(A=(0,8)\), \(B=(2,0)\); \(a=2\), \(b=-8\), \(c=1\); \(f(x)=\frac{2x^2-8}{x^2-1}\). c) X-intercepts: \((-2,0)\), \((2,0)\); y-intercept: \((0,8)\).

- Which vertical and horizontal lines does the graph approach? - How do their locations determine the shifts in the given formula? - Read the point's coordinates from the graph, then substitute them to find the remaining parameter.

1. The vertical asymptote is \(x=-2\), so \(d=-2\). 2. The horizontal asymptote is \(y=3\), so \(e=3\). 3. Substitute the plotted point \(P=(2,2)\): \(2=\frac{a}{2+2}+3\). 4. Thus, \(-1=\frac{a}{4}\), so \(a=-4\), giving \(f(x)=-\frac{4}{x+2}+3\).

a) \(d=-2\), \(e=3\). b) \(a=-4\); \(f(x)=-\frac{4}{x+2}+3\).

- Which vertical and horizontal lines does the graph approach? - How do these lines determine the shifts in the proposed function form? - Read point \(P\) on the graph, then substitute its coordinates. - Before combining terms, rewrite the whole-number term with the denominator of the fraction.

1. The vertical asymptote is \(x=2\), and the horizontal asymptote is \(y=1\). 2. Hence, \(d=2\), \(c=1\), and \(f(x)=\frac{a}{x-2}+1\). 3. Read \(P=(3,3)\) from the graph. Substitution gives \(3=\frac{a}{3-2}+1\), so \(a=2\). 4. Combine fractions: \(f(x)=\frac{2}{x-2}+\frac{x-2}{x-2}=\frac{x}{x-2}\), where \(x\ne2\).

a) Vertical asymptote: \(x=2\); horizontal asymptote: \(y=1\). b) \(d=2\), \(c=1\), \(a=2\); \(f(x)=\frac{2}{x-2}+1\). c) \(f(x)=\frac{x}{x-2}\), with \(x\ne2\).

- Use the vertical asymptote to find the value that makes the denominator zero. - Use the horizontal asymptote to determine the vertical translation. - After finding \(b\) and \(c\), substitute the coordinates of a clearly marked point to solve for \(a\).

1. The graph approaches the vertical line \(x = -2\). Since the denominator is zero when \(x = b\), it follows that \(b = -2\). 2. The graph approaches the horizontal line \(y = -1\), so \(c = -1\). 3. The function now has the form \(f(x) = \frac{a}{x + 2} - 1\). 4. The point \((1, 0)\) lies on the graph. Substitute it: \(0 = \frac{a}{1 + 2} - 1\). 5. Solving \(0 = \frac{a}{3} - 1\) gives \(a = 3\). 6. Therefore, \(a = 3\), \(b = -2\), and \(c = -1\), and the function is \(f(x) = \frac{3}{x + 2} - 1\).

\(a = 3\), \(b = -2\), and \(c = -1\); \(f(x) = \frac{3}{x + 2} - 1\)

- Compare the opening direction of the two rational graphs. - Locate the new vertical and horizontal asymptotes. - Use one point with integer coordinates to determine the vertical scale factor.

1. The vertical asymptote moves from \(x=0\) to \(x=2\), so the graph shifts right \(2\) units. The horizontal asymptote moves from \(y=0\) to \(y=3\), so the graph shifts up \(3\) units. 2. The branches approach negative infinity near \(x=2\), so the graph is reflected across the x-axis. 3. Write \(g(x)=\frac{a}{(x-2)^2}+3\). The point \((1, 1)\) lies on \(g\), so \(1=\frac{a}{(1-2)^2}+3=a+3\). Thus, \(a=-2\). 4. Therefore, \(g(x)=-\frac{2}{(x-2)^2}+3\). The factor \(-2\) represents a reflection across the x-axis and a vertical stretch by a factor of \(2\).

Reflect across the x-axis, stretch vertically by a factor of \(2\), shift right \(2\) units, and shift up \(3\) units; \(g(x)=-\frac{2}{(x-2)^2}+3\)

- The intersection of the asymptotes gives the translation parameters. - After finding \(x_0\) and \(y_0\), choose a point with integer coordinates and substitute it. - The placement of the branches relative to the asymptotes indicates the sign of \(a\).

1. The graph approaches the vertical line \(x = -1\), so \(x_0 = -1\). It approaches the horizontal line \(y = -3\), so \(y_0 = -3\). 2. The function now has the form \(f(x) = \frac{a}{x + 1} - 3\). 3. The point \((1, -1)\) lies on the graph. Substitute it: \(-1 = \frac{a}{1 + 1} - 3\). 4. Solving gives \(2 = \frac{a}{2}\), so \(a = 4\). 5. Therefore, the complete function rule is \(f(x) = \frac{4}{x + 1} - 3\).

a) Asymptotes: \(x = -1\) and \(y = -3\); \(x_0 = -1\) and \(y_0 = -3\) b) Using \((1, -1)\) gives \(a = 4\); \(f(x) = \frac{4}{x + 1} - 3\)

- In the form \(\frac{a}{x - h} + k\), the asymptotes are \(x = h\) and \(y = k\). - Substitute the coordinates of the origin after finding \(h\) and \(k\). - To find an input for a given output, set the function equal to that output and solve for \(x\). - Compare your solution with the graph.

1. The graph approaches the vertical line \(x = -2\) and the horizontal line \(y = -1\). 2. In the form \(g(x) = \frac{a}{x - h} + k\), the asymptotes are \(x = h\) and \(y = k\). Therefore, \(h = -2\) and \(k = -1\). 3. Substitute \((0, 0)\) into \(g(x) = \frac{a}{x + 2} - 1\): \(0 = \frac{a}{2} - 1\). Thus, \(a = 2\), and \(g(x) = \frac{2}{x + 2} - 1\). 4. Set \(g(x) = -2\): \(-2 = \frac{2}{x + 2} - 1\). Then \(-1 = \frac{2}{x + 2}\), so \(-(x + 2) = 2\) and \(x = -4\). 5. The point \((-4, -2)\) lies on the graph, confirming the result.

a) Vertical asymptote: \(x = -2\); horizontal asymptote: \(y = -1\) b) \(h = -2\) and \(k = -1\) c) \(a = 2\); \(g(x) = \frac{2}{x + 2} - 1\) d) \(x = -4\)

- Use the vertical and horizontal asymptotes to find \(b\) and \(c\). - Remember that a right shift produces a minus sign in the denominator. - Substitute one exact point from each graph to find \(a\).

1. Graph A has vertical asymptote \(x = 2\), so \(b = -2\), and horizontal asymptote \(y = -1\), so \(c = -1\). Using \((3, 0)\): \(0 = \frac{a}{3 - 2} - 1\), so \(a = 1\). 2. Graph B has vertical asymptote \(x = -3\), so \(b = 3\), and horizontal asymptote \(y = 2\), so \(c = 2\). Using \((-2, 4)\): \(4 = \frac{a}{-2 + 3} + 2\), so \(a = 2\). 3. Graph C has vertical asymptote \(x = 1\), so \(b = -1\), and horizontal asymptote \(y = -2\), so \(c = -2\). Using \((2, -3)\): \(-3 = \frac{a}{2 - 1} - 2\), so \(a = -1\).

A: \(a = 1\), \(b = -2\), \(c = -1\) B: \(a = 2\), \(b = 3\), \(c = 2\) C: \(a = -1\), \(b = -1\), \(c = -2\)

- Read the lines each curve approaches and use them to fill in the translation parameters. - Read each marked point and substitute its coordinates into the partly completed rule. - Check whether the signs of the numerator coefficients fit the branch orientations.

1. For \(f\), the vertical asymptote is \(x=1\), and the horizontal asymptote is \(y=2\). Thus, \(f(x)=\frac{a}{x-1}+2\). 2. Read \(P=(2,3)\). Substituting gives \(3=\frac{a}{2-1}+2\), so \(a=1\) and \(f(x)=\frac{1}{x-1}+2\). 3. For \(g\), the vertical asymptote is \(x=-2\), and the horizontal asymptote is \(y=-1\). Thus, \(g(x)=\frac{a}{x+2}-1\). 4. Read \(Q=(0,-1.5)\). Substituting gives \(-1.5=\frac{a}{2}-1\), so \(a=-1\) and \(g(x)=-\frac{1}{x+2}-1\).

a) For \(f\): vertical asymptote \(x=1\), horizontal asymptote \(y=2\). For \(g\): vertical asymptote \(x=-2\), horizontal asymptote \(y=-1\). b) \(f(x)=\frac{1}{x-1}+2\); \(g(x)=-\frac{1}{x+2}-1\).

- Identify the horizontal and vertical lines that the graph approaches. - Substitute the marked point to determine \(a\). - Set the function equal to \(0\) to find the x-intercept.

1. The graph approaches the vertical line \(x = 3\) and the horizontal line \(y = 1\). 2. Therefore, \(h(x)\) has the form \(h(x) = \frac{a}{x - 3} + 1\). 3. The point \((5, 2)\) lies on the graph. Substitution gives \(2 = \frac{a}{5 - 3} + 1\), so \(1 = \frac{a}{2}\) and \(a = 2\). 4. Thus, \(h(x) = \frac{2}{x - 3} + 1\). 5. To find the x-intercept, set \(h(x) = 0\): \(0 = \frac{2}{x - 3} + 1\). Then \(-1 = \frac{2}{x - 3}\), so \(-(x - 3) = 2\) and \(x = 1\). 6. The x-intercept is \((1, 0)\).

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

- Use the vertical and horizontal asymptotes to find \(h\) and \(k\). - Remember that a vertical asymptote at a negative input produces a plus sign in the denominator. - Substitute a point from the graph to solve for \(a\). - Verify the rule with a second point if possible.

1. The vertical asymptote is \(x = -1.5\), so \(h = -1.5\). 2. The horizontal asymptote is \(y = 2.5\), so \(k = 2.5\). 3. The function has the form \(f(x) = \frac{a}{x + 1.5} + 2.5\). 4. The point \((-1, 3.5)\) lies on the graph. Substitute it: \(3.5 = \frac{a}{-1 + 1.5} + 2.5\). 5. This gives \(1 = \frac{a}{0.5}\), so \(a = 0.5\). 6. Therefore, \(f(x) = \frac{0.5}{x + 1.5} + 2.5\).

\(f(x) = \frac{0.5}{x + 1.5} + 2.5\)

- Use the vertical asymptote to determine the expression in the denominator. - Use the horizontal asymptote to determine the constant outside the fraction. - Substitute a point on the graph and solve for \(a\). - The locations of the branches relative to the asymptotes indicate whether \(a\) is positive or negative.

1. The graph approaches the vertical line \(x = -3\) and the horizontal line \(y = -2\). 2. Therefore, the function has the form \(g(x) = \frac{a}{x + 3} - 2\). 3. The point \((-4, 0)\) lies on the graph. Substitute it: \(0 = \frac{a}{-4 + 3} - 2\). 4. This gives \(2 = \frac{a}{-1}\), so \(a = -2\). 5. Thus, \(g(x) = -\frac{2}{x + 3} - 2\).

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

- Read the two asymptotes to find the shifts in the reciprocal form. - Use the marked point to find the coefficient, then calculate the intercepts from the rule. - Which input is excluded by the denominator, and which output can the function never reach?

1. The vertical asymptote is \(x=-2\) and the horizontal asymptote is \(y=1\). Thus, \(f(x)=\frac{a}{x+2}+1\). 2. Read the point \((-1,2)\). Substitute: \(2=\frac{a}{1}+1\), so \(a=1\). Therefore, \(f(x)=\frac{1}{x+2}+1\). 3. Set \(f(x)=0\): \(\frac{1}{x+2}+1=0\), so \(x=-3\). The x-intercept is \((-3,0)\). 4. Compute \(f(0)=\frac{1}{2}+1=\frac{3}{2}\). The y-intercept is \(\left(0,\frac{3}{2}\right)\). 5. The denominator excludes \(x=-2\), and the nonzero fractional term prevents \(y=1\). The domain is \(( -\infty,-2)\cup(-2,\infty)\), and the range is \((-\infty,1)\cup(1,\infty)\).

a) \(f(x)=\frac{1}{x+2}+1\). b) X-intercept: \((-3,0)\); y-intercept: \(\left(0,\frac{3}{2}\right)\). c) Domain: \((-\infty,-2)\cup(-2,\infty)\); range: \((-\infty,1)\cup(1,\infty)\).

- Which axis or line is the vertical asymptote, and what is the horizontal asymptote? - Use a clearly readable point to determine the remaining coefficient. - Find each intercept from the equation, but first check that the required input is allowed.

1. The graph has vertical asymptote \(x=0\) and horizontal asymptote \(y=-2\), so \(g(x)=\frac{a}{x}-2\). 2. A readable point is \((1,2)\). Substitute to get \(2=a-2\), so \(a=4\) and \(g(x)=\frac{4}{x}-2\). 3. Set \(g(x)=0\): \(\frac{4}{x}-2=0\). Then \(2x=4\), giving \(x=2\), so the x-intercept is \((2,0)\). 4. A y-intercept would require \(x=0\), which is excluded from the domain, so there is no y-intercept.

a) \(g(x)=\frac{4}{x}-2\). b) X-intercept: \((2,0)\); no y-intercept because \(x=0\) is excluded.

- Identify the vertical and horizontal lines that the graph approaches. - Substitute the marked point after using the asymptotes to write the function with an unknown \(a\). - A translation right changes the input; a translation up changes the output. - Check the sign inside the denominator carefully.

1. The graph approaches the vertical line \(x = 2\) and the horizontal line \(y = -2\). 2. Therefore, \(g(x) = \frac{a}{x - 2} - 2\). 3. The marked point is \((3, -1)\). Substitution gives \(-1 = \frac{a}{3 - 2} - 2\), so \(a = 1\). Thus, \(g(x) = \frac{1}{x - 2} - 2\). 4. Translating \(2\) units right replaces \(x\) with \(x - 2\), and translating \(3\) units up adds \(3\). Therefore, \(p(x) = g(x - 2) + 3 = \frac{1}{(x - 2) - 2} - 2 + 3 = \frac{1}{x - 4} + 1\).

a) Vertical asymptote: \(x = 2\); horizontal asymptote: \(y = -2\) b) \(g(x) = \frac{1}{x - 2} - 2\) c) \(p(x) = \frac{1}{x - 4} + 1\)

- Identify the values approached by the graph vertically and horizontally. - In \(\frac{a}{x - h} + k\), the asymptotes are \(x = h\) and \(y = k\). - Substitute an integer-coordinate point to solve for \(a\). - At an x-intercept, the function value is \(0\).

1. The graph approaches the vertical line \(x = -3\) and the horizontal line \(y = -1\). 2. Therefore, \(h = -3\) and \(k = -1\), so \(f(x) = \frac{a}{x + 3} - 1\). 3. The point \((-2, 1)\) lies on the graph. Substitution gives \(1 = \frac{a}{-2 + 3} - 1\), so \(a = 2\). Thus, \(f(x) = \frac{2}{x + 3} - 1\). 4. To find the x-intercept, solve \(0 = \frac{2}{x + 3} - 1\). Then \(1 = \frac{2}{x + 3}\), so \(x + 3 = 2\) and \(x = -1\). 5. The x-intercept is \((-1, 0)\).

a) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = -1\) b) \(h = -3\) and \(k = -1\) c) \(a = 2\); \(f(x) = \frac{2}{x + 3} - 1\) d) \((-1, 0)\)

- Read the vertical and horizontal asymptotes before using the marked point. - Read \(P\)'s coordinates from the graph, then substitute them into the partly known formula. - At an intersection, both graphs have the same input and output. - To test \(Q\), compare the function output with its given y-coordinate.

1. The vertical asymptote is \(x=1\), so \(h=1\). The horizontal asymptote is \(y=3\), so \(k=3\). Thus, \(f(x)=\frac{a}{x-1}+3\). 2. The marked point is \(P=(2,5)\). Substituting gives \(5=\frac{a}{2-1}+3\), so \(a=2\) and \(f(x)=\frac{2}{x-1}+3\). 3. The two graphed intersections are \((0,1)\) and \((3,4)\). Both also satisfy the line \(g(x)=x+1\) and the rule for \(f\). 4. Compute \(f(5)=\frac{2}{5-1}+3=3.5\). Therefore, \(Q(5,3.5)\) is on the graph of \(f\).

a) \(f(x)=\frac{2}{x-1}+3\). b) \((0,1)\) and \((3,4)\). c) Yes, since \(f(5)=3.5\).

- What do the two asymptotes tell you about the unknown shifts? - How can a point on the curve help you determine the remaining coefficient? - Which input do you use to calculate the y-intercept?

1. The vertical asymptote is \(x=2\), giving \(d=2\); the horizontal asymptote is \(y=1\), giving \(c=1\). 2. The marked point is \(P=(3,3)\). Substitution gives \(3=\frac{k}{3-2}+1\), so \(k=2\) and \(f(x)=\frac{2}{x-2}+1\). 3. At \(x=0\), \(f(0)=\frac{2}{-2}+1=0\). Therefore, the y-intercept is \((0,0)\).

a) Vertical asymptote: \(x=2\); horizontal asymptote: \(y=1\); \(d=2\), \(c=1\). b) \(k=2\); \(f(x)=\frac{2}{x-2}+1\). c) \((0,0)\).

- Read the horizontal asymptote to determine the vertical shift. - Choose a point with integer coordinates from the graph. - Substitute that point into the function rule to find \(a\).

1. The graph approaches \(y = 2\) as \(|x|\) becomes large, so the horizontal asymptote is \(y = 2\). Therefore, \(c = 2\). 2. The point \((1, 3)\) lies on the graph. Substitute it into \(f(x) = \frac{a}{x^2} + 2\): \(3 = \frac{a}{1^2} + 2\). 3. Thus, \(3 = a + 2\), so \(a = 1\).

\(a = 1\) and \(c = 2\)

- Identify the vertical and horizontal asymptotes of the transformed graph. - Use them to write \(g(x)=\frac{a}{x-h}+k\). - Substitute a clear point on the graph to find \(a\).

1. The parent function has asymptotes \(x=0\) and \(y=0\). The graph of \(g\) has asymptotes \(x=2\) and \(y=1\), so \(g(x)=\frac{a}{x-2}+1\). 2. The graph passes through \((3, 3)\). Substitute: \(3=\frac{a}{3-2}+1\), so \(a=2\). 3. Therefore, \(g(x)=\frac{2}{x-2}+1\).

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

- Identify the asymptotes from the curve and coordinate grid. - Substitute an intercept with exact coordinates to find \(a\). - Evaluate the completed function at \(x=6\). - Check the axis scale before reading coordinates.

1. The graph shows vertical asymptote \(x=2\), so \(x_0=2\), and horizontal asymptote \(y=1\), so \(c=1\). Thus, \(g(x)=\frac{a}{x-2}+1\). 2. The graph passes through \((0, -1)\). Substituting gives \(-1=\frac{a}{-2}+1\), so \(a=4\). Therefore, \(g(x)=\frac{4}{x-2}+1\). 3. \(g(6)=\frac{4}{6-2}+1=2\).

a) Vertical asymptote: \(x=2\), so \(x_0=2\); horizontal asymptote: \(y=1\), so \(c=1\) b) \(a=4\), so \(g(x)=\frac{4}{x-2}+1\) c) \(g(6)=2\)

- Read the vertical and horizontal asymptotes. - Use the horizontal asymptote to determine the ratio of leading coefficients. - Use the x-intercept to find the remaining numerator constant.

1. The graph has vertical asymptote \(x=1\), so use denominator \(x-1\). 2. The horizontal asymptote is \(y=2\), so use a numerator with leading coefficient \(2\): \(f(x)=\frac{2x+b}{x-1}\). 3. The x-intercept is \(x=-2\), so \(2(-2)+b=0\). Therefore, \(b=4\). 4. Thus, \(f(x)=\frac{2x+4}{x-1}\). As a check, \(f(0)=-4\), which matches the graph.

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

- Use the vertical asymptotes to determine the denominator constant. - Use the y-intercept to determine the numerator constant. - Check the horizontal asymptote using the leading coefficients.

1. The graph has vertical asymptotes at \(x=-1\) and \(x=1\). Since the denominator is \(x^2+b\), the equation \(1+b=0\) gives \(b=-1\). 2. The graph passes through \((0, -1)\). Therefore, \(-1=f(0)=\frac{a}{-1}\), so \(a=1\). 3. Thus, \(f(x)=\frac{x^2+1}{x^2-1}\). The equal leading coefficients also give horizontal asymptote \(y=1\), matching the graph.

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

- Which lines does the rational curve approach, and what do they tell you about its rule? - Read the marked point and both line intercepts directly from the graph. - Equate the function rules and clear the denominator, but exclude values that make it zero.

1. The asymptotes of \(f\) are \(x=1\) and \(y=1\), so \(f(x)=\frac{a}{x-1}+1\). 2. The plotted point \(P=(2,5)\) gives \(5=\frac{a}{2-1}+1\), hence \(a=4\) and \(f(x)=\frac{4}{x-1}+1\). 3. The line crosses the axes at \((0,7)\) and \((7,0)\). Its slope is \(-1\), so \(g(x)=-x+7\). 4. Set the two expressions equal: \(\frac{4}{x-1}+1=-x+7\) with \(x\ne1\). Multiply to get \(4=(-x+6)(x-1)=-x^2+7x-6\). 5. This gives \(x^2-7x+10=(x-2)(x-5)=0\), so the input values are \(2\) and \(5\). Substitution into \(g\) gives the points \((2,5)\) and \((5,2)\).

a) \(f(x)=\frac{4}{x-1}+1\). b) \(g(x)=-x+7\). c) \((2,5)\) and \((5,2)\).

- Where does the curve get very high or low near one vertical line? - What does the open circle tell you about the graph? - Find a numerator factor for the x-intercept and a common factor for the hole.

1. The graph grows sharply up and down near \(x=1\), so the vertical asymptote is \(x=1\). 2. The open circle marks a hole at \((3,2)\). 3. The graph has x-intercept \(x=-1\) and horizontal asymptote \(y=1\). 4. Use \(x+1\) in the numerator for the zero, \(x-1\) in the denominator for the vertical asymptote, and \(x-3\) as a common factor to produce the hole. 5. One possible rule is \(f(x)=\frac{(x+1)(x-3)}{(x-1)(x-3)}\), with \(x\ne1,3\). The simplified expression has value \(\frac{3+1}{3-1}=2\) at the hole's x-coordinate, confirming its height.

a) Vertical asymptote: \(x=1\); hole: \((3,2)\). b) \(f(x)=\frac{(x+1)(x-3)}{(x-1)(x-3)}\), with \(x\ne1,3\).

- Find the vertical and horizontal lines the branches approach. - Use these lines to write a translated reciprocal rule with one unknown coefficient. - Read either marked point to find that coefficient; use the other to check.

1. The graph has vertical asymptote \(x=2\), so the domain excludes \(2\). Its horizontal asymptote is \(y=2\), which this translated reciprocal curve never reaches, so the range excludes \(2\). 2. Thus, the domain and range are both \((-\infty,2)\cup(2,\infty)\). 3. Write \(f(x)=\frac{a}{x-2}+2\). Read \(A=(0,1.5)\); substitution gives \(1.5=\frac{a}{-2}+2\), so \(a=1\). 4. Therefore, \(f(x)=\frac{1}{x-2}+2=\frac{2x-3}{x-2}\). The other marked point \(B=(1.5,0)\) confirms the result because \(f(1.5)=0\).

a) Domain and range: \((-\infty,2)\cup(2,\infty)\). b) Vertical asymptote: \(x=2\); horizontal asymptote: \(y=2\). c) \(f(x)=\frac{2x-3}{x-2}\).

- Read the two excluded inputs and the horizontal value approached by the branches. - What kind of powers keep a function unchanged when \(x\) is replaced by \(-x\)? - Read the marked y-intercept to determine the remaining numerator constant.

1. The vertical asymptotes are \(x=-2\) and \(x=2\), so the domain is \((-\infty,-2)\cup(-2,2)\cup(2,\infty)\). The horizontal asymptote is \(y=2\). 2. Because the graph is symmetric about the y-axis, use even powers: \(g(x)=\frac{2x^2+b}{x^2-4}\). The ratio of leading coefficients is \(2\), as required. 3. Read \(S_y=(0,-0.5)\). Substitution gives \(-0.5=\frac{b}{-4}\), so \(b=2\). 4. Thus, one possible rule is \(g(x)=\frac{2x^2+2}{x^2-4}\). The numerator is nonzero at both excluded inputs, confirming the vertical asymptotes.

a) Domain: \((-\infty,-2)\cup(-2,2)\cup(2,\infty)\); vertical asymptotes: \(x=-2\) and \(x=2\); horizontal asymptote: \(y=2\). b) \(g(x)=\frac{2x^2+2}{x^2-4}\).

- Identify the vertical asymptote of each reciprocal graph. - Use a horizontal shift of \(\frac{1}{x}\). - Substitute a clearly visible point to find the numerator.

1. A horizontally translated reciprocal function has the form \(f_k(x)=\frac{c}{x-k}\), with vertical asymptote \(x=k\). 2. The shown asymptotes are \(x=-3\), \(x=0\), and \(x=3\), which correspond to different values of \(k\). 3. The graph with asymptote \(x=0\) passes through \((1, 2)\), so \(2=\frac{c}{1}\) and \(c=2\). 4. Therefore, the family is \(f_k(x)=\frac{2}{x-k}\).

\(f_k(x)=\frac{2}{x-k}\)

- Read the two asymptotes and use them to find the shifts in the reciprocal rule. - Read the marked point to find the coefficient. - Compare the branch directions with the parent graph \(\frac{1}{x}\).

1. The vertical asymptote is \(x=2\), so \(b=2\), and the horizontal asymptote is \(y=1\), so \(c=1\). 2. The marked point is \(P=(3,2)\). Thus, \(2=\frac{a}{3-2}+1\), giving \(a=1\) and \(f(x)=\frac{1}{x-2}+1\). 3. The domain is \((-\infty,2)\cup(2,\infty)\). The asymptotes are \(x=2\) and \(y=1\). 4. The parent function \(\frac{1}{x}\) decreases on each side of its vertical asymptote. Translations do not change this, so \(f\) decreases on \((-\infty,2)\) and \((2,\infty)\). It has no increasing intervals.

a) \(f(x)=\frac{1}{x-2}+1\). b) Domain: \((-\infty,2)\cup(2,\infty)\); vertical asymptote \(x=2\); horizontal asymptote \(y=1\). c) Decreasing on \((-\infty,2)\) and \((2,\infty)\); no increasing intervals.

- The vertical asymptote determines the denominator zero. - The horizontal asymptote determines the leading-coefficient ratio. - Substitute the y-intercept to find the remaining constant.

1. The vertical asymptote is \(x=2\), so \(c=2\). 2. The horizontal asymptote is \(y=3\). Since the numerator and denominator have the same degree, the ratio of leading coefficients is \(a=3\). 3. The y-intercept is \((0, 2)\). Therefore, \(2=f(0)=\frac{b}{-2}\), so \(b=-4\). 4. Thus, \(a=3\), \(b=-4\), and \(c=2\).

\(a=3\), \(b=-4\), \(c=2\)

- Read the vertical and horizontal asymptotes first. - Use the horizontal asymptote to determine the leading numerator coefficient. - Use the x-intercept to determine \(b\). - Substitute \(x=0\) to verify the y-intercept.

1. The vertical asymptote is \(x=-1\), so \(c=1\). 2. The horizontal asymptote is \(y=2\), so the leading-coefficient ratio gives \(a=2\). 3. The x-intercept is \(x=2\). Therefore, \(2\cdot 2+b=0\), so \(b=-4\). Thus, \(f(x)=\frac{2x-4}{x+1}\). 4. Check: \(f(0)=\frac{-4}{1}=-4\), which matches the y-intercept shown on the graph.

a) \(a=2\), \(b=-4\), \(c=1\) b) \(f(0)=-4\), matching the graph.

- Read the vertical and horizontal asymptotes from the graph. - Choose a graph point whose coordinates are easy to read, then substitute it into the function rule. - How does shifting a graph vertically change its horizontal asymptote?

1. The vertical asymptote is \(x=1\), so \(b=1\). The horizontal asymptote is \(y=2\), so \(c=2\). 2. The graph passes through the origin. Substitute \((0, 0)\): \(0=\frac{a}{0-1}+2=-a+2\). Therefore, \(a=2\), and \(f(x)=\frac{2}{x-1}+2\). 3. Translating the graph 3 units down also translates the horizontal asymptote 3 units down. Its new equation is \(y=2-3=-1\).

a) \(b=1\), \(c=2\) b) \(a=2\) c) \(y=-1\)

- How would you express the total cost as a fixed setup charge plus a cost for each mug? - Use the average cost for \(15\) mugs to find the unknown cost per mug. - When you solve for the number of mugs, remember that it must be a positive whole number.

1. If glazing costs \(c\) dollars per mug, the total cost for \(15\) mugs is \(180+15c\). Thus \(\frac{180+15c}{15}=24\), giving \(c=12\). 2. For \(n>0\), total cost is \(180+12n\), so \(C(n)=\frac{180+12n}{n}=12+\frac{180}{n}\). 3. Solve \(12+\frac{180}{n}\le18\). Since \(n>0\), this gives \(180\le6n\), or \(n\ge30\). The fewest whole mugs is \(30\).

a) \(\$12\) per mug b) \(C(n)=12+\frac{180}{n}\), for positive whole numbers \(n\) c) \(30\) mugs

- How do you find the common point of two lines? - What does the intersection of the symmetry lines reveal about the center of the hyperbola? - Use the center to identify \(x_0\) and \(y_0\). - Substitute the known point to find \(k\).

1. Set the line equations equal: \(x - 2 = -x + 6\). Then \(2x = 8\), so \(x = 4\) and \(y = 2\). The intersection is \((4, 2)\). 2. This point is the center of the hyperbola and the intersection of its vertical and horizontal asymptotes. Thus, \(x_0 = 4\) and \(y_0 = 2\). 3. Write \(f(x) = \frac{k}{x - 4} + 2\). 4. Substitute \(P(2, 1)\): \(1 = \frac{k}{2 - 4} + 2\). Then \(-1 = \frac{k}{-2}\), so \(k = 2\). 5. Therefore, \(f(x) = \frac{2}{x - 4} + 2\).

a) The lines intersect at \((4, 2)\). This point is the center of the hyperbola and the intersection of its asymptotes. b) \(f(x) = \frac{2}{x - 4} + 2\)

- Use numerator factors for the two required zeros. - How does the power on a denominator factor affect the signs on opposite sides of its zero? - Match the degrees and leading coefficients to obtain the horizontal asymptote \(y=4\).

1. Use numerator factors \((x+5)\) and \((x-3)\) to create the required zeros. 2. Use an odd power of \(x\) in the denominator to create a vertical asymptote with a sign change at \(x=0\). 3. The numerator and denominator must have the same degree, with leading-coefficient ratio \(4\), so the graph approaches \(y=4\) at both ends. 4. One example is \(f(x)=\frac{4(x-3)(x+5)^2}{x^3}\). Its only zeros are \(-5\) and \(3\), its denominator has an odd-multiplicity zero at \(0\), and its leading-coefficient ratio is \(4\).

One possible formula is \(f(x)=\frac{4(x-3)(x+5)^2}{x^3}\).

- A nonzero horizontal asymptote can be obtained with equal numerator and denominator degrees. - A repeated denominator zero can create exactly one vertical asymptote. - Choose two distinct numerator zeros that do not cancel the denominator factor. - Set the leading-coefficient ratio equal to \(-0.5\).

1. Choose numerator and denominator polynomials of degree \(2\) so the function can have two zeros and a nonzero horizontal asymptote. 2. To make \(x=3\) the only vertical asymptote, choose the denominator \((x-3)^2\). 3. The horizontal asymptote \(y=-0.5\) requires a numerator leading coefficient of \(-0.5\). 4. Choose two numerator zeros different from \(3\), such as \(0\) and \(1\). Then the numerator is \(-0.5x(x-1)=-0.5x^2+0.5x\). 5. Therefore, one possible function is \(h(x)=\frac{-0.5x^2+0.5x}{(x-3)^2}\).

One possible formula is \(h(x)=\frac{-0.5x^2+0.5x}{(x-3)^2}\).

- Sketch the printed rectangle and all four margins. - Use the fixed printed area to express the printed height in terms of the width. - Add both horizontal and vertical margins to obtain the total dimensions. - Rewrite the area function as a linear expression plus a rational remainder.

1. If the printed height is \(y\), then \(xy=18\), so \(y=\frac{18}{x}\). Including the top and bottom margins, the total height is \(H(x)=\frac{18}{x}+2\). 2. The total width is \(x+1\). Therefore, \(A(x)=(x+1)\left(\frac{18}{x}+2\right)=2x+20+\frac{18}{x}=\frac{2x^2+20x+18}{x}\). 3. Since \(A(x)=2x+20+\frac{18}{x}\), the vertical asymptote is \(x=0\), and the slant asymptote is \(y=2x+20\). 4. Solve \(45=\frac{2x^2+20x+18}{x}\). This gives \(2x^2-25x+18=0\), so \(x=\frac{25\pm\sqrt{481}}{4}\). 5. Both solutions are positive: \(x\approx 0.77\) or \(x\approx 11.73\).

a) \(H(x)=\frac{18}{x}+2\), for \(x>0\) b) \(A(x)=\frac{2x^2+20x+18}{x}\) c) Vertical asymptote: \(x=0\); slant asymptote: \(y=2x+20\) d) \(x\approx 0.77\,\text{ft}\) or \(x\approx 11.73\,\text{ft}\)

- Identify the vertical and horizontal lines the rational graph approaches. - The vertical asymptote corresponds to the excluded x-value in \(x-b\). - Use two visible points to determine the line. - Set the functions equal and clear the denominator. - Factor the resulting quadratic equation.

1. The vertical asymptote is \(x=2\), so \(b=2\). The horizontal asymptote is \(y=1\), so \(c=1\). Therefore, \(f(x)=\frac{2}{x-2}+1\). 2. The line passes through \((0,-2)\) and \((2,0)\). Its slope is \(1\), and its y-intercept is \(-2\), so \(g(x)=x-2\). 3. Set the functions equal: \(\frac{2}{x-2}+1=x-2\), with \(x\ne2\). 4. Rearrange and clear the denominator: \(\frac{2}{x-2}=x-3\), so \(2=(x-3)(x-2)\). 5. Expand and simplify: \(x^2-5x+4=0\). Factor: \((x-1)(x-4)=0\), so \(x=1\) or \(x=4\). 6. Substituting into \(g\) gives \((1,-1)\) and \((4,2)\).

a) \(b=2\), \(c=1\), and \(f(x)=\frac{2}{x-2}+1\) b) \(g(x)=x-2\) c) \((1,-1)\) and \((4,2)\)

- Use each vertical asymptote to form the reciprocal denominator. - Use the horizontal or slant asymptote as the non-rational part of the rule. - Substitute one marked point to find the remaining coefficient.

1. For \(f\), the vertical asymptote is \(x=-3\), and the horizontal asymptote is \(y=2\). Use \(f(x)=2+\frac{k}{x+3}\). 2. The marked point \((-2, 3)\) lies on the graph, so \(3=2+\frac{k}{1}\), giving \(k=1\). Thus, \(f(x)=2+\frac{1}{x+3}\). 3. For \(g\), the vertical asymptote is \(x=4\). The slant asymptote passes through \((0, 1)\) and \((2, 0)\), so its slope is \(-\frac{1}{2}\), and its equation is \(y=-\frac{1}{2}x+1\). 4. Use \(g(x)=-\frac{1}{2}x+1+\frac{c}{x-4}\). The marked point \((5, -2)\) gives \(-2=-\frac{5}{2}+1+c\), so \(c=-\frac{1}{2}\). 5. Therefore, one possible rule is \(g(x)=-\frac{1}{2}x+1-\frac{1}{2(x-4)}\).

\(f(x)=2+\frac{1}{x+3}\) \(g(x)=-\frac{1}{2}x+1-\frac{1}{2(x-4)}\)

- Compare the leading terms to find the horizontal asymptote. - Read where the curve meets that horizontal level and substitute the point. - To create a hole at the excluded input, what must also be zero in the numerator?

1. Both polynomials have degree \(2\) and leading coefficient \(1\), so the horizontal asymptote is \(y=1\). The displayed graph intersects that line at \((4,1)\). 2. Substitute the intersection: \(1=\frac{16+4k}{16-4}\), so \(16+4k=12\), giving \(k=-1\). 3. A hole at \(x=2\) requires the numerator and denominator both to contain the factor \(x-2\). Therefore, \(4+2k=0\), so \(k=-2\). 4. Indeed, for \(k=-2\), \(h_{-2}(x)=\frac{x(x-2)}{(x-2)(x+2)}\), which agrees with \(\frac{x}{x+2}\) except at the excluded input \(x=2\).

a) Horizontal asymptote: \(y=1\); crossing: \((4,1)\). b) \(k=-1\). c) \(k=-2\).

- Distinguish the open point from the lines where the curve rises or falls sharply. - How could the original numerator and denominator produce those different exclusions and the shown x-intercept? - What does the marked point tell you about any overall multiplier in your possible rule? - After canceling a factor, remember that its original input restriction still applies.

1. The graph has an open circle at \(x=-3\) and vertical asymptotes at \(x=-1\) and \(x=2\). All three inputs are excluded. 2. Its x-intercept is \((1,0)\), and \(P=(0,\frac{1}{2})\). A possible rule is \(r(x)=\frac{a(x+3)(x-1)}{(x+3)(x-2)(x+1)}\). Substituting the marked point gives \(\frac{a}{2}=\frac{1}{2}\), so \(a=1\). 3. For allowed inputs, \(r(x)=\frac{x-1}{(x-2)(x+1)}\). At \(x=-3\), the simplified rule gives \(-\frac{2}{5}\), so the hole is \((-3,-\frac{2}{5})\). The simplified denominator has higher degree, so the horizontal asymptote is \(y=0\).

a) Excluded: \(x=-3,-1,2\). Hole at \(x=-3\); vertical asymptotes \(x=-1\) and \(x=2\). b) X-intercept: \((1,0)\), \(P=(0,\frac{1}{2})\). One rule is \(r(x)=\frac{(x+3)(x-1)}{(x+3)(x-2)(x+1)}\). c) Hole: \((-3,-\frac{2}{5})\); horizontal asymptote: \(y=0\).