- Find where the curve actually crosses the horizontal axis. - What is the y-coordinate of any x-intercept?

1. The curve crosses the x-axis where the output is \(0\). 2. The crossing occurs at \(x=2\), so the x-intercept is \((2,0)\).

\((2,0)\).

- Look for the vertical gap that separates the branches in each graph. - Use the horizontal scale to read each gap's x-coordinate. - Compare whether the two sides of each gap move in the same or opposite vertical directions.

1. In panel a), the curve splits at \(x=2\), so that input is excluded. One branch falls while the other rises near it. 2. In panel b), both branches rise near \(x=-1\), so \(-1\) is excluded and panel b) has the same upward behavior on both sides.

a) Missing input: \(x=2\). b) Missing input: \(x=-1\); both branches rise in panel b).

- Find the open circle rather than a filled point. - Read its horizontal and vertical coordinates separately. - Decide whether the missing point belongs to the graph.

1. The open circle lies at \((1,2)\). 2. An open circle indicates a missing point, so the function is not defined at \(x=1\), even though the nearby curve approaches height \(2\).

Hole: \((1,2)\). The function is not defined at \(x=1\).

- Find the x-value where the branches are separated. - Compare each branch with the horizontal axis. - Write one open interval for each branch.

1. To the left of \(x=1\), the branch is above the x-axis, so outputs are positive for \(x<1\). 2. To the right of \(x=1\), the branch is below the x-axis, so outputs are negative for \(x>1\). 3. The input \(x=1\) is excluded.

Positive on \((-\infty,1)\); negative on \((1,\infty)\).

- Find every part of the curve above the horizontal axis. - Notice both its x-intercept and the place where the curve is undefined. - Give a separate interval for each positive region.

1. The curve crosses the x-axis at \(x=-2\) and has a vertical asymptote at \(x=1\). 2. The graph is above the x-axis to the left of \(-2\) and to the right of \(1\). 3. Neither boundary is included because the question asks where the output is strictly positive.

\((-\infty,-2)\) and \((1,\infty)\).

- Read the height of the dashed line from the vertical scale. - Look for any intersection between the curve and that line. - What does an intersection with the dashed line tell you about the height the function reaches?

1. The dashed line is \(y=2\). 2. The curve crosses that line at the point \((0,2)\), so the asymptote's height is attained.

Yes, at \((0,2)\).

- Locate each place where the curve heads sharply up or down along a vertical line. - On each side of that line, is the curve above or below the x-axis? - Compare the two sides of both asymptotes before choosing.

1. At \(x=-2\), the nearby branches lie below the x-axis on both sides, so the sign does not switch. 2. At \(x=1\), the branch just to the left lies below the x-axis and the branch to the right lies above it. 3. The sign switches at \(x=1\).

\(x=1\).

- An x-intercept occurs where the function's output is zero. - When can a fraction equal zero? - Check the original denominator at any candidate input.

1. Set the numerator equal to zero: \(2x-6=0\), so \(x=3\). 2. The denominator at that input is \(3+2=5\), which is nonzero. 3. The x-intercept is \((3,0)\).

\((3,0)\); the denominator equals \(5\), so the point is allowed.

- Check the height of each graph at the first input in the table. - Do the same at the second input, on the other side of the vertical gap. - Both table entries must match the same graph.

1. In panel a), the left branch is below the x-axis at \(x=1\) and the right branch is above it at \(x=3\). This disagrees with the table. 2. In panel b), the left branch has output \(1\) at \(x=1\) and the right branch has output \(-1\) at \(x=3\). 3. Therefore, panel b) matches.

Panel b).

- First find the panel whose branch separation is at the stated x-value. - More than one panel may pass the first check. - Then compare the horizontal level the curves approach far away.

1. Panels a) and c) have vertical asymptote \(x=-1\), while panel b) has vertical asymptote \(x=1\). 2. Panel a) approaches \(y=2\) far from the vertical asymptote, whereas panel c) approaches \(y=-2\). 3. Panel a) has both required asymptotes.

Panel a).

- Find where the denominator is zero. - Substitute \(x=0\) directly to find the y-intercept value. - A rational function equals zero only when its numerator is zero. - Can a real number have a negative square?

1. Factor the denominator: \(x^2-6x+9=(x-3)^2\). Therefore, \(D_f=\mathbb{R}\setminus\{3\}\). 2. Evaluate the function at zero: \(f(0)=\frac{18}{9}=2\). 3. An x-intercept would require \(2x^2+18=0\), or \(x^2=-9\). 4. This equation has no real solution, so \(f\) has no x-intercepts.

a) \(D_f=\mathbb{R}\setminus\{3\}\), and \(f(0)=2\) b) The function has no x-intercepts because \(2x^2+18=0\) has no real solutions.

- Compare corresponding y-values at the same x-values. - Check whether the x-coordinates of key features remain unchanged. - A vertical scaling multiplies the entire function output.

1. The maximum value of \(f\) is \(2\), while the maximum value of \(g\) is \(1\). 2. At every x-value, the output of \(g\) is half the output of \(f\). For example, \(f(1)=1\) and \(g(1)=0.5\). 3. Therefore, the graph is vertically compressed by a factor of \(\frac{1}{2}\), and \(g(x)=\frac{1}{2}f(x)\).

Vertical compression by a factor of \(\frac{1}{2}\); \(g(x)=\frac{1}{2}f(x)\)

- Substitute \(t=0\) to find each liquid's starting temperature. - Can you write each function as a constant plus a fraction that gets smaller as time passes? - What would the model predict after a very long time?

1. At \(t=0\), \(g(0)=\frac{180}{2}=90\,\text{°C}\) and \(h(0)=\frac{160}{2}=80\,\text{°C}\). 2. Graph \(k\) begins at \(90\,\text{°C}\), so it represents \(g\). Graph \(m\) begins at \(80\,\text{°C}\), so it represents \(h\). 3. Rewrite \(g(t)=20+\frac{140}{t+2}\) and \(h(t)=10+\frac{140}{t+2}\). 4. As \(t\) becomes large, the fractions approach zero, so the horizontal asymptotes are \(T=20\,\text{°C}\) and \(T=10\,\text{°C}\). These represent the two room temperatures in the models.

a) Graph \(k\): \(g\); graph \(m\): \(h\). b) For \(g\): \(T=20\,\text{°C}\); for \(h\): \(T=10\,\text{°C}\). Each is the room temperature approached by that liquid.

- Find the x-values where each graph gets very high or low. - What does an open circle mark? - Read the x-coordinates carefully to distinguish the three descriptions.

1. In panel a), graph \(f\) has a vertical asymptote at \(x=2\) and a hole at \((1,-1)\). Thus, it matches description 2. 2. In panel b), graph \(g\) has a vertical asymptote at \(x=-2\) and a hole at \((-1,-1)\). Thus, it matches description 1. 3. In panel c), graph \(h\) has vertical asymptotes at \(x=-2\) and \(x=2\) and a hole at \((0,0)\). Thus, it matches description 3.

\(f\) (panel a): 2; \(g\) (panel b): 1; \(h\) (panel c): 3.

- Read the x-value of the vertical gap between branches. - Look for the horizontal height the graph approaches far from that gap. - The y-intercept is where the curve meets the vertical axis.

1. The graph has a vertical gap at \(x=2\), so \(x=2\) is excluded. 2. Far to either side, the graph approaches height \(1\), so the horizontal asymptote is \(y=1\). 3. The curve crosses the y-axis at the origin, so the y-intercept is \((0,0)\).

a) \(x=2\). b) \(y=1\). c) \((0,0)\).

- Check where each part of the curve lies relative to the x-axis. - Find the x-values where the sign changes. - At which values does the curve cross the axis, and at which is there a break?

1. Reading from left to right, the graph is below the x-axis on \((-\infty,-2)\), above it on \((-2,1)\), below it on \((1,3)\), and above it on \((3,\infty)\). 2. The graph crosses the x-axis at \(x=-2\) and \(x=3\). At those inputs, the output is zero. 3. At \(x=1\), the curve splits into branches that move sharply up and down. That input is excluded because it is a vertical asymptote.

a) Positive: \((-2,1)\) and \((3,\infty)\). Negative: \((-\infty,-2)\) and \((1,3)\). b) \(x=1\) is excluded: the graph has a vertical asymptote there. The other two values are x-intercepts.

- Find the lowest horizontal level the graph approaches. - Check whether the plotted curve reaches that level and whether its outputs seem bounded above. - Can you be certain about every input using only a finite drawing?

1. The plotted branches lie above \(y=2\) and appear to get closer to it without touching it. 2. The branches rise sharply near \(x=1\), suggesting no upper bound. The suggested range is \((2,\infty)\), with \(2\) excluded. 3. A finite graph cannot prove what happens at every input. A function rule or further reasoning is needed to confirm that the suggested range is exact.

Suggested range: \((2,\infty)\); \(2\) is excluded. No, a finite graph alone cannot prove the full range.

- Read the vertical gap in the displayed graph. - Predict the vertical gap from the proposed denominator. - Compare the distant horizontal level with the proposed vertical shift.

1. The displayed graph has vertical asymptote \(x=2\), but \(f\) would have vertical asymptote \(x=-2\). 2. The displayed graph approaches the horizontal line \(y=2\), but \(f\) would approach \(y=1\). 3. The student is incorrect because both asymptotes disagree.

The graph's vertical asymptote is \(x=2\), not \(x=-2\); its horizontal asymptote is \(y=2\), not \(y=1\).

- Which values make the denominator zero? - A rational function equals zero when its numerator is zero and its denominator is nonzero. - Check every numerator zero against the domain.

1. Factor the denominator: \(x^2-1=(x-1)(x+1)\). Therefore, the domain is \(D_f=\mathbb{R}\setminus\{-1, 1\}\). 2. Factor the numerator: \(x^2-4x+3=(x-1)(x-3)\). Its zeros are \(x=1\) and \(x=3\). 3. An x-intercept must be in the domain. Since \(x=1\) is excluded, it is not an x-intercept. The only x-intercept occurs at \(x=3\), giving the point \((3, 0)\).

The domain is \(D_f=\mathbb{R}\setminus\{-1, 1\}\). The only x-intercept is \((3, 0)\).

- When is a fraction equal to zero? - Which values are excluded by each denominator? - What happens when the same value makes both the numerator and denominator zero? - Factor each quadratic numerator when helpful.

1. For part a, set the numerator equal to zero: \(2x+6=0\), so \(x=-3\). The denominator is nonzero there, so the x-intercept is \((-3, 0)\). 2. For part b, factor the numerator: \(x^2-16=(x-4)(x+4)\). Its zeros are \(x=4\) and \(x=-4\), but \(x=-4\) is excluded because it makes the denominator zero. Thus, the only x-intercept is \((4, 0)\). 3. For part c, factor the numerator: \(x^2-3x-10=(x-5)(x+2)\). The zeros are \(x=5\) and \(x=-2\). The denominator is nonzero at both values, so the x-intercepts are \((5, 0)\) and \((-2, 0)\).

a) \((-3, 0)\) b) \((4, 0)\) c) \((5, 0)\) and \((-2, 0)\)

- Rewrite each equation so that one side equals zero. - When is a product equal to zero? - Check every candidate against the function’s domain. - What does it mean when a numerator zero is excluded by the denominator?

1. For part a, solve \(\frac{5}{x+2}-1=0\). Then \(\frac{5}{x+2}=1\), so \(x=3\). The denominator is nonzero, giving the x-intercept \((3, 0)\). 2. For part b, the numerator is zero when \(x-1=0\) or \(3x+2=0\). This gives \(x=1\) and \(x=-\frac{2}{3}\). Neither value is excluded by the denominator, so both are x-intercepts. 3. For part c, factor the numerator and denominator: \(3x^2-12x+12=3(x-2)^2\) and \(x^2-4=(x-2)(x+2)\). The only numerator zero is \(x=2\), but this value is excluded from the domain. Therefore, \(r\) has no x-intercepts.

a) \((3, 0)\) b) \((1, 0)\) and \(\left(-\frac{2}{3}, 0\right)\) c) No x-intercepts

- First find the zeros of the denominator. - What happens when the same value makes both the numerator and denominator zero? - An x-intercept must be a point where the function is defined and has output zero.

1. Factor the denominator: \(x^2-5x+6=(x-2)(x-3)\). Therefore, \(D_g=\mathbb{R}\setminus\{2, 3\}\). 2. Set the numerator equal to zero: \(2x-6=0\), so the only candidate is \(x=3\). 3. Since \(x=3\) is excluded from the domain, it cannot be an x-intercept. 4. Therefore, \(g\) has no x-intercepts.

\(D_g=\mathbb{R}\setminus\{2, 3\}\). The function has no x-intercepts because the only zero of the numerator, \(x=3\), is not in the domain.

- A rational function is undefined where its denominator is zero. - Is a zero of the numerator automatically an x-intercept? - Compare the denominator zeros of the two functions. - Every x-intercept must lie in the domain.

1. For \(f\), the denominator is \((x-1)^2\), so \(x=1\) is excluded. For \(g\), the denominator is \(x-1\), so \(x=1\) is also excluded. Thus, both domains are \(\mathbb{R}\setminus\{1\}\). 2. Solving \(x^2-1=0\) gives the numerator zeros \(x=1\) and \(x=-1\). 3. The value \(x=1\) is not in the domain of \(f\), so it is not an x-intercept. Only \(x=-1\) gives an x-intercept. 4. Therefore, part a is true, while part b is false because numerator zeros must still be checked against the domain.

a) True. Both functions have domain \(\mathbb{R}\setminus\{1\}\). b) False. Solving the numerator equation gives \(x=1\) and \(x=-1\), but \(x=1\) is excluded from the domain. The only x-intercept is \((-1, 0)\).

- Factor the quadratic denominator to find every excluded value. - For an x-intercept, the numerator must be zero and the denominator must be nonzero. - Simplifying the expression does not restore values excluded from the original domain. - Check each numerator zero individually.

1. Factor the denominator: \(x^2-16=(x-4)(x+4)\). Therefore, \(D_h=\mathbb{R}\setminus\{-4, 4\}\), so statement a is false. 2. Factor the numerator: \(x^2-4x=x(x-4)\). Its zeros are \(x=0\) and \(x=4\). 3. The value \(x=0\) is in the domain, so \((0, 0)\) is an x-intercept. Statement b is true. 4. The value \(x=4\) is excluded from the domain, so it is not an x-intercept. The function has only one x-intercept, making statement c false.

a) False. \(D_h=\mathbb{R}\setminus\{-4, 4\}\). b) True. \((0, 0)\) is an x-intercept. c) False. The function has exactly one x-intercept because \(x=4\) is excluded from the domain.

- Find the minimum possible value of the denominator. - Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - Use \(x=0\) for the y-intercept and set the numerator equal to \(0\) for x-intercepts. - Compare leading coefficients for the end behavior.

1. Since \(0.5x^2+2\ge 2\) for every real \(x\), the denominator is never \(0\). Thus, the domain is all real numbers and there are no vertical asymptotes. 2. Because \(f(-x)=f(x)\), the function is even and the graph is symmetric about the y-axis. 3. The y-intercept is \(f(0)=-8\), so it is \((0, -8)\). For x-intercepts, solve \(4x^2-16=0\), giving \(x=\pm 2\). The x-intercepts are \((-2, 0)\) and \((2, 0)\). 4. For large positive or negative \(x\), the highest-degree terms control the fraction. Their coefficient ratio is \(\frac{4}{0.5}=8\), so the graph approaches the horizontal asymptote \(y=8\) at both ends.

a) Domain: \(\mathbb{R}\); no vertical asymptotes b) Symmetric about the y-axis c) y-intercept: \((0, -8)\); x-intercepts: \((-2, 0)\) and \((2, 0)\) d) Horizontal asymptote: \(y=8\)

- Set the denominator equal to \(0\) to find excluded values. - Compare \(f(-x)\) and \(f(x)\). - An x-intercept requires the numerator to equal \(0\). - Use one-sided sign analysis near each excluded value and compare leading coefficients for the end behavior.

1. Solve \(0.25x^2-4=0\). This gives \(x=\pm 4\), so the domain is \(\mathbb{R}\setminus\{-4, 4\}\). 2. Since \(f(-x)=f(x)\), the function is even and its graph is symmetric about the y-axis. 3. An x-intercept would require \(x^2+1=0\), which has no real solution. Therefore, there are no x-intercepts. 4. The numerator and denominator have the same degree, so the graph approaches the horizontal asymptote \(y=\frac{1}{0.25}=4\) at both ends. 5. The numerator is always positive. Near \(x=-4\), the denominator is positive on the left and negative on the right, so the graph rises without bound on the left and falls without bound on the right. 6. Near \(x=4\), the denominator is negative on the left and positive on the right, so the graph falls without bound on the left and rises without bound on the right.

a) \(\mathbb{R}\setminus\{-4, 4\}\) b) Symmetric about the y-axis c) No x-intercepts because \(x^2+1>0\) for all real \(x\) d) At \(x=-4\): rises without bound from the left and falls without bound from the right. At \(x=4\): falls without bound from the left and rises without bound from the right. As \(x\to\pm\infty\), \(f(x)\) approaches \(4\), so \(y=4\) is the horizontal asymptote.

- Can the denominator ever equal \(0\)? - Compare \(k(-x)\) with \(k(x)\). - Set \(x=0\) for the y-intercept and set the numerator equal to \(0\) for x-intercepts. - Compare leading coefficients for the end behavior.

1. Since \(x^2+2>0\) for all real \(x\), the domain is \(\mathbb{R}\). Also, \(k(-x)=k(x)\), so the graph is symmetric about the y-axis. 2. The y-intercept is \(k(0)=-9\), so it is \((0, -9)\). For x-intercepts, solve \(2x^2-18=0\), giving \(x=\pm 3\). The x-intercepts are \((-3, 0)\) and \((3, 0)\). 3. The numerator and denominator have the same degree and leading-coefficient ratio \(\frac{2}{1}=2\), so \(k(x)\) approaches \(2\) as \(x\) becomes very large in either direction. The horizontal asymptote is \(y=2\). 4. \(k(1)=\frac{-16}{3}\), and \(k(4)=\frac{14}{18}=\frac{7}{9}\).

a) Domain: \(\mathbb{R}\); symmetric about the y-axis b) y-intercept: \((0, -9)\); x-intercepts: \((-3, 0)\) and \((3, 0)\) c) \(k(x)\to 2\) as \(x\to\pm\infty\); horizontal asymptote: \(y=2\) d) \(k(1)=-\frac{16}{3}\), \(k(4)=\frac{7}{9}\)

- How can you determine whether a quadratic is ever zero? - What is the \(x\)-coordinate of every point on the y-axis? - What must be true of the numerator for a rational function to equal zero? - Can you find a positive lower bound for the numerator?

1. Complete the square in the denominator: \(x^2+4x+6=(x+2)^2+2\), which is positive for every real \(x\). Thus, \(D_g=\mathbb{R}\). 2. Evaluate \(g(0)\): \(g(0)=\frac{5}{6}\). Therefore, the y-intercept is \(\left(0, \frac{5}{6}\right)\). 3. The numerator satisfies \(x^2+5\geq 5>0\) for every real \(x\). Since it is never zero, the graph has no x-intercepts.

a) \(D_g=\mathbb{R}\) b) \(\left(0, \frac{5}{6}\right)\) c) There are no x-intercepts because \(x^2+5\) is positive for every real \(x\).

- Factor the common denominator. - A rational function is zero when its numerator is zero and its denominator is nonzero. - Compare \(g(x)\) directly with \(f(x)\) to identify the transformation.

1. The common denominator factors as \((x-2)(x+2)\). No numerator cancels either factor, so the domain is \(\mathbb{R}\setminus\{-2, 2\}\), and the vertical asymptotes are \(x=-2\) and \(x=2\). 2. \(f(3)=\frac{3}{5}=0.6\), \(g(3)=\frac{12}{5}=2.4\), and \(h(3)=2\). Also, \(f(0)=0\), \(g(0)=0\), and \(h(0)=-\frac{1}{4}\). 3. Since \(g(x)=4f(x)\), the graph of \(g\) is a vertical stretch of the graph of \(f\) by a factor of \(4\). 4. The functions \(f\) and \(g\) have the x-intercept \((0, 0)\). The numerator \(x^2+1\) has no real zeros, so \(h\) has no x-intercepts.

a) Domain: \(\mathbb{R}\setminus\{-2, 2\}\); vertical asymptotes: \(x=-2\) and \(x=2\) b) \(f(3)=0.6\), \(g(3)=2.4\), \(h(3)=2\); \(f(0)=0\), \(g(0)=0\), \(h(0)=-\frac{1}{4}\) c) Vertical stretch by a factor of \(4\) d) \(f\) and \(g\): \((0, 0)\); \(h\): no x-intercepts

- What happens when \(x\) is replaced by \(-x\) in an even power? - Compare the degrees and leading coefficients for the end behavior. - Multiplying all function values by \(-1\) produces which reflection?

1. Each formula contains only even powers of \(x\), so \(p(-x)=p(x)\), \(q(-x)=q(x)\), and \(r(-x)=r(x)\). All three graphs are symmetric about the y-axis. 2. For \(p\) and \(q\), the denominator has greater degree than the numerator, so both functions approach \(0\). Their horizontal asymptote is \(y=0\). 3. For \(r\), the numerator and denominator have the same degree, and the ratio of leading coefficients is \(2\). Its horizontal asymptote is \(y=2\). 4. Since \(q(x)=-p(x)\), the graph of \(q\) is the reflection of the graph of \(p\) across the x-axis.

a) All three graphs are symmetric about the y-axis. b) \(p\) and \(q\): \(y=0\); \(r\): \(y=2\) c) The graph of \(q\) is the reflection of the graph of \(p\) across the x-axis.

- Compare \(f(-x)\) with \(f(x)\). - Substitute the given x-values directly. - What happens when a constant numerator is divided by a denominator that grows without bound? - When can a rational expression equal \(0\)?

1. Since \(f(-x)=f(x)\), the function is even and its graph is symmetric about the y-axis. 2. \(f(-3)=1\), \(f(0)=4\), and \(f(3)=1\). 3. The denominator has greater degree than the numerator, so the function approaches \(0\) as \(x\) becomes very large positive or negative. The horizontal asymptote is \(y=0\). 4. A rational function can equal \(0\) only when its numerator is \(0\). The numerator is the nonzero constant \(12\), so there are no x-intercepts.

a) Symmetric about the y-axis b) \(f(-3)=1\), \(f(0)=4\), \(f(3)=1\) c) \(f(x)\to 0\) as \(x\to\pm\infty\); horizontal asymptote: \(y=0\) d) No x-intercepts

- Compare \(g(-x)\) with \(g(x)\). - Find the smallest possible denominator. - Use the signs of the numerator and denominator to determine whether the function can be positive or zero. - Consider what happens as the denominator becomes very large.

1. \(g(-x)=\frac{-10}{(-x)^2+2}=\frac{-10}{x^2+2}=g(x)\), so the graph is symmetric about the y-axis. 2. \(g(0)=-5\), so the y-intercept is \((0, -5)\). Also, \(g(2)=\frac{-10}{6}=-\frac{5}{3}\). 3. Since \(x^2+2\ge 2\), the denominator is positive and has its smallest value at \(x=0\). Thus, the minimum function value is \(-5\). All values are negative, and they approach but never equal \(0\). Therefore, the range is \([-5, 0)\). 4. As \(x\to\pm\infty\), the denominator grows without bound, so \(g(x)\to 0\).

a) \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) y-intercept: \((0, -5)\); \(g(2)=-\frac{5}{3}\) c) Range: \([-5, 0)\) d) \(g(x)\to 0\) as \(x\to\pm\infty\)

- Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - Set the denominator equal to \(0\). - The multiplicity of a denominator zero helps determine whether the sign changes. - Compare the degrees for the end behavior.

1. Since \(f(-x)=-f(x)\), the function is odd and its graph is symmetric about the origin. 2. \(f(0)=0\), and \(f(1)=\frac{1}{1-2.25}=-0.8\). 3. Solve \(x^2-2.25=0\), giving \(x=\pm 1.5\). Thus, the domain is \(\mathbb{R}\setminus\{-1.5, 1.5\}\). 4. Each excluded value is a simple zero of the denominator and is not a zero of the numerator. Therefore, \(x=-1.5\) and \(x=1.5\) are vertical asymptotes, and the function changes sign across each one. 5. The denominator has greater degree than the numerator, so \(f(x)\to 0\) as \(x\to\pm\infty\).

a) Symmetric about the origin; \(f(0)=0\), \(f(1)=-0.8\) b) Domain: \(\mathbb{R}\setminus\{-1.5, 1.5\}\); vertical asymptotes: \(x=-1.5\) and \(x=1.5\), with a sign change at each c) \(f(x)\to 0\) as \(x\to\pm\infty\)

- Replace \(x\) with \(-x\) to test symmetry. - Set the denominator equal to \(0\) to find excluded values. - Compare leading coefficients for the horizontal asymptote. - Set the numerator equal to \(0\) for x-intercepts.

1. Since \(g(-x)=g(x)\), the function is even and its graph is symmetric about the y-axis. 2. The denominator is zero when \(9-x^2=0\), so \(x=\pm 3\). The domain is \(\mathbb{R}\setminus\{-3, 3\}\), and the vertical asymptotes are \(x=-3\) and \(x=3\). 3. The numerator and denominator have the same degree. The ratio of leading coefficients is \(\frac{2}{-1}=-2\), so the horizontal asymptote is \(y=-2\). 4. Solve \(2x^2-2=0\), giving \(x=\pm 1\). Thus, the x-intercepts are \((-1, 0)\) and \((1, 0)\).

a) \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) Domain: \(\mathbb{R}\setminus\{-3, 3\}\); vertical asymptotes: \(x=-3\) and \(x=3\) c) Horizontal asymptote: \(y=-2\) d) x-intercepts: \((-1, 0)\) and \((1, 0)\)

- Compare \(f(-x)\) with \(f(x)\). - Set the function equal to \(0\) to find x-intercepts. - Substitute \(x=0\) to check the given point. - What happens to the rational term as \(\lvert x\rvert\) becomes large?

1. Since \(f(-x)=f(x)\), the function is even and its graph is symmetric about the y-axis. 2. Solve \(\frac{4}{x^2-25}+0.2=0\). Then \(4=-0.2(x^2-25)\), so \(x^2=5\), giving \(x=\pm\sqrt{5}\). 3. Evaluate \(f(0)=\frac{4}{-25}+0.2=-0.16+0.2=0.04\), so \((0, 0.04)\) lies on the graph. 4. As \(x\) becomes very large positive or negative, \(\frac{4}{x^2-25}\) approaches \(0\). Thus \(f(x)\) approaches \(0.2\) at both ends, and the horizontal asymptote is \(y=0.2\).

a) Symmetric about the y-axis b) x-intercepts: \((-\sqrt{5}, 0)\) and \((\sqrt{5}, 0)\); \(f(0)=0.04\) c) \(f(x)\to 0.2\) as \(x\to\pm\infty\); horizontal asymptote: \(y=0.2\)

- Set each denominator factor equal to \(0\). - Compare the leading coefficients after expanding the denominator. - An x-intercept requires the numerator to equal \(0\). - Compare \(f(-x)\) with \(f(x)\), and use \(x=0\) for the y-intercept.

1. The denominator is zero at \(x=-2\) and \(x=2\). The numerator is nonzero at both values, so the domain is \(\mathbb{R}\setminus\{-2, 2\}\), and the vertical asymptotes are \(x=-2\) and \(x=2\). 2. Since \((2x-4)(2x+4)=4x^2-16\), the numerator and denominator have the same degree. The horizontal asymptote is \(y=\frac{3}{4}\). 3. An x-intercept would require \(3x^2+9=0\), or \(x^2=-3\), which has no real solution. 4. Because \(f(-x)=f(x)\), the graph is symmetric about the y-axis. Also, \(f(0)=\frac{9}{-16}=-\frac{9}{16}\), so the y-intercept is \(\left(0, -\frac{9}{16}\right)\).

a) Domain: \(\mathbb{R}\setminus\{-2, 2\}\); vertical asymptotes: \(x=-2\) and \(x=2\) b) Horizontal asymptote: \(y=\frac{3}{4}\) c) No x-intercepts d) Symmetric about the y-axis; y-intercept: \(\left(0, -\frac{9}{16}\right)\)

- Test the proposed output \(y=3\) by solving the corresponding equation. - Split the domain into intervals on which \(x^2-1\) has the same sign. - On each interval, identify which output values are possible and which boundary values are approached but not reached.

1. Set \(g(x)=3\): \(\frac{3x^2}{x^2-1}=3\). Multiplying by the nonzero denominator gives \(3x^2=3x^2-3\), a contradiction. Thus the graph never reaches \(y=3\). 2. For \(|x|>1\), rewrite \(g(x)=3+\frac{3}{x^2-1}\). The fraction is positive, so \(g(x)>3\). It approaches \(3\) for large \(|x|\) and grows without bound near \(x=\pm1\) from outside. This gives outputs \((3,\infty)\). 3. For \(-1<x<1\), the denominator is negative and the numerator is nonnegative. Thus \(g(x)\le0\), with \(g(0)=0\); the outputs decrease without bound near the excluded inputs from inside. This gives \((-\infty,0]\). 4. Combining these sets, the range is \((-\infty,0]\cup(3,\infty)\).

a) The equation leads to \(0=-3\), so there is no intersection. b) \((-\infty,0]\cup(3,\infty)\)

- Use \(x^2\ge 0\) and \(t>0\) to determine the signs of the numerator and denominator. - Substitute \(x=0\). - Compare the leading coefficients or divide by \(x^2\).

1. Since \(t>0\), both \(x^2+2t\) and \(2x^2+t\) are positive for every real \(x\). Thus, the domain is \(\mathbb{R}\), and there are no x-intercepts. 2. Evaluate \(g_t(0)=\frac{t}{2t}=\frac{1}{2}=0.5\), so every graph passes through \((0, 0.5)\). 3. Divide the numerator and denominator by \(x^2\) for nonzero \(x\): \(g_t(x)=\frac{2+\frac{t}{x^2}}{1+\frac{2t}{x^2}}\). 4. As \(x\) becomes very large positive or negative, the terms containing \(\frac{1}{x^2}\) approach \(0\). Therefore, the function approaches \(2\) at both ends. This is independent of \(t\) because the leading coefficients are always \(2\) and \(1\).

a) Domain: \(\mathbb{R}\); no x-intercepts b) \(g_t(0)=0.5\) c) \(g_t(x)\to 2\) as \(x\to\pm\infty\); horizontal asymptote: \(y=2\), independent of \(t\)

- Compare leading coefficients for the horizontal asymptote. - Analyze when \(kx^2=18\) can have real solutions. - A hole occurs when the same factor appears in the numerator and denominator. - Evaluate the simplified expression at the excluded x-value to locate the hole.

1. The numerator and denominator have the same degree and leading-coefficient ratio \(k\), so the horizontal asymptote is \(y=k\). 2. X-intercepts require \(kx^2-18=0\). If \(k=0\), this becomes \(-18=0\). If \(k<0\), then \(x^2=\frac{18}{k}<0\). Thus, there are no real x-intercepts for \(k\le 0\). 3. For \(x=3\) to be a hole rather than a vertical asymptote, the numerator must also be zero there: \(9k-18=0\), so \(k=2\). 4. For \(k=2\), \(f_2(x)=\frac{2(x-3)(x+3)}{(x-3)(x+2)}=\frac{2(x+3)}{x+2}\) for \(x\neq 3,-2\). The graph approaches the height \(\frac{2(6)}{5}=\frac{12}{5}\) as \(x\) approaches \(3\). The hole is \(\left(3, \frac{12}{5}\right)\).

a) \(y=k\) b) No x-intercepts for \(k\le 0\) c) \(k=2\); hole: \(\left(3, \frac{12}{5}\right)\)

- Compare leading coefficients. - Determine when \(x^2=-\frac{a}{2}\) has two real solutions. - Check whether either denominator zero also makes the numerator zero. - For \(a=0\), test the numerator at \(x=\pm 4\).

1. The numerator and denominator have the same degree, and the ratio of leading coefficients is \(2\). Therefore, every graph has horizontal asymptote \(y=2\). 2. X-intercepts satisfy \(2x^2+a=0\), or \(x^2=-\frac{a}{2}\). There are exactly two real solutions when \(a<0\). 3. The denominator is zero at \(x=\pm 4\). To keep the expression fully reduced, the numerator cannot also be zero there. Since \(2(4^2)+a=32+a\), require \(a\neq-32\). Thus, \(a<0\) and \(a\neq-32\). 4. For \(a=0\), the numerator is nonzero at \(x=\pm 4\), so the vertical asymptotes are \(x=-4\) and \(x=4\).

a) \(y=2\) b) \(a<0\) and \(a\neq-32\) c) \(x=-4\) and \(x=4\)

- Set the two function expressions equal. - Subtract the common constant term first. - Clear the denominators while keeping track of excluded values. - Substitute each valid x-value into either function.

1. Set the functions equal: \(\frac{1}{x-2}+2=2-\frac{1}{x^2}\), with \(x\neq 0, 2\). 2. Subtract \(2\): \(\frac{1}{x-2}=-\frac{1}{x^2}\). 3. Clear the denominators to get \(x^2=-(x-2)\), so \(x^2+x-2=0\). 4. Factor: \((x+2)(x-1)=0\), giving \(x=-2\) and \(x=1\). Both are in the common domain. 5. Evaluate \(f(1)=1\) and \(f(-2)=\frac{7}{4}\).

The intersection points are \((1, 1)\) and \(\left(-2, \frac{7}{4}\right)\).

- Find denominator zeros and compare degrees. - Set the two function expressions equal. - Clear denominators and solve the resulting quadratic equation. - Verify that each solution lies in both domains.

1. For \(f\), the vertical asymptote is \(x=3\), and the horizontal asymptote is \(y=0\). 2. For \(g\), the vertical asymptote is \(x=1\), and the horizontal asymptote is \(y=0\). 3. Set the functions equal: \(\frac{4}{x-3}=\frac{-4}{(x-1)^2}\), where \(x\neq 1, 3\). 4. Clear the denominators to get \((x-1)^2=-(x-3)\). This simplifies to \(x^2-x-2=0\), so \(x=2\) or \(x=-1\). 5. Evaluate \(f(2)=-4\) and \(f(-1)=-1\).

a) \(f\): vertical \(x=3\), horizontal \(y=0\); \(g\): vertical \(x=1\), horizontal \(y=0\) b) \((2, -4)\) and \((-1, -1)\)

- Set the function expressions equal. - Clear denominators while keeping track of excluded values. - Use the discriminant to determine the number of real solutions to a quadratic equation. - A contradiction means the graphs do not intersect.

1. For a), set the functions equal and clear the denominator: \(x^2-2x+3=(2x-1)(x-2)\). This simplifies to \(x^2-3x-1=0\). 2. The discriminant is \(13>0\), so the equation has two distinct real solutions. Neither solution is the excluded value \(x=2\). Therefore, there are two intersection points. 3. For b), rewrite the left side as \(\frac{x+1}{x-1}\), then set \(\frac{x+1}{x-1}=\frac{x+3}{x+1}\). 4. Cross-multiplication gives \((x+1)^2=(x+3)(x-1)\), which simplifies to the contradiction \(1=-3\). Therefore, there are no intersection points.

a) \(2\) intersection points b) \(0\) intersection points

- Set the two functions equal. - Clear the denominator and solve the resulting quadratic equation. - Check each solution against the domain. - Substitute valid x-values into the line to find the y-coordinates.

1. Set the functions equal, with \(x\neq 2\): \(\frac{x^2-4x+3}{x-2}=3x-9\). 2. Clear the denominator: \(x^2-4x+3=(3x-9)(x-2)\). 3. Simplify to \(2x^2-11x+15=0\), which factors as \((2x-5)(x-3)=0\). 4. Thus, \(x=\frac{5}{2}\) or \(x=3\). Both values are in the domain. 5. Evaluate \(g\left(\frac{5}{2}\right)=-\frac{3}{2}\) and \(g(3)=0\).

The intersection points are \(\left(\frac{5}{2}, -\frac{3}{2}\right)\) and \((3, 0)\).

- Identify all excluded x-values first. - Multiply by the least common denominator. - Check every algebraic solution against the original domains. - Substitute the valid x-value into either function.

1. The common domain excludes \(x=-1\) and \(x=1\). 2. Set the functions equal: \(\frac{2}{x^2-1}=\frac{1}{x-1}-1\). 3. Multiply by \((x-1)(x+1)\): \(2=x+1-(x^2-1)\). 4. Simplify to \(x^2-x=0\), so \(x=0\) or \(x=1\). 5. The value \(x=1\) is excluded. For \(x=0\), \(g(0)=-2\).

The only intersection point is \((0, -2)\).

- Factor the denominator and compare leading coefficients. - Combine the terms in the proposed expression over a common denominator. - Compare \(k(-x)\) with \(k(x)\). - Use the sign of the remainder term to compare the graph with the asymptote.

1. The denominator is zero at \(x=\pm 1\), and the numerator is nonzero there. Thus, the domain is \(\mathbb{R}\setminus\{-1, 1\}\), and the vertical asymptotes are \(x=-1\) and \(x=1\). 2. The numerator and denominator have the same degree, so the horizontal asymptote is \(y=1\). 3. Also, \(1-\frac{3}{x^2-1}=\frac{x^2-1-3}{x^2-1}=\frac{x^2-4}{x^2-1}\). 4. Since \(k(-x)=k(x)\), the graph is symmetric about the y-axis. 5. The graph is above \(y=1\) when \(-\frac{3}{x^2-1}>0\). This occurs when \(x^2-1<0\), or \(-1<x<1\).

a) Domain: \(\mathbb{R}\setminus\{-1, 1\}\); vertical asymptotes: \(x=-1\), \(x=1\); horizontal asymptote: \(y=1\) b) \(k(x)=1-\frac{3}{x^2-1}\) c) Symmetric about the y-axis d) \((-1, 1)\)

- Use the least common denominator. - Compare \(f(-x)\) with \(f(x)\). - Find denominator zeros and compare degrees. - Substitute \(x=0\) for the y-intercept.

1. Use the common denominator \((x-2)(x+2)\): \(f(x)=\frac{x+2-(x-2)}{(x-2)(x+2)}=\frac{4}{x^2-4}\). 2. Since \(f(-x)=\frac{4}{(-x)^2-4}=f(x)\), the graph is symmetric about the y-axis. 3. The denominator is zero at \(x=-2\) and \(x=2\), so these are vertical asymptotes. Since the denominator has greater degree than the numerator, the horizontal asymptote is \(y=0\). 4. Evaluate \(f(0)=\frac{4}{-4}=-1\), so the y-intercept is \((0, -1)\).

1) \(f(x)=\frac{4}{x^2-4}\) 2) Symmetric about the y-axis 3) Vertical asymptotes: \(x=-2\) and \(x=2\); horizontal asymptote: \(y=0\) 4) y-intercept: \((0, -1)\)

- Use a common denominator to subtract the fractions. - Complete the square in the quadratic denominator. - A rational expression is zero only when its numerator is zero. - Find denominator zeros and compare degrees.

1. Use the common denominator \(x(x+4)\): \(g(x)=\frac{x+4-x}{x(x+4)}=\frac{4}{x^2+4x}\). 2. Since \((x+2)^2-4=x^2+4x\), the second form is equivalent. 3. A rational expression equals \(0\) only when its numerator is \(0\). The numerator is the nonzero constant \(4\), so there are no x-intercepts. 4. The denominator is zero at \(x=-4\) and \(x=0\), so these are vertical asymptotes. Since the denominator has greater degree than the numerator, the horizontal asymptote is \(y=0\).

1) \(g(x)=\frac{4}{x^2+4x}\) 2) \(g(x)=\frac{4}{(x+2)^2-4}\) 3) No x-intercepts 4) Vertical asymptotes: \(x=-4\) and \(x=0\); horizontal asymptote: \(y=0\)

- Compare the outside coefficient and the input in the two rules. - Decide whether a vertical stretch moves a vertical symmetry line. - A horizontal translation moves every vertical feature by the same amount. - Track the location of the original symmetry line.

1. Multiplying \(f\) by \(3\) gives \(3f(x)=\frac{18}{x^2+3}\), a vertical stretch by a factor of \(3\). 2. Replacing \(x\) with \(x+2\) shifts the graph left \(2\) units, producing \(g(x)=\frac{18}{(x+2)^2+3}\). 3. A vertical stretch does not move a vertical line of symmetry, so the symmetry line remains \(x=0\) after the stretch. 4. Shifting the graph left \(2\) units moves the symmetry line from \(x=0\) to \(x=-2\).

1. Vertically stretch by a factor of \(3\), then shift left \(2\) units. 2. The vertical stretch leaves the symmetry line at \(x=0\); the left shift moves it to \(x=-2\).

- Find the denominator zero and compare leading coefficients. - Use \(x=0\) for the y-intercept and set the numerator equal to \(0\) for the x-intercept. - Rewrite \(f\) as a constant plus a rational remainder before adding \(x\).

1. The denominator is zero at \(x=-1\), so the domain is \(\mathbb{R}\setminus\{-1\}\), and the vertical asymptote is \(x=-1\). 2. The numerator and denominator have the same degree, so the horizontal asymptote is \(y=-2\). 3. Evaluate \(f(0)=6\), so the y-intercept is \((0, 6)\). Solve \(6-2x=0\), giving the x-intercept \((3, 0)\). 4. Rewrite \(f(x)=-2+\frac{8}{x+1}\). Then \(h(x)=x-2+\frac{8}{x+1}\). 5. Therefore, \(h\) has vertical asymptote \(x=-1\) and slant asymptote \(y=x-2\).

a) Domain: \(\mathbb{R}\setminus\{-1\}\); vertical asymptote: \(x=-1\); horizontal asymptote: \(y=-2\) b) x-intercept: \((3, 0)\); y-intercept: \((0, 6)\) c) Vertical asymptote: \(x=-1\); slant asymptote: \(y=x-2\)

- Compare \(f(-x)\) with \(f(x)\). - Divide the numerator and denominator by \(x^2\). - Clear the denominator and solve the resulting quadratic equation.

1. \(f(-x)=\frac{2(-x)^2}{(-x)^2+1}=f(x)\), so the graph is symmetric about the y-axis. 2. Divide by \(x^2\) for nonzero \(x\): \(f(x)=\frac{2}{1+\frac{1}{x^2}}\). As \(x\) becomes very large, the denominator approaches \(1\), so the function approaches \(2\). The horizontal asymptote is \(y=2\). 3. Solve \(\frac{2x^2}{x^2+1}=1.8\). This gives \(2x^2=1.8x^2+1.8\), so \(0.2x^2=1.8\), \(x^2=9\), and \(x=\pm 3\).

1) Symmetric about the y-axis 2) \(f(x)\to 2\) as \(x\to\infty\); horizontal asymptote: \(y=2\) 3) \(x=-3\) and \(x=3\)

- Rewrite the numerator using the denominator. - Determine the possible values of \(x^2+2\). - Track which endpoint values are reached and which are only approached.

1. Rewrite the function: \(g(x)=\frac{x^2+2-6}{x^2+2}=1-\frac{6}{x^2+2}\). 2. Since \(x^2\ge0\), the denominator \(x^2+2\) takes all values in \([2, \infty)\). 3. Therefore, \(\frac{6}{x^2+2}\) takes all values in \((0, 3]\). The value \(3\) occurs at \(x=0\), while \(0\) is approached but never reached. 4. Subtracting these values from \(1\) gives \(-2\le g(x)<1\). Hence, the range is \([-2, 1)\).

\([-2, 1)\)

- A rational function equals zero when its numerator is zero and its denominator is nonzero. - Check the domain before accepting numerator zeros. - In part c, notice that the numerator contains only even powers. Look for a factorization in terms of \(x^2\). - Factor each numerator completely before deciding which candidate zeros survive the domain check.

1. For part a, the denominator is positive for every real \(x\). Factor the numerator: \(x^4-5x^3=x^3(x-5)\). The x-intercepts are \((0,0)\) and \((5,0)\). 2. For part b, \(x=2\) is excluded from the domain. Factor the numerator: \(x^2-7x+10=(x-2)(x-5)\). The candidate \(x=2\) is excluded, so the only x-intercept is \((5,0)\). 3. For part c, the denominator is positive for every real \(x\). The numerator contains only even powers: \(x^4-17x^2+16=(x^2-1)(x^2-16)\). Therefore, \(x=\pm1\) or \(x=\pm4\).

a) \((0,0)\) and \((5,0)\) b) \((5,0)\) c) \((-4,0)\), \((-1,0)\), \((1,0)\), and \((4,0)\)

- Set the two function expressions equal. - Clear the denominator, while keeping the excluded value in mind. - After finding each x-value, substitute it into either function to find the corresponding y-value. - Check that each x-value is in the domain.

1. Set the function values equal: \(\frac{10}{x-3}=x\), with \(x\neq 3\). 2. Multiply by \(x-3\): \(10=x(x-3)=x^2-3x\). 3. Rewrite in standard form: \(x^2-3x-10=0\). 4. Factor: \((x-5)(x+2)=0\), so \(x=5\) or \(x=-2\). 5. Since \(g(x)=x\), the corresponding y-values are \(5\) and \(-2\). Both x-values are in the domain of \(f\).

The intersection points are \((5, 5)\) and \((-2, -2)\).

- Find the zeros of the denominator. - Rewrite the function as a constant plus a rational term. - Analyze the cases \(|x|>1\) and \(|x|<1\) separately. - Check the behavior near the excluded values and at infinity.

1. The denominator is zero at \(x=-1\) and \(x=1\), so the domain is \(\mathbb{R}\setminus\{-1, 1\}\). 2. Rewrite the function as \(f(x)=1-\frac{3}{x^2-1}\). For \(|x|>1\), the denominator \(x^2-1\) is positive, so \(f(x)<1\). The function approaches \(1\) as \(x\to\pm\infty\) and approaches \(-\infty\) from the outer sides of the vertical asymptotes. Thus, this part of the domain produces \((-\infty, 1)\). 3. For \(|x|<1\), write \(f(x)=1+\frac{3}{1-x^2}\). Since \(0<1-x^2\leq 1\), \(\frac{3}{1-x^2}\geq 3\), so \(f(x)\geq 4\). 4. The minimum value \(4\) occurs at \(x=0\), and \(f(x)\to\infty\) as \(x\to-1^+\) or \(x\to 1^-\). Therefore, the inner interval produces \([4, \infty)\). 5. Combining the two parts gives the range \((-\infty, 1)\cup[4, \infty)\).

Domain: \(\mathbb{R}\setminus\{-1, 1\}\) Range: \((-\infty, 1)\cup[4, \infty)\)

- Complete the square in the denominator. - Compare the function values at \(1+h\) and \(1-h\). - Use the fact that \((-h)^2=h^2\).

1. Complete the square in the denominator: \(x^2-2x+5=(x-1)^2+4\). Thus, \(f(x)=\frac{8}{(x-1)^2+4}\). 2. To verify the symmetry algebraically, compare inputs equally far from \(1\): \(f(1+h)=\frac{8}{h^2+4}\) and \(f(1-h)=\frac{8}{(-h)^2+4}=\frac{8}{h^2+4}\). 3. Since \(f(1+h)=f(1-h)\) for every real \(h\), the graph is symmetric about \(x=1\).

\(f(1+h)=f(1-h)=\frac{8}{h^2+4}\) for every real \(h\), so the graph is symmetric about \(x=1\).

- Divide each numerator term by the denominator. - Compare the two function values at the same allowed input. - How do the size and sign of the remaining fraction change when \(\lvert x\rvert\) grows?

1. Divide each numerator term by \(x^2\): \(f(x)=x^2-2+\frac{5}{x^2}\). Thus, \(p(x)=x^2-2\) and \(r(x)=\frac{5}{x^2}\). 2. The vertical distance between the graphs is \(\lvert f(x)-p(x)\rvert=\frac{5}{x^2}\). As \(\lvert x\rvert\) grows, the denominator grows and the distance approaches \(0\). 3. For every \(x\neq 0\), \(\frac{5}{x^2}>0\). Thus, \(f(x)>p(x)\), so the graph of \(f\) lies above the parabola throughout its domain.

1) \(p(x)=x^2-2\), \(r(x)=\frac{5}{x^2}\) 2) The vertical distance \(\frac{5}{x^2}\) approaches \(0\) as \(\lvert x\rvert\) becomes very large. 3) The graph of \(f\) lies above the graph of \(p\) for every \(x\neq 0\).

- A hole requires a common numerator and denominator zero. - Use polynomial division for the slant asymptote. - Determine which parameter value removes the quadratic term.

1. For \(x=3\) to be a hole, the numerator must also be zero there: \(9k+18=0\). Thus, \(k=-2\). 2. For \(k=1\), divide \(x^2+6x\) by \(x-3\): \(f_1(x)=x+9+\frac{27}{x-3}\). Therefore, the slant asymptote is \(y=x+9\). 3. A horizontal asymptote is possible when the quadratic numerator term disappears, so set \(k=0\). 4. Then \(f_0(x)=\frac{6x}{x-3}\). The numerator and denominator have the same degree, so the horizontal asymptote is \(y=6\).

a) \(k=-2\) b) \(y=x+9\) c) For \(k=0\), the horizontal asymptote is \(y=6\).

- Compare the vertical asymptotes first. - Which curve stays above the x-axis on both sides of its vertical asymptote? - Use an intercept or a horizontal asymptote to distinguish the remaining two curves.

1. Graph a has vertical asymptote \(x=1\), horizontal asymptote \(y=1\), and x-intercept \((-2, 0)\), matching \(f_1\). 2. Graph b has vertical asymptote \(x=-2\), horizontal asymptote \(y=-1\), and x-intercept \((0, 0)\), matching \(f_2\). 3. Graph c has vertical asymptote \(x=1\), horizontal asymptote \(y=0\), and stays above the x-axis, matching \(f_3\). 4. Function \(f_4\) is unused. It has vertical asymptote \(x=-1\) and x-intercept \((2, 0)\).

Graph a: \(f_1\); graph b: \(f_2\); graph c: \(f_3\); unused: \(f_4\).

- Compare each function's numerator zero with the x-intercepts on the graph. - Find which x-value makes each denominator zero. - Rewrite each function as a constant plus a fraction to determine the horizontal asymptote and possible output restrictions.

1. Function \(f\) has x-intercept \((-2, 0)\) and vertical asymptote \(x=1\), matching \(g_1\). Function \(g\) has x-intercept \((3, 0)\) and vertical asymptote \(x=-2\), matching \(g_2\). 2. Rewrite \(f(x)=2+\frac{6}{x-1}\). Its vertical asymptote is \(x=1\), and its horizontal asymptote is \(y=2\). 3. Rewrite \(g(x)=1-\frac{5}{x+2}\). Its vertical asymptote is \(x=-2\), and its horizontal asymptote is \(y=1\). 4. The domain of \(f\) excludes \(x=1\), and its range excludes \(y=2\). The domain of \(g\) excludes \(x=-2\), and its range excludes \(y=1\).

a) \(g_1\) is \(f\); \(g_2\) is \(g\). b) For \(f\): \(x=1\), \(y=2\). For \(g\): \(x=-2\), \(y=1\). c) Domain of \(f\): \(\mathbb{R}\setminus\{1\}\); range of \(f\): \(\mathbb{R}\setminus\{2\}\). Domain of \(g\): \(\mathbb{R}\setminus\{-2\}\); range of \(g\): \(\mathbb{R}\setminus\{1\}\).

- What does the denominator of the rational function tell you about its graph? - Look for every place where the two curves cross, not only where they meet an axis. - Before multiplying an equation by a variable, what must be true about that variable?

1. The denominator of \(h\) is zero at \(x=0\), so \(x=0\) is not allowed. As \(x\) approaches \(0\) from the positive side, \(2/x\) grows large and positive; from the negative side, it becomes large and negative. 2. The graph has three intersections, at about \(x=-2.2\), \(x=0.5\), and \(x=1.7\). 3. Set the function values equal: \(\frac{2}{x}=4-x^2\), with \(x\ne0\). 4. Multiply by nonzero \(x\): \(2=4x-x^3\). Rearranging gives \(x^3-4x+2=0\).

a) \(x=0\). From the right, \(h(x)\) grows positive without bound; from the left, it becomes negative without bound. b) Approximately \(x=-2.2\), \(x=0.5\), and \(x=1.7\). c) \(\frac{2}{x}=4-x^2\), with \(x\ne0\); clearing the denominator gives \(x^3-4x+2=0\).

- Which inputs make the denominator zero? - Compare the behavior of the fraction for large positive and negative inputs. - Find the zeros algebraically, then use them to distinguish curves with similar asymptotes.

1. The denominator is zero at \(x=-1\) and \(x=1\), so the domain excludes these values. 2. The numerator of the fractional term is nonzero, giving vertical asymptotes \(x=-1\) and \(x=1\). 3. For large positive or negative \(x\), the fraction approaches zero, so the horizontal asymptote is \(y=1\). 4. Set \(1-\frac{4}{x^2-1}=0\). Then \(x^2-1=4\), so the zeros are \(x=\pm\sqrt{5}\). 5. Only graph \(g_1\) has these asymptotes and zeros. It also passes through \((0, 5)\), as required by the formula.

a) Domain: \(\mathbb{R}\setminus\{-1,1\}\); vertical asymptotes: \(x=-1\) and \(x=1\); horizontal asymptote: \(y=1\). b) \((-\sqrt{5}, 0)\) and \((\sqrt{5}, 0)\). c) \(g_1\).

- Read which vertical and horizontal lines each curve approaches. - Examine the branches very close to, but not at, their vertical asymptotes. - Compare the first two formulas using an intercept, then compare the last two using the direction of their branches.

1. Panel a) has vertical asymptote \(x=1\) and horizontal asymptote \(y=1\). Near \(x=1\), the outputs become very large negative from the left and very large positive from the right. 2. Panel b) has vertical asymptote \(x=-2\) and horizontal asymptote \(y=2\). Near \(x=-2\), the outputs become very large negative from both sides. 3. Panel a) matches \(t_1\) or \(t_2\) by its asymptotes. Its y-intercept is \((0,-1)\); since \(t_1(0)=-1\) and \(t_2(0)=0\), it matches \(t_1\). 4. Panel b) matches \(t_3\) or \(t_4\) by its asymptotes. Since it drops on both sides of \(x=-2\), it matches \(t_3\).

a) Panel a): \(x=1\), \(y=1\). Panel b): \(x=-2\), \(y=2\). b) Panel a): negative on the left, positive on the right near \(x=1\). Panel b): negative on both sides near \(x=-2\); each branch falls without bound. c) Panel a): \(t_1\); panel b): \(t_3\).

- What happens to the function rule when you replace \(x\) with \(-x\)? - Can you rewrite the numerator using a multiple of the denominator? - To find the requested input values, set the function equal to \(2.4\) and clear the denominator.

1. Since \(g(-x)=\frac{3(-x)^2-12}{(-x)^2+4}=g(x)\), the graph is symmetric about the y-axis. 2. Rewrite \(g(x)=3-\frac{24}{x^2+4}\). For large positive or negative \(x\), the fraction gets close to zero, so the graph approaches \(y=3\) at both ends. 3. Set \(\frac{3x^2-12}{x^2+4}=2.4\). Multiplying by \(x^2+4\), which is always positive, gives \(3x^2-12=2.4x^2+9.6\). 4. Then \(0.6x^2=21.6\), so \(x^2=36\) and \(x=-6\) or \(x=6\).

a) \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) The outputs approach \(3\) at both ends; horizontal asymptote: \(y=3\). c) \(x=-6\) and \(x=6\).

- Factor each numerator and denominator before simplifying. - Keep the original excluded inputs even when a factor cancels. - Substitute an excluded input into the simplified expression to locate its hole. - Compare the remaining asymptotes and hole coordinates with the graphs.

1. Factor \(f(x)=\frac{(x-2)(x+1)}{(x-2)(x+2)}\). Its domain excludes \(-2\) and \(2\), and it agrees with \(\frac{x+1}{x+2}\) on the domain. Thus, the vertical asymptote is \(x=-2\), and the horizontal asymptote is \(y=1\). 2. The canceled factor leaves a hole at \(x=2\), where the simplified expression equals \(\frac{2+1}{2+2}=\frac{3}{4}\). 3. Factor \(g(x)=\frac{2x(x-2)}{(x-2)(x+1)}\). Its domain excludes \(-1\) and \(2\), and it agrees with \(\frac{2x}{x+1}\) on the domain. The vertical asymptote is \(x=-1\) and the horizontal asymptote is \(y=2\). 4. The canceled factor leaves a hole at \(x=2\), where the simplified expression equals \(\frac{4}{3}\). 5. Panel a) has \(f\)'s asymptotes and hole; panel b) has \(g\)'s asymptotes and hole.

a) Domain of \(f\): \(\mathbb{R}\setminus\{-2,2\}\); asymptotes \(x=-2\), \(y=1\). Domain of \(g\): \(\mathbb{R}\setminus\{-1,2\}\); asymptotes \(x=-1\), \(y=2\). b) Hole in \(f\): \(\left(2,\frac{3}{4}\right)\). Hole in \(g\): \(\left(2,\frac{4}{3}\right)\). c) Panel a): \(f\); panel b): \(g\).

- When is a reciprocal undefined? - What happens to \(\frac{1}{t}\) as positive \(t\) becomes very large? - Check whether the graphed values just to the right of \(x=2\) are positive or negative. - If two input values give the same output for \(g\), what happens after you take their reciprocals?

1. Statement 1 is true. The graph shows that \(g(x)=0\) at \(x=-2\) and \(x=2\), so \(f\) is undefined at those inputs. 2. Statement 2 is true. As \(x\) becomes very large positive or negative, \(g(x)\) becomes very large positive, and its reciprocal approaches zero. 3. Statement 3 is false. Just to the right of \(x=2\), \(g(x)\) is positive and approaches zero, so \(f(x)\) becomes very large positive. 4. Statement 4 is true. The graph shows \(g(-x)=g(x)\). Hence \(f(-x)=\frac{1}{g(-x)}=\frac{1}{g(x)}=f(x)\) wherever defined.

1) True. 2) True. 3) False; \(f(x)\) becomes very large positive when \(x\) approaches \(2\) from the right. 4) True.

- How many vertical asymptotes does each function have? - Compare the horizontal asymptotes and intercepts. - Which function has the same output for \(x\) and \(-x\)?

1. The denominator of \(h\) is always positive, so its graph has no vertical asymptotes. Its y-intercept is \((0,2)\) and its horizontal asymptote is \(y=0\). Thus, \(h\) matches Graph 1. 2. Function \(f\) has vertical asymptote \(x=1\), horizontal asymptote \(y=2\), and x-intercept \((2,0)\), matching Graph 2. 3. Function \(g\) has vertical asymptotes \(x=-1\) and \(x=1\), horizontal asymptote \(y=2\), and passes through \((0,0)\). It is also even because \(g(-x)=g(x)\), matching Graph 3.

\(f\): Graph 2; \(g\): Graph 3; \(h\): Graph 1.

- Rewrite the sum and the two differences as single fractions. - Which numerator can never be zero for a real input? - How does changing the sign of every output affect the graph? - Try one positive input in each difference to tell their graphs apart.

1. Combine the sum: \(s(x)=x+\frac{2}{x}=\frac{x^2+2}{x}\), for \(x\ne0\). It has no real zeros because \(x^2+2>0\). This matches panel b). 2. The first difference is \(d_1(x)=x-\frac{2}{x}=\frac{x^2-2}{x}\), which has zeros \(x=\pm\sqrt{2}\). For large positive \(x\), the graph rises. This matches panel c). 3. The second difference is \(d_2(x)=\frac{2}{x}-x=-d_1(x)\). It has the same zeros as \(d_1\), but its graph is reflected across the x-axis and falls for large positive \(x\). This matches panel a).

a) \(d_2(x)=g(x)-f(x)\); it falls on the right and has zeros \(\pm\sqrt{2}\). b) \(s(x)=f(x)+g(x)\); it has no real zeros. c) \(d_1(x)=f(x)-g(x)\); it rises on the right and has zeros \(\pm\sqrt{2}\).

- Consider what happens to the fraction when \(x^2+5\) becomes very large. - Does a small positive coefficient prevent \(x^4\) from becoming large? - Compare what you can see in the window with the behavior for much larger inputs.

1. For large positive \(x\), the fraction \(-\frac{40}{x^2+5}\) gets closer to \(0\). 2. However, the term \(0.00001x^4\) grows without bound, even though its coefficient is small. 3. The positive fourth-power term eventually dominates the constant and rational terms. Thus, \(f(x)\to\infty\) as \(x\to\infty\). 4. Leon's claim is false. The limited viewing window makes the graph appear to level off before its rapid increase becomes visible.

Leon is incorrect. \(f(x)\to\infty\) as \(x\to\infty\) because the positive fourth-power term eventually dominates.

- Look for the x-coordinate of the graph’s highest point. - Substitute inputs the same distance to the left and right of that x-value. - Expand and simplify each denominator.

1. The graph reaches its maximum at \(x=3\) and appears mirrored across that vertical line, so \(x_0=3\). 2. Evaluate the two symmetric inputs: \(h(3-a)=\frac{2}{(3-a)^2-6(3-a)+10}=\frac{2}{a^2+1}\). 3. Similarly, \(h(3+a)=\frac{2}{(3+a)^2-6(3+a)+10}=\frac{2}{a^2+1}\). 4. Therefore, \(h(3-a)=h(3+a)\) for every real \(a\), proving symmetry about \(x=3\).

a) \(x=3\) b) Both \(h(3-a)\) and \(h(3+a)\) simplify to \(\frac{2}{a^2+1}\), so the graph is symmetric about \(x=3\).

- Compare the locations of the maximum points. - Compare horizontal distances from each graph center at the same y-value. - Remember that multiplying the input by a number greater than \(1\) makes a graph narrower. - The horizontal scale factor is the reciprocal of \(b\).

1. The parameter \(c\) controls the horizontal shift. Because the maximum moves from \(x=0\) to \(x=2\), the graph shifts right \(2\) units, so \(c=-2\). 2. The parameter \(b\) controls horizontal scaling. If \(b>1\), the graph is horizontally compressed; if \(0<b<1\), it is horizontally stretched. 3. The blue graph has value \(2\) at \(x=-1\) and \(x=1\), one unit from its center. The red graph has value \(2\) at \(x=0\) and \(x=4\), two units from its center. 4. The horizontal distances doubled, so the horizontal stretch factor is \(2\). Therefore, \(b=\frac{1}{2}\), and \(0<b<1\).

\(c=-2\), so the graph shifts right \(2\) units; the red graph is wider, so \(0<b<1\), with \(b=\frac{1}{2}\) and horizontal stretch factor \(2\).

- Compare denominator zeros, numerator zeros, and the number of vertical asymptotes. - Which rule is unchanged when \(x\) is replaced by \(-x\)? - Rewrite \(f\) as a constant plus a fraction to determine both end directions.

1. Function \(f\) has vertical asymptote \(x=-1\), horizontal asymptote \(y=2\), and x-intercept \((1,0)\). These features match graph a. 2. Function \(g\) has vertical asymptotes \(x=-2\) and \(x=2\), horizontal asymptote \(y=1\), and x-intercepts \((-1,0)\) and \((1,0)\). It is symmetric about the y-axis because \(g(-x)=g(x)\), matching graph b. 3. Rewrite \(f(x)=2-\frac{4}{x+1}\). For large positive \(x\), the fraction is small and positive, so \(f(x)\) approaches \(2\) from below. For large negative \(x\), the fraction is small and negative, so \(f(x)\) approaches \(2\) from above. 4. In both directions the graph approaches its horizontal asymptote \(y=2\).

1) \(f\): graph a; \(g\): graph b. 2) The outputs approach \(2\) at both ends, from below on the right and from above on the left; horizontal asymptote \(y=2\).

- Compare symmetry and vertical asymptotes. - How small or large can the fraction in \(h\) become? - For the x-intercept of \(k\), check when its numerator is zero and its denominator is nonzero.

1. Graph a is symmetric about the y-axis and has no vertical asymptote, so it matches \(h\). Graph b has vertical asymptote \(x=2\), so it matches \(k\). 2. The denominator \(x^2+1\) is always positive, so the domain of \(h\) is \(\mathbb{R}\). 3. The largest value is \(h(0)=3\). For large positive or negative \(x\), the positive fraction approaches zero, so \(h(x)\) approaches \(-1\) from above but never equals \(-1\). Thus, its range is \((-1,3]\). 4. Set the numerator of \(k\) equal to zero: \(x-1=0\), giving the x-intercept \((1,0)\), which is in the domain.

1) \(h\): graph a; \(k\): graph b. 2) Domain of \(h\): \(\mathbb{R}\); range of \(h\): \((-1,3]\). 3) X-intercept of \(k\): \((1,0)\).

- First compare the horizontal asymptotes. - Then compare the numerator zeros with the graph x-intercepts. - All three functions have the same vertical asymptote.

1. All three functions have vertical asymptote \(x=3\). 2. The horizontal asymptotes of \(f_1\) and \(f_3\) are \(y=1\). The horizontal asymptote of \(f_2\) is \(y=2\). Since graph \(g\) approaches \(y=2\), it matches \(f_2\). Its x-intercept \((1, 0)\) confirms the match. 3. The x-intercept of \(f_1\) is \((2, 0)\), so graph \(f\) matches \(f_1\). 4. The x-intercept of \(f_3\) is \((4, 0)\), so graph \(h\) matches \(f_3\).

Graph \(f\): \(f_1(x)=\frac{x-2}{x-3}\); graph \(g\): \(f_2(x)=\frac{2x-2}{x-3}\); graph \(h\): \(f_3(x)=\frac{x-4}{x-3}\)

- Locate the \(x\)-value where each graph increases or decreases without bound. - Compare the direction of the two branches on opposite sides of each vertical asymptote. - Identify the horizontal line each graph approaches as \(x\) becomes very large in magnitude.

1. For \(f\), the graph approaches a vertical line at \(x=-4\). It approaches \(-\infty\) from the left and \(+\infty\) from the right, so it changes sign across the vertical asymptote. As \(|x|\) increases, the graph approaches \(y=-3\). 2. For \(g\), the graph approaches a vertical line at \(x=3\). It approaches \(+\infty\) from both sides, so it does not change sign across the vertical asymptote. As \(|x|\) increases, the graph approaches \(y=1\).

For \(f\), the vertical asymptote is \(x=-4\), the graph changes sign across it, and the horizontal asymptote is \(y=-3\). For \(g\), the vertical asymptote is \(x=3\), the graph does not change sign across it, and the horizontal asymptote is \(y=1\).

- Find the x-values that make the denominator zero. - Compare \(f(-x)\) with \(f(x)\). - Compare the degrees and leading coefficients of the numerator and denominator. - A rational expression is zero when its numerator is zero and its denominator is not zero.

1. The denominator is zero when \(x^2-9=0\), so \(x=-3\) or \(x=3\). Therefore, the domain is \((-\infty, -3)\cup(-3, 3)\cup(3, \infty)\). Also, \(f(-x)=\frac{2(-x)^2}{(-x)^2-9}=\frac{2x^2}{x^2-9}=f(x)\). Thus, \(f\) is even, and its graph is symmetric about the y-axis. 2. The vertical asymptotes are \(x=-3\) and \(x=3\). The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of leading coefficients, \(y=2\). 3. Since \(f(0)=0\), the y-intercept is \((0, 0)\). For x-intercepts, set the numerator equal to zero: \(2x^2=0\), so \(x=0\). Therefore, the only intercept is \((0, 0)\).

1. Domain: \((-\infty, -3)\cup(-3, 3)\cup(3, \infty)\); \(f\) is even, so the graph is symmetric about the y-axis. 2. Vertical asymptotes: \(x=-3\) and \(x=3\); horizontal asymptote: \(y=2\) 3. The only intercept is \((0, 0)\).

- Factor the numerator and check whether a denominator factor cancels. - What happens on a graph when a canceled factor still excludes an input? - Use the simplified expression to find the height of the missing point.

1. Both expressions are undefined at \(x=3\). 2. Factor \(f(x)=\frac{x^2(x-3)}{x-3}\). For allowed \(x\), this equals \(x^2\), so its graph is a parabola with a hole at \(x=3\); this is Figure 2. 3. At \(x=3\), the numerator of \(g\) equals \(4\), so the denominator factor does not cancel. The graph has a vertical asymptote there and is Figure 1. 4. The hole in \(f\) has the y-coordinate \(3^2=9\), giving \((3,9)\).

a) Figure 1: \(g\), with vertical asymptote \(x=3\); Figure 2: \(f\), with a hole at \(x=3\). b) \((3,9)\).

- Rewrite \(f\) in terms of \(h\). - A positive fraction with a fixed numerator is largest when its denominator is smallest. - To prove symmetry about \(x=a\), compare \(f(a-z)\) and \(f(a+z)\).

1. Since \(f(x)=3h(x-8)\), shift the graph of \(h\) right \(8\) units and stretch it vertically by a factor of \(3\). 2. The denominator \((x-8)^2+3\) is minimized when \(x=8\). Then \(f(8)=\frac{12}{3}=4\), so the maximum point is \((8,4)\). 3. For any real number \(z\), \(f(8-z)=\frac{12}{(-z)^2+3}=\frac{12}{z^2+3}\). 4. Also, \(f(8+z)=\frac{12}{z^2+3}\). Therefore, \(f(8-z)=f(8+z)\) for every real \(z\), proving symmetry about \(x=8\).

1) Shift right \(8\) units and stretch vertically by a factor of \(3\). 2) Maximum point: \((8,4)\) 3) \(f(8-z)=f(8+z)=\frac{12}{z^2+3}\), so the graph is symmetric about \(x=8\).

- Compare the locations and heights of the maximum points. - Determine what happens to the fraction as \(|x|\) becomes large. - Compare \(f(5-h)\) and \(f(5+h)\).

1. Since \(f(x)=2g(x-5)+1\), translate the graph of \(g\) 5 units right, stretch it vertically by a factor of \(2\), and translate it 1 unit up. 2. As \(x\to\pm\infty\), the fraction \(\frac{2}{1+(x-5)^2}\) approaches \(0\). Therefore, \(f(x)\to1\), so the horizontal asymptote is \(y=1\). 3. For every real number \(h\), \(f(5-h)=1+\frac{2}{1+h^2}=f(5+h)\). Therefore, the graph is symmetric about the line \(x=5\).

1. Translate 5 units right, stretch vertically by a factor of \(2\), and translate 1 unit up. 2. \(y=1\) 3. The graph is symmetric about \(x=5\).

- Can you rewrite the numerator as a multiple of the denominator plus a constant? - Find intercepts by evaluating at \(x=0\) or setting the numerator to zero. - To find the stated outputs, clear the denominator, which is always positive.

1. Rewrite \(g(x)=2-\frac{6}{x^2+1}\). When \(x\) is very large positive or negative, the fraction gets close to zero, so \(g(x)\) approaches \(2\) from below at both ends. The horizontal asymptote is \(y=2\). 2. The y-intercept is \(g(0)=-4\), giving \((0,-4)\). 3. To find x-intercepts, solve \(2x^2-4=0\), giving \(x=\pm\sqrt{2}\) and intercepts \((-\sqrt{2},0)\), \((\sqrt{2},0)\). 4. Solve \(\frac{2x^2-4}{x^2+1}=1.4\): \(2x^2-4=1.4x^2+1.4\), so \(0.6x^2=5.4\) and \(x^2=9\). Thus, \(x=-3\) or \(x=3\).

a) Both ends approach \(2\) from below; horizontal asymptote \(y=2\). b) Y-intercept: \((0,-4)\); x-intercepts: \((-\sqrt{2},0)\), \((\sqrt{2},0)\). c) \(x=-3\) and \(x=3\).

- Set the function equal to an arbitrary output \(y\). - Solve the equation for \(x\). - Identify the output that makes the resulting expression undefined. - Verify that the calculated input is never the excluded domain value.

1. Let \(y=\frac{3x}{x+4}\) and solve for \(x\): \(y(x+4)=3x\). 2. Rearranging gives \(x(y-3)=-4y\), so \(x=\frac{-4y}{y-3}\). 3. This expression is defined for every real \(y\) except \(y=3\). Substituting \(y=3\) into the original equation gives \(3x+12=3x\), a contradiction. 4. For \(y\ne3\), the resulting input can never equal the excluded value \(-4\). Otherwise, \(-4=\frac{-4y}{y-3}\) would imply \(12=0\). 5. Therefore, every real value except \(3\) occurs, and the range is \(\mathbb{R}\setminus\{3\}\).

\(\mathbb{R}\setminus\{3\}\)

- Use zeros and extrema to identify the unshifted graph. - Determine each horizontal shift. - Replace \(x\) with \(x-t\) to represent a horizontal translation.

1. The function \(g\) has its only zero at \(x=0\) and a local maximum at \((1, 2)\), so it is graph q. 2. Graph p is shifted \(3\) units left, giving \(\frac{4(x+3)}{(x+3)^2+1}\). 3. Graph r is shifted \(2\) units right, giving \(\frac{4(x-2)}{(x-2)^2+1}\). 4. A family describing horizontal shifts is \(f_t(x)=\frac{4(x-t)}{(x-t)^2+1}\).

a) Graph q b) p: \(\frac{4(x+3)}{(x+3)^2+1}\); r: \(\frac{4(x-2)}{(x-2)^2+1}\) c) \(f_t(x)=\frac{4(x-t)}{(x-t)^2+1}\)

- Can you write the new output rule using the original function \(h\)? - Find one quarter of the maximum power, then solve for the input time. - The squared equation gives two times. Which one is in the afternoon? - Convert the fraction of an hour into minutes.

1. Since \(P(x)=6h(x-14)\), shift \(h\) right \(14\) units and stretch vertically by a factor of \(6\). 2. Function \(h\) is even, so the translated graph is symmetric about \(x=14\). 3. The maximum occurs at \(x=14\): \(P(14)=\frac{24}{2}=12\,\text{kW}\). One quarter is \(3\,\text{kW}\). 4. Solve \(\frac{24}{(x-14)^2+2}=3\): \((x-14)^2+2=8\), giving \(x=14\pm\sqrt{6}\). 5. The afternoon input is \(x=14+\sqrt{6}\approx16.449\). The decimal part is approximately \(0.449\cdot60\approx27\) minutes, giving about 4:27 p.m.

a) Shift right \(14\) units and stretch vertically by \(6\); symmetry line \(x=14\). b) Approximately 4:27 p.m.

- Compare \(g(-x)\) with \(g(x)\) and \(-g(x)\). - Read where the curve changes from rising to falling or from falling to rising. - Compare the degrees of the numerator and denominator to explain the end behavior.

1. Since \(g(-x)=\frac{-3x}{x^2+1}=-g(x)\), the function is odd, so its graph is symmetric about the origin. 2. The graph has a local maximum at \((1,1.5)\) and a local minimum at \((-1,-1.5)\). 3. The denominator has a higher degree than the numerator. For large positive or negative inputs, \(\frac{3x}{x^2+1}\) approaches zero. Thus, the horizontal asymptote is \(y=0\).

a) Symmetric about the origin. b) Local maximum: \((1,1.5)\); local minimum: \((-1,-1.5)\). c) Horizontal asymptote: \(y=0\).

- Find the zeros of each denominator to identify possible vertical asymptotes. - Compare the functions’ end behavior to identify horizontal or slant asymptotes. - Find intercepts that distinguish functions with similar asymptotes. - Check whether each graph changes sign across a vertical asymptote.

1. Graph a matches \(g_1\). It has a vertical asymptote at \(x=0\), x-intercepts at \(x=-1\) and \(x=1\), and slant asymptote \(y=x\). 2. Graph b matches \(g_2\). It is symmetric about the y-axis, has no vertical asymptotes, has a maximum at \((0, 2)\), and approaches \(y=0\). 3. Graph c matches \(g_3\). It has a vertical asymptote at \(x=-2\), an x-intercept at \((2, 0)\), and horizontal asymptote \(y=1\). 4. Graph d matches \(g_4\). It stays above the x-axis and has a vertical asymptote at \(x=1\), approaching positive infinity from both sides.

\(g_1\to\text{a}\); \(g_2\to\text{b}\); \(g_3\to\text{c}\); \(g_4\to\text{d}\)

- Denominator zeros identify possible vertical asymptotes. - Compare leading coefficients for horizontal asymptotes. - When solving the intersection equation, do not cancel \(x+1\) before considering \(x=-1\). - Check each solution against both domains.

1. For \(h\), the vertical asymptote is \(x=1\), and the horizontal asymptote is \(y=1\). 2. For \(k\), the vertical asymptote is \(x=-2\), and the horizontal asymptote is \(y=2\). 3. Set the functions equal: \(\frac{x+1}{x-1}=\frac{2(x+1)}{x+2}\). 4. One solution occurs when \(x+1=0\), giving \(x=-1\) and \(y=0\). 5. If \(x+1\neq 0\), cancel it and solve \(\frac{1}{x-1}=\frac{2}{x+2}\). Then \(x+2=2x-2\), so \(x=4\). 6. Evaluate \(h(4)=\frac{5}{3}\).

a) \(h\): vertical \(x=1\), horizontal \(y=1\); \(k\): vertical \(x=-2\), horizontal \(y=2\) b) \((-1, 0)\) and \(\left(4, \frac{5}{3}\right)\)

- Set the two function expressions equal and obtain a quadratic equation in \(x\). - Use the discriminant to classify the number of real roots. - Check whether the excluded value \(x=-1\) appears as a root for any parameter value.

1. Set the functions equal: \(\frac{2a}{x+1}=x-2\), with \(x\neq-1\). 2. Clear the denominator: \(2a=(x-2)(x+1)=x^2-x-2\). Thus, \(x^2-x-2-2a=0\). 3. The discriminant is \(\Delta=1+4(2+2a)=9+8a\). 4. If \(a<-\frac{9}{8}\), then \(\Delta<0\), so there are no intersections. If \(a=-\frac{9}{8}\), then \(\Delta=0\), so there is one intersection. If \(a>-\frac{9}{8}\), the quadratic has two real roots before domain restrictions are checked. 5. The excluded value \(x=-1\) is a root of the quadratic only when \(a=0\). For \(a=0\), the other root is \(x=2\), so there is one valid intersection. 6. Therefore, there are two intersections when \(a>-\frac{9}{8}\) and \(a\neq 0\).

If \(a<-\frac{9}{8}\): \(0\) intersection points If \(a=-\frac{9}{8}\) or \(a=0\): \(1\) intersection point If \(a>-\frac{9}{8}\) and \(a\neq 0\): \(2\) intersection points

- Set the two function expressions equal and clear the denominator. - A quadratic has one repeated real solution when its discriminant is \(0\). - A second possibility occurs when one algebraic solution is excluded from the rational function's domain.

1. Set the functions equal, with \(x\neq 2\): \(\frac{x+k}{x-2}=x-4\). 2. Clear the denominator to obtain \(x^2-7x+8-k=0\). 3. One intersection occurs when the quadratic has one repeated root. Its discriminant is \(49-4(8-k)=17+4k\). Setting this equal to \(0\) gives \(k=-\frac{17}{4}\). The repeated root is \(x=\frac{7}{2}\), which is valid. 4. One intersection can also occur when the quadratic has two roots but one is the excluded value \(x=2\). Substituting \(x=2\) gives \(k=-2\). 5. For \(k=-2\), the quadratic factors as \((x-2)(x-5)=0\). Only \(x=5\) is valid, so there is exactly one intersection.

\(k=-\frac{17}{4}\) or \(k=-2\)

- Treat an arbitrary output \(y\) as a fixed number and solve for \(x\). - Separate the case \(y=0\). - Use the discriminant of the resulting quadratic equation. - Check that the solutions cannot equal the excluded inputs.

1. Let \(y\) be an arbitrary real number and solve \(y=\frac{3x}{x^2-4}\) for \(x\). Clearing the denominator gives \(y(x^2-4)=3x\), or \(yx^2-3x-4y=0\). 2. If \(y=0\), the equation becomes \(-3x=0\), so \(x=0\), which is in the domain. 3. If \(y\ne0\), this is a quadratic equation in \(x\). Its discriminant is \(\Delta=(-3)^2-4\cdot y\cdot(-4y)=9+16y^2>0\). 4. Therefore, the equation has real solutions for every nonzero real \(y\). 5. Neither excluded input can be a solution: substituting \(x=2\) into \(y(x^2-4)=3x\) would give \(0=6\), and substituting \(x=-2\) would give \(0=-6\). 6. Thus, every real \(y\) has at least one allowed preimage, so the range is \(\mathbb{R}\).

For every \(y\in\mathbb{R}\), the equation \(y=\frac{3x}{x^2-4}\) has an allowed real solution. Therefore, the range is \(\mathbb{R}\).

- Determine the original domain before accepting any numerator zeros. - The numerator contains only even powers of \(x\). Look for a factorization built from expressions in \(x^2\). - A rational function equals zero only when its numerator is zero and its denominator is nonzero. - Compare the numerator and denominator factors before deciding which zeros remain.

1. Factor the denominator: \(x^2-2x-3=(x-3)(x+1)\). Thus, the domain excludes \(x=3\) and \(x=-1\). 2. The numerator contains only even powers. Factor it as a quadratic in \(x^2\): \(x^4-10x^2+9=(x^2-1)(x^2-9)\). 3. Thus the numerator zeros are \(x=\pm1\) and \(x=\pm3\). 4. The values \(x=3\) and \(x=-1\) are excluded from the domain. Factoring the full expression gives \(\frac{(x-3)(x+3)(x-1)(x+1)}{(x-3)(x+1)}=(x+3)(x-1)\) on the original domain. 5. Therefore, the zeros are \(x=-3\) and \(x=1\), each with multiplicity \(1\).

\(x=-3\) has multiplicity \(1\), and \(x=1\) has multiplicity \(1\).

- Set the functions equal and clear the denominator. - Look for a difference-of-squares structure in the resulting quartic. - Use discriminants to count the real roots of the quadratic factors. - Check the excluded value.

1. Set the functions equal: \(\frac{1}{x}+2=x^3-x\), with \(x\neq 0\). 2. Multiply by \(x\): \(1+2x=x^4-x^2\), so \(x^4-x^2-2x-1=0\). 3. Rewrite this as \(x^4-(x+1)^2=0\), then factor the difference of squares: \((x^2-x-1)(x^2+x+1)=0\). 4. The equation \(x^2-x-1=0\) has two real solutions because its discriminant is \(5\). The equation \(x^2+x+1=0\) has no real solutions because its discriminant is \(-3\). 5. Neither real solution is \(0\), so the graphs have exactly two intersection points.

The graphs have exactly \(2\) intersection points.

- Check whether every output you can see is above the x-axis. How far does the x-axis extend in the drawing? - Think about the long-term behavior of each part of the formula separately. - Could a smaller coefficient eventually be outweighed by a higher power of x? - Test an input beyond the graph's window to challenge your first impression.

1. All displayed outputs are positive. A graph drawn over a limited window cannot prove that every output is positive. 2. The numerator and denominator of the rational term have the same degree, so that term approaches \(4\) for large positive \(x\). 3. The term \(-0.00005x^4\) falls without bound and eventually dominates the rational term. Thus, \(g(x)\to-\infty\) as \(x\to\infty\). 4. For example, \(g(50)\approx-308.50\), outside the displayed window. The full graph eventually has negative outputs.

a) No negative outputs are shown. No, the displayed window does not prove \(g(x)>0\) for every real \(x\). b) \(g(x)\to-\infty\) as \(x\to\infty\). The negative fourth-power term eventually dominates the bounded rational part.

- Find the denominator zeros and compare leading coefficients. - Set the function equal to the horizontal-asymptote value. - Use the middle and outer branches separately to determine the range.

1. The denominator is zero at \(x=-3\) and \(x=3\), so the vertical asymptotes are \(x=-3\) and \(x=3\). 2. The numerator and denominator have the same degree, and the ratio of leading coefficients is \(3\). Thus, the horizontal asymptote is \(y=3\). 3. To test for an intersection, set \(g(x)=3\): \(\frac{3x^2}{x^2-9}=3\). This gives \(3x^2=3x^2-27\), which is impossible. Therefore, the graph does not intersect \(y=3\). 4. On \((-3, 3)\), the graph has a maximum at \((0, 0)\) and approaches \(-\infty\) near both vertical asymptotes, producing all values in \((-\infty, 0]\). 5. On the outer intervals, the graph approaches \(\infty\) near the vertical asymptotes and approaches \(3\) from above as \(|x|\to\infty\), producing \((3, \infty)\). 6. Therefore, the range is \((-\infty, 0]\cup(3, \infty)\).

a) Vertical asymptotes: \(x=-3\) and \(x=3\); horizontal asymptote: \(y=3\) b) The graph does not intersect \(y=3\). c) \((-\infty, 0]\cup(3, \infty)\)

- Establish a lower bound using the signs of the numerator and denominator. - Rewrite the function as a constant minus a positive fraction. - To prove every value in the interval occurs, set the function equal to an arbitrary \(y\) in that interval and solve for \(x^2\).

1. Since \(x^2\ge0\) and \(x^2+1>0\), \(k(x)\ge0\). The value \(0\) occurs at \(x=0\). 2. Rewrite the function as \(k(x)=2-\frac{2}{x^2+1}\). The fraction is always positive, so \(k(x)<2\). As \(|x|\to\infty\), the fraction approaches \(0\), so the outputs approach \(2\) without reaching it. 3. To show that every value between the bounds occurs, let \(0\le y<2\). Solving \(y=\frac{2x^2}{x^2+1}\) gives \(x^2=\frac{y}{2-y}\). 4. Because \(\frac{y}{2-y}\ge0\), a real solution exists, for example \(x=\sqrt{\frac{y}{2-y}}\). Thus, every \(y\in[0,2)\) is attained.

\([0,2)\)