- Compare \(g(x)\) with \(f(x)\) and focus on the constant added outside the exponential expression. - An added constant changes y-values without changing x-values.

Adding \(5\) to the function value shifts every point on the graph upward by \(5\) units.

Shift the graph of \(f\) up \(5\) units.

- A reflection across the x-axis changes the sign of every y-value. - Apply that sign change to the entire output of \(f(x)\).

Reflection across the x-axis multiplies every output by \(-1\). Therefore the reflected function is \(g(x)=-3^x\).

\(g(x)=-3^x\)

- Does a horizontal shift change the input or the output of the function? - Think about which input makes the shifted graph reach the original height at \(x=0\).

1. A shift \(3\) units right replaces \(x\) with \(x-3\). 2. Therefore, \(g(x)=f(x-3)=2^{x-3}\).

\(g(x)=2^{x-3}\)

- Find the change to \(x\) inside the exponent. - Compare where \(f\) and \(h\) have the same function value. How does this show the direction of the shift?

1. We have \(h(x)=f(x+2)\). 2. Adding \(2\) inside the input shifts the graph \(2\) units left.

\(2\) units left.

- A vertical stretch multiplies the output of the parent function. - Which value does any positive exponential base have at input \(0\)?

1. At \(x=0\), \(5^0=1\), so \(g(0)=a\cdot1=a\). 2. The given point requires \(g(0)=7\), so \(a=7\).

\(a=7\)

- How are the x-coordinates of matching points related after reflection across the y-axis? - Which change to the input of \(f\) produces that reflection?

1. Reflecting across the y-axis replaces the input \(x\) with \(-x\). 2. Thus \(g(x)=f(-x)=3^{-x}\).

\(g(x)=3^{-x}\)

- Which part of the equation shifts every point upward? - As \(x\) becomes very negative, what happens to the exponential part of \(g(x)\)?

1. The graph of \(f\) has horizontal asymptote \(y=0\). 2. Adding \(4\) shifts the graph and its asymptote up \(4\) units, giving \(y=4\).

\(y=4\)

- How can you find where a graph intersects an axis? - What happens to \(e^x\) as \(x\) becomes very negative? - How does reflection across the x-axis affect whether a graph increases or decreases? - What values can \(5-e^x\) approach but never reach? - Recall the value of \(e^0\).

1. Set \(f(x)=0\): \(5-e^x=0\), so \(e^x=5\) and \(x=\ln(5)\). Statement a is true. 2. As \(x\to-\infty\), \(e^x\to0\), so \(5-e^x\to5\). Statement b is true. 3. The graph of \(e^x\) is strictly increasing. Reflecting it across the x-axis makes \(-e^x\) strictly decreasing, and shifting it upward does not change monotonicity. Statement c is false. 4. Since \(e^x>0\) for every real \(x\), \(f(x)<5\). Also, \(f(x)\to-\infty\) as \(x\to\infty\), so the range is \((-\infty, 5)\). Statement d is true. 5. Since \(f(0)=5-e^0=5-1=4\), the y-intercept is \((0, 4)\). Statement e is true.

a) True; solving \(5-e^x=0\) gives \(x=\ln(5)\). b) True; \(f(x)\to5\) as \(x\to-\infty\). c) False; \(f\) is strictly decreasing. d) True; the range is \((-\infty, 5)\). e) True; \(f(0)=4\).

- How does a horizontal shift change the input of a function, while a vertical shift changes its output? - How is a reflection across the x-axis represented in a function equation? - Apply the transformations in the stated order.

1. Reflecting across the x-axis gives \(-e^x\). 2. A vertical stretch by a factor of \(2.5\) gives \(-2.5e^x\). 3. Shifting \(3\) units left replaces \(x\) with \(x+3\), giving \(-2.5e^{x+3}\). 4. Shifting \(1\) unit down gives \(g(x)=-2.5e^{x+3}-1\).

\(g(x)=-2.5e^{x+3}-1\)

- Which values can an exponential function take, and which can it approach without reaching? - Where does a graph cross the y-axis? - What would be true of a point and its opposite if a graph had origin symmetry? - How does changing the input affect the graph's position?

1. As \(x\) becomes more negative, \(e^x\) approaches \(0\), so a) is true. 2. Since \(e^x>0\), we have \(e^x+2>2\). The graph does not cross the x-axis, so b) is false. 3. Since \(f(0)=e^0=1\), c) is true. 4. Origin symmetry would require \(f(-x)=-f(x)\), but \(e^x\) is always positive. Thus d) is false. 5. Replacing \(x\) by \(x-2\) shifts the graph \(2\) units right, so e) is true.

a) True: \(e^x\) approaches \(0\) as \(x\) decreases without bound. b) False: \(e^x+2>2\) for every real \(x\). c) True: \(f(0)=e^0=1\). d) False: \(f\) is always positive and is not symmetric about the origin. e) True: replacing \(x\) with \(x-2\) shifts the graph right \(2\) units.

- What value does \(e^x\) approach as \(x\to-\infty\)? - An added constant produces a vertical translation. - How does the sign of \(t\) affect the direction of the shift? - An x-intercept has y-coordinate \(0\), and \(e^0=1\).

1. As \(x\to-\infty\), \(e^x\to0\), so \(g_t(x)\to2t\). The horizontal asymptote is \(y=2t\). 2. The graph is shifted vertically by \(2t\) units. It shifts up when \(t>0\), down when \(t<0\), and is unchanged when \(t=0\). 3. Crossing the x-axis at \(x=0\) requires \(g_t(0)=0\). Thus, \(1+2t=0\), so \(t=-\frac{1}{2}\).

1. \(y=2t\) 2. A vertical shift of \(2t\) units. 3. \(t=-\frac{1}{2}\)

- Use a transformed exponential function of the form \(ae^x+c\). - The vertical shift determines the horizontal asymptote. - Substitute \(x=0\) to use the y-intercept. - Since \(e^0=1\), the intercept gives a simple equation for \(a\).

1. Use the form \(F(x)=ae^x+c\), whose horizontal asymptote is \(y=c\). 2. For part a, \(c=5\). The y-intercept condition gives \(a+5=6\), so \(a=1\). One solution is \(F(x)=e^x+5\). 3. For part b, \(c=-2\). The y-intercept condition gives \(a-2=3\), so \(a=5\). One solution is \(F(x)=5e^x-2\).

a) \(F(x)=e^x+5\) b) \(F(x)=5e^x-2\)

- Recall the end behavior of \(e^x\). - Consider the effect of multiplying by a negative number. - The vertical shift determines the horizontal asymptote. - Set \(x=0\) to find the y-intercept.

1. As \(x\to\infty\), \(e^x\to\infty\), so \(f(x)\to-\infty\). As \(x\to-\infty\), \(e^x\to0\), so \(f(x)\to4\). Thus, \(y=4\) is a horizontal asymptote. 2. Since \(e^x\) is strictly increasing, \(-2e^x\) is strictly decreasing. Adding \(4\) does not change monotonicity, so \(f\) is strictly decreasing. The function approaches but never reaches \(4\), and it is unbounded below, so its range is \((-\infty, 4)\). 3. \(f(0)=4-2e^0=2\), so the y-intercept is \((0, 2)\).

1. As \(x\) increases without bound, \(f(x)\) decreases without bound. As \(x\) becomes very negative, \(f(x)\) approaches \(4\) from below. 2. Strictly decreasing; range \((-\infty, 4)\). 3. \((0, 2)\).

- How does a left or right shift change the input of a function? - Compare the shifted function with \(f(x)\) using a ratio. - Which power of \(3\) gives the requested smaller output?

1. A shift \(5\) units left gives \(g(x)=f(x+5)=3^{x+5}=3^5f(x)=243f(x)\). 2. A shift \(k\) units right gives \(h(x)=f(x-k)=3^{-k}f(x)\). 3. Since \(3^{-k}=\frac{1}{9}=3^{-2}\), we get \(k=2\). The graph shifts \(2\) units right.

a) A factor of \(243\). b) \(2\) units right.

- How do changes inside the input differ from constants added to the entire function? - Start with the horizontal asymptote of \(e^x\) and apply the vertical shift. - What is the x-coordinate of every point on the y-axis?

1. Shifting \(2\) units right replaces \(x\) with \(x-2\), and shifting \(5\) units down subtracts \(5\). Thus, \(g(x)=e^{x-2}-5\). 2. The horizontal asymptote of \(e^x\) is \(y=0\). Shifting the graph down \(5\) units changes the asymptote to \(y=-5\). 3. At the y-intercept, \(x=0\). Therefore, \(g(0)=e^{-2}-5\), so the y-intercept is \((0, e^{-2}-5)\).

a) \(g(x)=e^{x-2}-5\) b) \(y=-5\) c) \((0, e^{-2}-5)\)

- Which parameter determines the horizontal asymptote? - Use the given point to write an equation involving \(a\). - Recall the value of \(e^0\).

1. For \(h(x)=ae^x+c\), the horizontal asymptote is \(y=c\). Therefore, \(c=-2\). 2. Since the graph passes through \((0, 1)\), \(h(0)=1\). 3. Substitute into \(h(x)=ae^x-2\): \(1=ae^0-2=a-2\). 4. Thus, \(a=3\), and \(h(x)=3e^x-2\).

\(a=3\), \(c=-2\), and \(h(x)=3e^x-2\)

- How does a vertical stretch affect the entire function expression? - What change to the input shifts a graph to the right? - What x-value identifies a y-intercept?

1. A vertical stretch by a factor of \(2\) gives \(2e^x\). 2. Shifting \(3\) units right replaces \(x\) with \(x-3\), so \(g(x)=2e^{x-3}\). 3. At the y-intercept, \(x=0\). Therefore, \(g(0)=2e^{-3}\), and the y-intercept is \((0, 2e^{-3})\).

a) \(g(x)=2e^{x-3}\) b) \((0, 2e^{-3})\)

- Evaluate each equation at \(x=0\) to compare y-intercepts. - Determine which graphs increase and which decrease. - Compare horizontal asymptotes and end behavior. - Identify reflections and vertical shifts.

1. Graph \(a\) increases, has y-intercept \((0, 1)\), and approaches \(y=0\) on the left. It matches (6), \(y=e^x\). 2. Graph \(b\) decreases, has y-intercept \((0, 1)\), and approaches \(y=0\) on the right. It matches (2), \(y=e^{-x}\). 3. Graph \(c\) increases, has y-intercept \((0, -2)\), and approaches \(y=-3\) on the left. It matches (5), \(y=e^x-3\). 4. Graph \(d\) decreases, has y-intercept \((0, 1)\), and approaches \(y=2\) on the left. It matches (3), \(y=-e^x+2\).

\(a\): (6); \(b\): (2); \(c\): (5); \(d\): (3)

- Read the y-value where the curve crosses the y-axis. - Substitute \(x=0\) into the given form after reading that intercept. - Use the center and total width of the band to determine its two boundaries. - Compare the calculated function values with those boundaries.

1. The graph crosses the y-axis at \(3\), so \(f(0)=3\). 2. Since \(f(0)=1+a(0.5)^0=1+a\), we get \(1+a=3\), so \(a=2\). Thus, \(f(x)=1+2(0.5)^x\). 3. A total band width of \(0.4\) means the boundaries are \(y=0.8\) and \(y=1.2\). 4. \(f(3)=1+2(0.5)^3=1.25\). Since \(1.25>1.2\), the graph is outside the band at \(x=3\). 5. \(f(4)=1+2(0.5)^4=1.125\). Since \(0.8\le1.125\le1.2\), the graph is inside the band at \(x=4\).

a) \(a=2\) b) At \(x=3\), outside the band; at \(x=4\), inside the band.

- Look for a height that both curves reach. - Compare the horizontal locations at that height before writing the shifted input.

1. The graph of \(f\) passes through \((0, 1)\), while \(g\) passes through \((2, 1)\). The same height occurs \(2\) units farther right. 2. So the graph shifts \(2\) units right, giving \(g(x)=f(x-2)=e^{x-2}\).

Shift \(2\) units right; \(g(x)=e^{x-2}\).

- Does each curve lie above or below the horizontal line it approaches? - Think about how an x-axis reflection changes output signs, compared with a y-axis reflection. - Read the height of the dashed line from the graph.

1. Curve \(g\) stays above the dashed line and falls toward it as \(x\) increases. This matches \(e^{-x}+1\): reflect \(e^x\) across the y-axis, then shift up \(1\). 2. Curve \(h\) lies below the dashed line and falls as \(x\) increases. This matches \(-e^x+1\): reflect across the x-axis, then shift up \(1\). 3. Both graphs approach the dashed line \(y=1\), their horizontal asymptote.

\(g\): reflect across the y-axis, then shift up \(1\). \(h\): reflect across the x-axis, then shift up \(1\). Common asymptote: \(y=1\).

- Which part of the formula determines the asymptote? - Substitute the given point after finding the vertical shift. - How can the sign of the remaining coefficient confirm that the graph is decreasing?

1. The horizontal asymptote of \(a\cdot2^x+c\) is \(y=c\), so \(c=3\). 2. Using \(g(0)=1\) and \(2^0=1\) gives \(a+3=1\), hence \(a=-2\). 3. Therefore, \(g(x)=-2\cdot2^x+3\). Since \(2^x\) increases and \(a<0\), \(g\) decreases.

\(a=-2\), \(c=3\), \(g(x)=-2\cdot2^x+3\); decreasing.

- Translate a left shift into a new input for \(f\). - Use an exponent rule to separate the added exponent from \(x\). - Compare the resulting multiplier with each person's claim.

1. Shifting the graph left \(1\) unit gives \(g(x)=f(x+1)=2^{x+1}\). 2. The exponent law gives \(2^{x+1}=2\cdot2^x=2f(x)\). 3. Mia is correct: the same graph is obtained by a vertical stretch of factor \(2\), not \(\frac{1}{2}\).

Mia is correct because \(2^{x+1}=2\cdot2^x\).

- Start by comparing the outputs of both functions at a familiar input. - What reflection results from replacing \(x\) with \(-x\)? - Compare \(e^x\) and \(e^{-x}\) when \(x\) is positive. - Use exponent rules for the product and consider what happens as \(x\) becomes large.

1. At \(x=0\), both functions equal \(1\), so (1) is true. 2. Since \(g(x)=f(-x)\), the graph is reflected across the y-axis, so (2) is false. 3. When \(x>0\), \(e^{-x}<1<e^x\), so (3) is false. 4. Since \(e^xe^{-x}=1\) for all real \(x\), (4) is true. 5. As \(x\to\infty\), \(e^x\) grows and \(e^{-x}\) approaches \(0\), so (5) is false.

(1) True (2) False (3) False (4) True (5) False

- First factor the coefficient of \(x\) from the entire exponent. - How does a coefficient multiplying \(x\) inside the exponent affect horizontal scaling? - What does the negative coefficient outside the exponential function do to the graph? - Which constant determines the horizontal asymptote?

1. Rewrite the exponent by factoring: \(0.5x-1=0.5(x-2)\), so \(g(x)=-2e^{0.5(x-2)}+4\). 2. Replacing \(x\) with \(0.5x\) stretches the graph horizontally by a factor of \(2\). 3. Replacing \(x\) with \(x-2\) after that stretch shifts the graph \(2\) units right, producing \(e^{0.5(x-2)}\). 4. Multiplying by \(-2\) stretches the graph vertically by a factor of \(2\) and reflects it across the x-axis. 5. Adding \(4\) shifts the graph \(4\) units up. The result is \(g(x)=-2e^{0.5(x-2)}+4\).

1. Stretch horizontally by a factor of \(2\). 2. Shift \(2\) units right. 3. Stretch vertically by a factor of \(2\). 4. Reflect across the x-axis. 5. Shift \(4\) units up.

- Rewrite the exponent so the horizontal shift is visible. - What transformation is represented by a negative sign on \(x\) inside the exponent? - How does the constant added outside the exponential expression affect the horizontal asymptote? - What does multiplying the entire exponential expression by \(-1\) do?

1. Rewrite the exponent as \(3-x=-(x-3)\). 2. Replacing \(x\) with \(-x\) reflects the graph of \(e^x\) across the y-axis, producing \(e^{-x}\). 3. Replacing \(x\) with \(x-3\) shifts that graph \(3\) units right, producing \(e^{-(x-3)}=e^{3-x}\). 4. Multiplying by \(-1\) reflects the graph across the x-axis, producing \(-e^{3-x}\). 5. Adding \(5\) shifts the graph \(5\) units up, producing \(k(x)=-e^{3-x}+5\). 6. As \(x\to\infty\), \(e^{3-x}\to0\), so \(k(x)\to5\). The horizontal asymptote is \(y=5\).

One possible sequence is: reflect across the y-axis, shift \(3\) units right, reflect across the x-axis, and shift \(5\) units up. The horizontal asymptote is \(y=5\).

- How does a horizontal shift change the input of the function? - Use the product rule for powers with the same base. - Rewrite the constant scale factor as a power of \(e\). - How does the sign in \(x-c\) determine the direction of a horizontal shift?

1. A shift \(3\) units left replaces \(x\) with \(x+3\), so \(g(x)=e^{x+3}\). 2. Using exponent rules, \(e^{x+3}=e^3e^x=e^3f(x)\). Therefore, the same graph is produced by a vertical stretch with scale factor \(k=e^3\). 3. For part b, \(h(x)=\frac{1}{e^2}e^x=e^{-2}e^x=e^{x-2}=f(x-2)\). 4. Replacing \(x\) with \(x-2\) shifts the graph \(2\) units right.

a) \(g(x)=e^3e^x\), so \(k=e^3\). b) Shift \(G_f\) \(2\) units to the right.

- Factor the coefficient of \(x\) from the entire exponent. - A coefficient multiplying \(x\) inside the exponent changes horizontal scale. - Factors and constants outside the exponential affect outputs. - Check the sequence by reconstructing the final exponent.

1. Factor the exponent: \(0.5x-2=0.5(x-4)\). 2. Horizontally stretch the graph of \(g\) by a factor of \(2\), producing \(e^{0.5x}\). 3. Shift right \(4\) units, producing \(e^{0.5(x-4)}\). 4. Vertically stretch by a factor of \(4\). 5. Shift down \(1\) unit.

Horizontally stretch by a factor of \(2\), shift right \(4\) units, vertically stretch by a factor of \(4\), and shift down \(1\) unit.

- How can a positive number be written using \(e\) and the natural logarithm? - What change to the input reflects a graph across the y-axis? - Recall the meaning of a negative exponent.

1. Since \(4=e^{\ln(4)}\), \(f(x)=\left(e^{\ln(4)}\right)^x=e^{x\ln(4)}\). 2. Reflection across the y-axis replaces \(x\) with \(-x\), so \(g(x)=f(-x)=4^{-x}\). 3. Using exponent rules, \(4^{-x}=\left(4^{-1}\right)^x=\left(\frac{1}{4}\right)^x\). 4. In base \(e\), \(g(x)=e^{-x\ln(4)}\).

1. \(f(x)=e^{x\ln(4)}\) 2. \(g(x)=\left(\frac{1}{4}\right)^x=e^{-x\ln(4)}\)

- Write \(f(x+3)\) as a multiple of \(f(x)\). - How does shifting a graph left differ from shifting it right in the function rule? - Use the exponent law for a power raised to another power to connect the two shift sizes.

1. Since \(f(x+3)=ca^{x+3}=a^3f(x)\), the condition gives \(a^3=4\). Thus \(a=\sqrt[3]{4}\approx1.5874\). 2. A shift \(1.5\) units left gives \(f(x+1.5)=a^{1.5}f(x)=\sqrt{a^3}f(x)=2f(x)\). 3. A shift \(1\) unit right gives \(f(x-1)=a^{-1}f(x)\), so the vertical factor is \(\frac{1}{\sqrt[3]{4}}\approx0.6300\). 4. If a shift \(\Delta x\) left has factor \(q=a^{\Delta x}\), then shifting \(2\Delta x\) left has factor \(a^{2\Delta x}=(a^{\Delta x})^2=q^2\).

a) \(a=\sqrt[3]{4}\approx1.5874\). b) \(2\). c) \(\frac{1}{\sqrt[3]{4}}\approx0.6300\). d) \(a^{2\Delta x}=(a^{\Delta x})^2\).

- How can you write a positive base as a power of \(e\)? - For a base less than \(1\), what is the sign of its natural logarithm? - What does a negative input do to the graph of \(e^x\), and how can you account for the size of the coefficient? - Can a positive number raised to a real power become zero?

1. Since \(b=e^{\ln(b)}\), we have \(b^x=e^{x\ln(b)}\), so \(k=\ln(b)\). 2. When \(0<b<1\), \(\ln(b)<0\). Let \(d=\frac{1}{|\ln(b)|}\). Then \(g(x)=f\!\left(-\frac{x}{d}\right)\). Reflect the graph of \(f\) across the y-axis and scale it horizontally by the positive factor \(d\). The resulting graph decreases as \(x\) increases. 3. For every real \(x\), \(b^x>0\). The range is \((0,\infty)\). The graph can approach \(y=0\), but cannot reach or cross the x-axis.

1. \(k=\ln(b)\). 2. Reflect across the y-axis, then scale horizontally by \(d=\frac{1}{|\ln(b)|}\). The graph is strictly decreasing because \(\ln(b)<0\). 3. Range \((0,\infty)\); \(b^x\) is always positive.

- Write the scale factor \(3\) as a power with base \(e\). - Combine powers with the same base. - What direction is represented by replacing \(x\) with \(x+c\)?

1. Write \(3=e^{\ln(3)}\). Then \(h(x)=e^{\ln(3)}e^x=e^{x+\ln(3)}\). 2. Since \(h(x)=f(x+\ln(3))\), the graph of \(f\) is shifted \(\ln(3)\) units to the left.

Shift the graph \(\ln(3)\) units to the left.

- How does reflection across the x-axis change each output? - Apply the second transformation to the entire result of the first. - Compare the final expressions for all x-values.

1. Reflecting \(f\) across the x-axis gives \(-e^x\). Shifting that graph \(5\) units down gives \(h_1(x)=-e^x-5\). 2. Reversing the order, first shift \(f\) down \(5\) units to obtain \(e^x-5\). 3. Reflecting the entire shifted function across the x-axis gives \(h_2(x)=-(e^x-5)=-e^x+5\). 4. Since \(-e^x-5\ne-e^x+5\), the two orders produce different functions.

a) \(h_1(x)=-e^x-5\) b) \(h_2(x)=-e^x+5\). The order matters because \(h_1\ne h_2\).

- Apply each transformation to the current function, not just to the original exponential term. - Use parentheses when a vertical stretch follows a vertical shift. - Recall the end behavior of \(e^x\) as \(x\to-\infty\).

1. For the first order, shift and then scale: \(g_1(x)=2(e^x+3)=2e^x+6\). 2. Since \(e^x\to0\) as \(x\to-\infty\), the horizontal asymptote of \(g_1\) is \(y=6\). Also, \(g_1(0)=2e^0+6=8\), so the y-intercept is \((0, 8)\). 3. For the second order, scale and then shift: \(g_2(x)=2e^x+3\). 4. The horizontal asymptote of \(g_2\) is \(y=3\). Also, \(g_2(0)=2e^0+3=5\), so the y-intercept is \((0, 5)\).

1) \(g_1(x)=2e^x+6\); horizontal asymptote: \(y=6\); y-intercept: \((0, 8)\) 2) \(g_2(x)=2e^x+3\); horizontal asymptote: \(y=3\); y-intercept: \((0, 5)\)

- A reflection across the x-axis multiplies the entire current output by \(-1\). - Pay attention to whether the vertical shift is inside or outside the reflection. - Use the fact that \(e^x\) is always positive.

1. Reflect first and then shift: \(g(x)=-e^x+2\). 2. Solve \(-e^x+2=0\): \(e^x=2\), so \(x=\ln(2)\). 3. Shift first and then reflect: \(h(x)=-(e^x+2)=-e^x-2\). 4. Since \(e^x>0\) for every real \(x\), \(-e^x-2<0\). Therefore, \(h\) has no real zeros.

\(g(x)=-e^x+2\) has the zero \(x=\ln(2)\). \(h(x)=-e^x-2\) has no real zeros.

- Evaluate the function at \(x=0\). - Read each y-intercept from the graph. - At the same x-value, compare each graph's vertical distance from \(y=-2\). - What does \(e^x\) approach as \(x\to-\infty\)?

1. The y-intercept is \(f_a(0)=a-2\), so \(a\) equals the y-intercept plus \(2\). 2. Graph p has y-intercept \(1\), so \(a=3\). Graph q has y-intercept \(0\), so \(a=2\). Graph r has y-intercept \(-1\), so \(a=1\). Graph s has y-intercept \(-1.5\), so \(a=0.5\). 3. Increasing \(a\) raises the y-intercept. At a fixed \(x\), the vertical distance from the asymptote \(y=-2\) is \(ae^x\), so increasing \(a\) increases that distance by the same multiplicative factor. 4. As \(x\to-\infty\), \(e^x\to0\), so \(ae^x-2\to-2\) for every \(a>0\). Therefore, every graph has horizontal asymptote \(y=-2\).

a) p: \(a=3\); q: \(a=2\); r: \(a=1\); s: \(a=0.5\) b) The y-intercept moves upward, and at any fixed \(x\), the vertical distance from \(y=-2\) increases in direct proportion to \(a\). c) \(y=-2\); the term \(ae^x\) approaches \(0\) as \(x\to-\infty\) for every \(a>0\).

- Compare the horizontal asymptotes of the three graphs. - Track the point \((0, 1)\) on the parent function. - Determine which transformed graph is decreasing. - Use the form \(ae^{x-d}+c\) to organize the transformations.

1. The graph of \(g\) has horizontal asymptote \(y=-2\), so \(f\) is shifted down \(2\) units. The point \((0, 1)\) on \(f\) corresponds to \((1, -1)\) on \(g\), showing a shift right \(1\) unit. 2. Therefore, \(g(x)=e^{x-1}-2\). 3. The graph of \(h\) is decreasing and has horizontal asymptote \(y=3\). Reflecting \(f\) across the x-axis gives \(-e^x\), and shifting up \(3\) units gives the displayed graph. 4. Therefore, \(h(x)=-e^x+3\).

\(g\): shift right \(1\) unit and down \(2\) units; \(g(x)=e^{x-1}-2\) \(h\): reflect across the x-axis and shift up \(3\) units; \(h(x)=-e^x+3\)

- Compare where the two graphs level off to identify the vertical translation. - Track the point \((0, 1)\) from the parent graph. - How does a horizontal translation change the input of a function?

1. The horizontal asymptote moves from \(y=0\) to \(y=-2\), so the graph shifts down \(2\) units. 2. The point \((0, 1)\) on \(f\) moves to \((-3, -1)\) on \(g\), so the graph shifts left \(3\) units. 3. Thus, \(g(x)=f(x+3)-2=e^{x+3}-2\). 4. At \(x=0\), \(g(0)=e^3-2\).

1) Shift left \(3\) units and down \(2\) units. 2) \(g(x)=e^{x+3}-2\) 3) \(g(0)=e^3-2\)

- Substitute \(x=0\) to compare y-intercepts. - Identify each horizontal asymptote. - Compare growth rates at the same positive x-value. - Use exponent rules to identify equivalent equations.

1. Graph \(a\) has y-intercept \(2\) and grows like a vertical scaling of \(e^x\). It matches \(y=2e^x\). 2. Graph \(b\) also has y-intercept \(2\), but its horizontal asymptote is \(y=1\). It matches \(y=e^x+1\). 3. Graph \(c\) passes through \((-1, 1)\) and grows like \(e^x\). It matches \(y=e^{x+1}\). Since \(e^{x+1}=e\cdot e^x\), both listed equations describe this graph. 4. Graph \(d\) has y-intercept \(-1\) and horizontal asymptote \(y=-2\). It matches \(y=e^x-2\).

Graph \(a\): \(y=2e^x\); graph \(b\): \(y=e^x+1\); graph \(c\): \(y=e^{x+1}\) and \(y=e\cdot e^x\); graph \(d\): \(y=e^x-2\)

- How does horizontal scaling change the input of a function? - Which exponent rule rewrites \(e^{x/2}\) as a power raised to \(x\)? - Express \(5\) as a power with base \(e\). - Use the inverse relationship between exponential and logarithmic functions.

1. A horizontal stretch by a factor of \(d\) replaces \(x\) with \(\frac{x}{d}\). For \(d=2\), \(g(x)=f\!\left(\frac{x}{2}\right)=e^{x/2}\). 2. Rewrite \(e^{x/2}=\left(e^{1/2}\right)^x=(\sqrt{e})^x\). Thus, \(b=\sqrt{e}\). 3. To produce \(h(x)=5^x\), require \(e^{x/d}=5^x\). Since \(5^x=e^{x\ln(5)}\), the exponents must satisfy \(\frac{1}{d}=\ln(5)\). 4. Therefore, \(d=\frac{1}{\ln(5)}\approx0.621\). Because \(d<1\), this is a horizontal compression.

a) \(g(x)=(\sqrt{e})^x\), so \(b=\sqrt{e}\). b) \(d=\frac{1}{\ln(5)}\approx0.621\)