- Decide whether each visible \(3\) acts on the input or on the output. - A horizontal change is made inside the function's input. - Compare each option with the form \(f(cx)\).

1. A horizontal compression by a factor of \(\frac{1}{3}\) is produced by replacing the input \(x\) with \(3x\). 2. Therefore, \(g(x)=f(3x)=2^{3x}\). 3. In this rule, the coefficient \(3\) multiplies the input \(x\), not the output of the exponential function.

A. \(g(x)=2^{3x}\). The coefficient \(3\) multiplies the input \(x\).

- Decide whether the \(-3\) changes the input or the output. - A constant added outside a function changes every y-value by that amount.

1. The \(-3\) is outside the exponential expression, so it subtracts \(3\) from every output value. 2. Therefore, the graph shifts down \(3\) units.

A vertical shift down \(3\) units.

- Compare the outputs of the two functions at the same input. - Does the added number change the input or the output?

1. Adding \(4\) outside the exponential increases every output by \(4\), so the graph shifts up \(4\) units.

Up \(4\) units.

- Compare where both curves have the same height. - Which direction must you move a point on \(f\) to reach the matching point on \(g\)?

1. The point \((0,1)\) on the graph of \(f\) corresponds to \((2,1)\) on the graph of \(g\). 2. Every point moves right \(2\) units, so the horizontal shift is right \(2\) units.

Right \(2\) units.

- Start with the point on the original graph where the input is \(1\). Where would it move after reflection? - What change to a function's input would produce the same reflection for every point?

1. Reflecting across the y-axis changes the input from \(x\) to \(-x\). 2. Therefore, \(g(x)=f(-x)=3^{-x}\).

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

- Compare outputs in the same row, not outputs in different rows. - Is the ratio of the two outputs the same in each row?

1. Each \(g\) output is three times the corresponding \(f\) output; for example, \(g(1)/f(1)=6/2=3\). 2. Hence \(a=3\).

\(a=3\)

- What horizontal line does the graph of \(2^x\) approach? - How does a vertical translation affect that line?

1. The parent exponential \(2^x\) approaches \(0\) as \(x\to-\infty\). 2. Subtracting \(5\) shifts its horizontal asymptote to \(y=-5\).

\(y=-5\)

- A reflection across the y-axis changes the sign of the input. - A vertical shift changes all output values and the horizontal asymptote. - Recall the domain and range of an exponential function. - Track the parent asymptote through the vertical shift.

1. Reflecting across the y-axis replaces \(x\) with \(-x\), giving \(2^{-x}\). 2. Shifting up \(3\) units gives \(g(x)=2^{-x}+3\). 3. Exponential functions are defined for every real input, so the domain is \((-\infty, \infty)\). 4. Since \(2^{-x}>0\), the range is \((3, \infty)\). 5. The parent horizontal asymptote \(y=0\) shifts up \(3\) units to \(y=3\).

a) \(g(x)=2^{-x}+3\) b) Domain: \((-\infty, \infty)\); range: \((3, \infty)\); horizontal asymptote: \(y=3\)

- Substitute \(x=0\) to find each y-intercept. - Analyze \(1.4^x\) as \(x\to-\infty\). - Compare the base with \(1\). - Decide whether multiplying by a positive constant reverses the graph's direction.

1. \(f(0)=2.5\cdot(1.4)^0=2.5\), so the y-intercept is \((0, 2.5)\). Also, \(g(0)=0.4\cdot(1.4)^0=0.4\), so the y-intercept is \((0, 0.4)\). 2. As \(x\to-\infty\), \(1.4^x\to0\). Multiplying by either positive constant still gives a limit of \(0\), so both graphs have horizontal asymptote \(y=0\). 3. Since the common base \(1.4\) is greater than \(1\) and both leading factors are positive, both functions are increasing.

1. \(f\): \((0, 2.5)\); \(g\): \((0, 0.4)\) 2. Horizontal asymptote: \(y=0\) 3. Both functions are increasing.

- Substitute \(x=3\). - Compare the base with \(1\). - Determine which horizontal line the graph approaches for large positive \(x\). - Examine the coefficient in front of the exponential expression.

1. \(h(3)=4\cdot(0.5)^3=4\cdot0.125=0.5\). 2. Since \(0<0.5<1\), the function is decreasing. 3. As \(x\to\infty\), \((0.5)^x\to0\), so the horizontal asymptote is \(y=0\). 4. Multiplying by \(4\) produces a vertical stretch by a factor of \(4\). Every output is four times as large, and the y-intercept changes from \((0, 1)\) to \((0, 4)\).

a) \(h(3)=0.5\) b) Decreasing because \(0<0.5<1\) c) \(y=0\) d) A vertical stretch by a factor of \(4\); the y-intercept changes from \((0, 1)\) to \((0, 4)\).

- Reflection across the y-axis replaces \(x\) with \(-x\). - Reflection across the x-axis changes the sign of every output. - Use the negative-exponent rule. - A positive exponential decay function has a base between \(0\) and \(1\).

1. Reflection across the y-axis replaces \(x\) with \(-x\): \(g(x)=f(-x)=5^{-x}=\left(\frac{1}{5}\right)^x=(0.2)^x\). 2. Reflection across the x-axis multiplies every output by \(-1\): \(h(x)=-f(x)=-5^x\). 3. Exponential decay with positive outputs has a base between \(0\) and \(1\). Thus \(g\) is the decay function, with decay factor \(0.2\).

1. \(g(x)=(0.2)^x\), or \(g(x)=5^{-x}\) 2. \(h(x)=-5^x\) 3. \(g\); decay factor \(0.2\)

- Rewrite a negative exponent as a reciprocal. - Recall how coordinates change under reflection across an axis. - Reflection across the x-axis changes the sign of the output.

1. Reflection across the y-axis replaces \(x\) with \(-x\): \(q(x)=p(-x)=\left(\frac{1}{4}\right)^{-x}=4^x\). 2. Reflection across the y-axis changes \((x, y)\) to \((-x, y)\). Thus \(A(1, 0.25)\) maps to \(A'(-1, 0.25)\). 3. Reflection across the x-axis multiplies every output by \(-1\), so \(r(x)=-q(x)=-4^x\).

1. \(q(x)=4^x\) 2. \(A'(-1, 0.25)\) 3. \(r(x)=-4^x\)

- Compare the horizontal asymptotes to determine the vertical translation. - Track a recognizable point to determine the horizontal translation. - Keep the description in terms of one horizontal and one vertical translation, as requested.

1. The horizontal asymptote of \(f\) is \(y=0\), while the horizontal asymptote of \(g\) is \(y=-2\). Therefore, the graph is shifted \(2\) units down. 2. The point \((0, 1)\) on \(f\) corresponds to \((1, -1)\) on \(g\), indicating a shift \(1\) unit right in addition to the downward shift. 3. Applying those translations gives \(g(x)=2^{x-1}-2\).

Shift the graph \(1\) unit right and \(2\) units down. The equation is \(g(x)=2^{x-1}-2\).

- Read the y-intercept of each graph. - Determine whether each graph increases or decreases. - Identify each horizontal asymptote. - Check for vertical translations.

1. Graph A is increasing, passes through \((0, 1)\), and has horizontal asymptote \(y=0\), so it matches (1). 2. Graph B is decreasing, passes through \((0, 1)\), and has horizontal asymptote \(y=0\), so it matches (2). 3. Graph C is increasing, has y-intercept \((0, -2)\), and has horizontal asymptote \(y=-3\), so it matches (3). 4. Equation (4) has y-intercept \((0, 2)\), which none of the graphs has, so it is unused.

Graph A: (1); Graph B: (2); Graph C: (3); Unused equation: (4)

- Compare whether each candidate should increase or decrease. - Evaluate each candidate at \(x=0\) to compare y-intercepts. - Track what an x-axis reflection and an upward shift do to the parent asymptote.

1. The graph is increasing and has y-intercept \((0,2)\). 2. Both \(t_1\) and \(t_2\) use base \(0.5\), but \(t_1\) is decreasing and has y-intercept \(4\), so it does not match. 3. The function \(t_3(x)=3-2^x\) has y-intercept \(2\), but it is decreasing, so it does not match. 4. For \(t_2(x)=3-(0.5)^x\), reflecting \((0.5)^x\) across the x-axis makes the graph increasing, and shifting it up \(3\) units moves the horizontal asymptote from \(y=0\) to \(y=3\). 5. Therefore, the graph is \(t_2(x)=3-(0.5)^x\).

\(h(x)=t_2(x)=3-(0.5)^x\). It is increasing, has y-intercept \((0,2)\), and has horizontal asymptote \(y=3\).

- Evaluate the family at \(x=0\). - Read each y-intercept from the graph. - The parameter controls the vertical stretch, reflection, and intercept.

1. Since \(e^0=1\), \(f_a(0)=a\). Therefore, \(a\) is the y-intercept of each graph. 2. Graph (1) has y-intercept \(2\), so \(a=2\). 3. Graph (2) has y-intercept \(0.5\), so \(a=0.5\). 4. Graph (3) has y-intercept \(-1\), so \(a=-1\).

(1) \(a=2\); (2) \(a=0.5\); (3) \(a=-1\)

- Evaluate each function at \(x=0\). - Identify the horizontal asymptote of each graph. - Determine whether each function increases or decreases, including the effect of a negative sign in front of the exponential term.

1. For A, the y-intercept is \(2\), the horizontal asymptote is \(y=0\), and the function decreases. This matches graph c). 2. For B, the y-intercept is \(-1\), the horizontal asymptote is \(y=-2\), and the function decreases. This matches graph a). 3. For C, the y-intercept is \(1\), the horizontal asymptote is \(y=2\), and the negative sign reflects the decreasing base function, making the result increasing. This matches graph b).

A: c); B: a); C: b)

- Rewrite the expression using a positive exponent. - Use the base to decide whether the function increases or decreases. - Evaluate the function at \(x=0\) and \(x=1\). - Identify the horizontal asymptote.

1. Rewrite the function: \(0.25^{-x}=\left(\frac{1}{4}\right)^{-x}=4^x\). 2. Since the base \(4\) is greater than \(1\), the graph is increasing. 3. The graph passes through \((0, 1)\) and \((1, 4)\). 4. Only graph a) has these features.

Graph a)

- Choose one y-value that both curves reach and compare the corresponding x-values. - Focus on horizontal distances rather than the y-intercepts, which are the same here. - An input multiplier changes horizontal scale; an outside multiplier changes vertical scale.

1. On the graph, \(g\) reaches a given positive output at half the x-value needed by \(f\). For example, \(g(1)=4\) while \(f(2)=4\). 2. Halving horizontal distances is a horizontal compression by a factor of \(\frac{1}{2}\), which is produced by replacing \(x\) with \(2x\). 3. Therefore, \(g(x)=2^{2x}\).

A. \(g(x)=2^{2x}\); its horizontal distances from the y-axis are half those of \(f\) for corresponding output values.

- Compare the horizontal levels the curves approach to the left. - How does a constant outside an exponential move its graph? - Check the y-intercepts to confirm the shift.

1. The graph of \(f\) approaches \(y=0\); the graph of \(g\) approaches \(y=2\). 2. The graph of \(g\) is shifted up \(2\), so \(c=2\) and \(g(x)=3^x+2\).

\(g(x)=3^x+2\)

- Find an input where the exponent of \(g\) equals zero. - Compare that input with where \(f\) equals \(1\). - Check your conclusion by moving a single point on the original curve.

1. At \(x=0\), the graph of \(f\) passes through \((0,1)\). 2. For \(g\), \(g(-3)=2^0=1\), so the same height occurs at \(x=-3\). 3. The graph shifts left \(3\) units, not right. Jordan is incorrect.

No. The shift is left \(3\) units; \((0,1)\) on \(f\) moves to \((-3,1)\) on \(g\).

- How are reflections across both coordinate axes related to a reflection across the origin? - What change to a function reflects its graph across the x-axis? Across the y-axis? - How do you find the y-intercept of a graph? - Recall the value of \(e^0\).

1. Reflecting a graph across the y-axis replaces \(x\) with \(-x\). Therefore, \(g(x)=f(-x)=0.5e^{-x}-2\). 2. Reflecting across the origin replaces \(x\) with \(-x\) and then multiplies the output by \(-1\). Thus, \(h(x)=-f(-x)=-(0.5e^{-x}-2)=-0.5e^{-x}+2\). 3. At the y-intercept, \(x=0\). Since \(e^0=1\), \(h(0)=-0.5(1)+2=1.5\). The y-intercept is \((0, 1.5)\).

a) \(g(x)=0.5e^{-x}-2\) b) \(h(x)=-0.5e^{-x}+2\) c) \((0, 1.5)\)

- What change inside the exponent moves the graph sideways? - How does the coefficient outside the exponential affect all output values? - Which part of the rule changes the graph's height without changing its shape?

1. The input \(x+2\) shifts the graph of \(q\) left \(2\) units. 2. Multiplying all outputs by \(-0.5\) reflects the graph across the x-axis and compresses it vertically by a factor of \(0.5\). 3. Subtracting \(3\) shifts the resulting graph down \(3\) units.

One way: shift left \(2\) units, reflect across the x-axis and vertically compress by a factor of \(0.5\), then shift down \(3\) units. Other equivalent sequences are possible.

- Express \(0.2\) as a power of \(5\). - Consider how replacing \(x\) with \(-x\) transforms a graph. - Determine whether each function increases or decreases for larger inputs.

1. \(f(-1)=0.2\), \(f(0)=1\), and \(f(1)=5\). Also, \(g(-1)=5\), \(g(0)=1\), and \(g(1)=0.2\). 2. Since \(0.2=\frac{1}{5}=5^{-1}\), \(g(x)=(5^{-1})^x=5^{-x}=f(-x)\). Therefore, the graph of \(g\) is the reflection of the graph of \(f\) across the y-axis. 3. As \(x\to\infty\), \(f(x)=5^x\to\infty\), while \(g(x)=0.2^x\to0\).

a) \(f(-1)=0.2\), \(f(0)=1\), \(f(1)=5\); \(g(-1)=5\), \(g(0)=1\), \(g(1)=0.2\) b) \(g(x)=5^{-x}=f(-x)\); the graphs are reflections across the y-axis. c) \(f(x)\to\infty\) and \(g(x)\to0\) as \(x\to\infty\).

- What is the horizontal asymptote of \(3^x\), and what does subtracting \(9\) do to it? - What equation describes an x-intercept? - How does replacing \(x\) with \(x+1\) affect the graph? - Does moving a graph sideways change which outputs are possible?

1. Subtracting \(9\) shifts the graph of \(f\) down \(9\) units, so the horizontal asymptote of \(g\) is \(y=-9\). 2. Solve \(3^x-9=0\). Since \(9=3^2\), the x-intercept is \((2, 0)\). 3. Since \(h(x)=f(x+1)\), its graph is the graph of \(f\) shifted left \(1\) unit. 4. A horizontal shift does not change the set of outputs, so the range stays \((0, \infty)\).

a) Asymptote: \(y=-9\); x-intercept: \((2, 0)\) b) Shift left \(1\) unit. c) No. Both ranges are \((0, \infty)\).

- Use \(x=0\) for the y-intercept and \(f(x)=0\) for the x-intercept. - Consider what happens to \(2^{x-2}\) as \(x\) becomes very negative. - Identify the horizontal and vertical shifts from the equation.

1. For the y-intercept, \(f(0)=2^{-2}-4=\frac{1}{4}-4=-3.75\), so the point is \((0, -3.75)\). 2. For the x-intercept, solve \(2^{x-2}-4=0\). Then \(2^{x-2}=4=2^2\), so \(x-2=2\) and \(x=4\). The point is \((4, 0)\). 3. Since \(2^{x-2}\to0\) as \(x\to-\infty\), \(f(x)\to-4\). Thus the horizontal asymptote is \(y=-4\), approached from above. 4. The graph of \(2^x\) is shifted right \(2\) units and down \(4\) units.

a) x-intercept: \((4, 0)\); y-intercept: \((0, -3.75)\) b) Horizontal asymptote: \(y=-4\); as \(x\to-\infty\), \(f(x)\to-4\) from above. c) Shift right \(2\) units and down \(4\) units.

- Compare each base with \(1\). - Evaluate both functions at \(x=0\). - Consider the end behavior of each exponential function. - Use the reciprocal relationship between \(1.25\) and \(0.8\).

1. Since \(1.25>1\), \(f\) is increasing. Since \(0<0.8<1\), \(g\) is decreasing. 2. Both functions equal \(1\) at \(x=0\), so the graphs intersect at \((0, 1)\). Both have horizontal asymptote \(y=0\). 3. Since \(1.25=\frac{5}{4}\) and \(0.8=\frac{4}{5}\), \(f(-x)=1.25^{-x}=\left(\frac{4}{5}\right)^x=0.8^x=g(x)\). Therefore, replacing \(x\) by \(-x\) reflects the graph across the y-axis.

a) \(f\) is increasing; \(g\) is decreasing. b) Intersection: \((0, 1)\); horizontal asymptote: \(y=0\) c) \(f(-x)=g(x)\), so the graphs are reflections across the y-axis.

- Compare examples such as \(a=1\) and \(a=2\). - Consider the effect of multiplying a positive function by a positive constant. - Determine all possible values of \((0.8)^x\).

1. \(f_a(0)=a(0.8)^0=a\), so the y-intercept \((0, a)\) depends on \(a\). 2. Since \((0.8)^x\to0\) as \(x\to\infty\), multiplying by any positive \(a\) still gives a limit of \(0\). The horizontal asymptote is always \(y=0\). 3. Because \(0<0.8<1\) and \(a>0\), every function in the family is decreasing. 4. Since \((0.8)^x>0\) and \(a>0\), every output is positive. The range is always \((0, \infty)\).

Depends on \(a\): the y-intercept, \((0, a)\). Independent of \(a\): the horizontal asymptote \(y=0\), decreasing behavior, and range \((0, \infty)\).

- Compare the bases and coefficients. - Evaluate each function at \(x=0\). - Substitute the function expressions into the quotient and simplify. - Interpret a constant ratio geometrically.

1. Both functions are increasing because \(1.2>1\), and both have horizontal asymptote \(y=0\). 2. The coefficient produces a vertical scaling and determines the y-intercept. Since \(5>2\), the graph of \(g\) lies above the graph of \(h\) for every \(x\), with y-intercepts \(5\) and \(2\), respectively. 3. \(\frac{g(x)}{h(x)}=\frac{5(1.2)^x}{2(1.2)^x}=\frac{5}{2}=2.5\). The ratio is constant for every \(x\).

a) Both are increasing and both have horizontal asymptote \(y=0\). b) The coefficient determines the vertical scale and y-intercept; the graph of \(g\) lies above the graph of \(h\). c) \(\frac{g(x)}{h(x)}=2.5\), a constant ratio.

- Express \(8\) as a power of \(2\). - Use the product rule for powers with the same base. - Recall how replacing \(x\) with \(x+c\) translates a graph. - Write a right shift using \(x-s\).

1. From \(g(x)=8(2)^x\), the vertical scale factor is \(8\). 2. Since \(8=2^3\), \(g(x)=2^3\cdot2^x=2^{x+3}=f(x+3)\). Therefore, the graph is shifted left \(3\) units. 3. For an exponential base \(b>0\), \(b\ne1\), a shift right by \(s\) units gives \(b^{x-s}=b^{-s}b^x\). Thus the equivalent vertical scale factor is \(b^{-s}=\frac{1}{b^s}\).

a) \(8\) b) \(g(x)=2^{x+3}\), so the graph is shifted left \(3\) units. c) \(b^{-s}\), or \(\frac{1}{b^s}\)

- Substitute the point into \(y=b^x\). - Determine whether a positive-base exponential function can equal zero or become negative. - Identify the horizontal line approached by the graph. - A reflection across the y-axis replaces \(x\) with \(-x\).

1. Substituting \((2, 9)\) gives \(9=b^2\). Since an exponential base is positive, \(b=3\). Because \(3>1\), the function represents growth. 2. The range of \(3^x\) is \((0, \infty)\), and its horizontal asymptote is \(y=0\). 3. Reflection across the y-axis replaces \(x\) with \(-x\): \(g(x)=3^{-x}=\left(\frac{1}{3}\right)^x\).

1. \(b=3\); exponential growth 2. Range: \((0, \infty)\); horizontal asymptote: \(y=0\) 3. \(g(x)=\left(\frac{1}{3}\right)^x\)

- Substitute the given point into the function equation. - What does multiplying every output by a constant do to a graph? - Consider the magnitude and sign of the coefficient separately. - What happens to every y-coordinate when the entire function is multiplied by \(-1\)?

1. Substitute \((3, 2)\): \(2=a\cdot2^3=8a\). Therefore, \(a=\frac{1}{4}=0.25\). 2. The magnitude \(|a|\) controls vertical scaling. If \(|a|>1\), the graph is stretched vertically. If \(0<|a|<1\), it is compressed vertically. If \(a<0\), the graph is also reflected across the x-axis. For \(a=-1\), the only transformation is reflection across the x-axis.

1. \(a=0.25\) 2. \(|a|>1\): vertical stretch; \(0<|a|<1\): vertical compression; \(a<0\): reflection across the x-axis in addition to scaling. For \(a=-1\), there is only a reflection across the x-axis.

- How do you find a y-intercept? - What roles do the sign and magnitude of the coefficient play? - Track how the y-values change as \(x\) increases. - Recall the behavior of an exponential function with a base between \(0\) and \(1\).

1. At \(x=0\), \(f(0)=0.5^0=1\), so the y-intercept is \((0, 1)\). Also, \(h(0)=-3\cdot0.5^0=-3\), so its y-intercept is \((0, -3)\). 2. Multiplying by \(3\) stretches the graph vertically by a factor of \(3\), and the negative sign reflects it across the x-axis. 3. Since \(0<0.5<1\), \(f\) is strictly decreasing. Multiplying all outputs by a negative number reverses their order, so the reflected graph \(h\) is strictly increasing.

1. The y-intercepts are \((0, 1)\) for \(f\) and \((0, -3)\) for \(h\). 2. Stretch vertically by a factor of \(3\), then reflect across the x-axis. 3. The reflection across the x-axis reverses the decreasing behavior of \(f\), so \(h\) is increasing.

- Which height does the curve approach far to the left? - Does this curve lie above or below its limiting horizontal line? - What does the y-intercept tell you about the vertical translation?

1. As \(x\) becomes very negative, the plotted curve levels off toward \(y=3\). This is its horizontal asymptote. 2. The curve decreases and lies below this asymptote, consistent with an x-axis reflection of \(2^x\). 3. The curve passes through \((0, 2)\), so \(g(x)=-2^x+d\) gives \(2=-1+d\). Hence \(d=3\). 4. Thus \(g(x)=3-2^x=-f(x)+3\): reflect the graph of \(f\) across the x-axis, then shift up \(3\) units.

a) \(y=3\) b) \(g(x)=3-2^x=-f(x)+3\); reflect across the x-axis, then shift up \(3\) units.

- Read key points such as the y-intercept and the value at \(x=1\). - Use the base to distinguish increasing and decreasing exponential functions. - Test candidate functions at \(x=0\), \(x=1\), or \(x=-1\). - Look for algebraically equivalent expressions involving negative exponents.

1. Graph p is increasing, passes through \((0, 1)\), and has value \(3\) at \(x=1\). Therefore, it matches (1), \(f(x)=3^x\). 2. Graph q is increasing, passes through \((0, 1)\), and has value \(1.5\) at \(x=1\). Therefore, it matches (2), \(f(x)=1.5^x\). 3. Graph r is decreasing, passes through \((0, 1)\), and has value \(2\) at \(x=-1\). Therefore, it matches (3), \(f(x)=0.5^x\). Since \(2^{-x}=(2^{-1})^x=(0.5)^x\), it also matches (6). 4. Graph s has horizontal asymptote \(y=-2\), y-intercept \((0, -1)\), and x-intercept \((1, 0)\). Therefore, it matches (4), \(f(x)=2^x-2\).

Graph p: (1); Graph q: (2); Graph r: (3) and (6); Graph s: (4)

- Write an equation for each claimed transformation. - Which exponent rule connects a sum in an exponent with multiplication? - Read point \(P\) from the graph. How do the two transformations affect its coordinates differently?

1. Shifting left \(2\) gives \(g(x)=f(x+2)=2^{x+2}\). A vertical stretch by \(4\) gives \(g(x)=4f(x)=4\cdot 2^x\). 2. Since \(2^{x+2}=2^2\cdot 2^x=4\cdot 2^x\), the two rules agree. 3. The plotted point is \(P(-1, 0.5)\). A left shift sends it to \((-3, 0.5)\), and \(g(-3)=2^{-1}=0.5\). 4. A vertical stretch sends it to \((-1, 2)\), and \(g(-1)=2^1=2\). Both points lie on \(g\).

a) \(2^{x+2}=4\cdot2^x\), so the two transformations give the same function. b) Left shift: \((-3, 0.5)\); vertical stretch: \((-1, 2)\). Both satisfy \(g(x)=2^{x+2}\).

- Recall the shape, y-intercept, and horizontal asymptote of \(2^x\). - Changes inside the exponent create horizontal transformations. - A negative input reflects a graph across the y-axis. - A constant added outside the exponential shifts the graph vertically.

1. The function \(h(x)=2^{-x}\) is the reflection of \(2^x\) across the y-axis, so it is the only decreasing graph. Therefore, \(h\) matches graph a. 2. The function \(g(x)=2^{x-3}\) is shifted right \(3\) units. Its point \((0, 1)\) moves to \((3, 1)\), so \(g\) matches graph b. 3. The function \(k(x)=2^x+2\) is shifted up \(2\) units. Its horizontal asymptote is \(y=2\), and its y-intercept is \((0, 3)\), so \(k\) matches graph e. 4. Graph c is the parent function, so its equation is \(y=2^x\). 5. Graph d is shifted left \(2\) units, so its equation is \(y=2^{x+2}\), equivalently \(y=4\cdot2^x\).

a) \(g\): graph b; \(h\): graph a; \(k\): graph e b) Graph c: \(y=2^x\); Graph d: \(y=2^{x+2}\), or equivalently \(y=4\cdot2^x\)

- Convert the decimal bases to fractions. - Use the negative-exponent rule. - Recall how reflection across the y-axis changes point coordinates.

1. Since \(2.5=\frac{5}{2}\) and \(0.4=\frac{2}{5}\), \(f(-x)=\left(\frac{5}{2}\right)^{-x}=\left(\frac{2}{5}\right)^x=(0.4)^x=g(x)\). 2. The relationship \(g(x)=f(-x)\) means that the graph of \(g\) is the reflection of the graph of \(f\) across the y-axis. 3. Both functions equal \(1\) at \(x=0\). Since one is increasing and the other is decreasing, their only intersection is \((0, 1)\).

a) \(g(x)=(0.4)^x=\left(\frac{5}{2}\right)^{-x}=f(-x)\) b) The graphs are reflections across the y-axis. c) \((0, 1)\)

- First compare the y-intercepts. - For graphs with the same y-intercept, compare their growth rates. - Evaluating each function at \(x=1\) can help compare the bases.

1. In \(y=ab^x\), the y-intercept is \(a\). Graph (1) has y-intercept \(2\), so it matches (e). Graph (2) has y-intercept \(0.5\), so it matches (d). 2. Graphs (3), (4), and (5) all have y-intercept \(1\). For \(x>0\), a larger base produces faster growth. 3. Graph (3) grows fastest, so it matches (c). Graph (4) matches (b), and graph (5), which grows slowest, matches (a).

(1) → (e); (2) → (d); (3) → (c); (4) → (b); (5) → (a)

- How does the graph behave as \(x\) becomes large and positive? - What sign change represents reflection across the x-axis? - Use the graph's y-intercept to check the vertical shift.

1. The curve levels off toward \(y=3\) as \(x\) increases, so its horizontal asymptote is \(y=3\). 2. Reflection across the x-axis gives \(-\left(\frac{1}{2}\right)^x\). Shifting it up \(3\) units gives \(g(x)=-\left(\frac{1}{2}\right)^x+3\). 3. The equation gives \(g(0)=-1+3=2\), which matches the plotted y-intercept.

\(g(x)=-\left(\frac{1}{2}\right)^x+3\)

- Use the horizontal asymptote to determine \(d\). - Substitute a clearly readable point to find \(a\). - The y-intercept is especially convenient.

1. The horizontal asymptote is \(y=-6\), so \(d=-6\). 2. The graph has y-intercept \((0, -3)\). Substituting gives \(a(2)^0-6=-3\), so \(a=3\). 3. Therefore, \(h(x)=3(2)^x-6\). The x-intercept \((1, 0)\) verifies the equation.

\(h(x)=3(2)^x-6\)

- Evaluate the exponential term at \(x=0\). - The horizontal asymptote gives \(d\). - A graph below its asymptote requires a negative value of \(a\). - Check your equation with a second point.

1. The horizontal asymptote is \(y=8\), so \(d=8\). 2. The point \((0, 6)\) lies on the graph. Substituting gives \(a\left(\frac{1}{2}\right)^0+8=6\), so \(a=-2\). 3. Therefore, \(k(x)=-2\left(\frac{1}{2}\right)^x+8\). The point \((-2, 0)\) verifies the equation.

\(k(x)=-2\left(\frac{1}{2}\right)^x+8\)

- Compare the levels approached by the two curves far to the right. - Track one point on the blue curve to the corresponding red point. - What change to the input represents a shift to the right? - How do you write an upward shift outside a function rule?

1. The blue graph approaches \(y=0\), while the red graph approaches \(y=1\). The vertical shift is up \(1\) unit. 2. The point \((0, 4)\) on \(f\) corresponds to \((2, 5)\) on \(g\), so the horizontal shift is right \(2\) units. 3. Thus \(g(x)=f(x-2)+1\). 4. Substitution gives \(g(x)=4\cdot0.5^{x-2}+1\).

1) Right \(2\) units and up \(1\) unit. 2) \(g(x)=f(x-2)+1\) 3) \(g(x)=4\cdot0.5^{x-2}+1\)

- Use each horizontal asymptote to identify the vertical shift. - Track the reference point \((0, 1)\) from the base graph to each transformed graph.

1. The graph of \(g\) is shifted right \(3\) units and up \(1\) unit from \(f\). Therefore, \(g(x)=2^{x-3}+1\). 2. The graph of \(h\) is shifted down \(3\) units from \(f\). Therefore, \(h(x)=2^x-3\).

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

- Evaluate the function at \(x=0\). - Determine the limit of \(2^x\) as \(x\to-\infty\). - Identify which side of each graph approaches the horizontal asymptote. - Check one additional input if needed.

1. The y-intercept is \(f(0)=3-2^0=2\). The correct graph must pass through \((0, 2)\). 2. As \(x\to-\infty\), \(2^x\to0\), so \(f(x)\to3\). Thus, the graph has horizontal asymptote \(y=3\) on its left side. 3. Graphs a) and d) have y-intercept \(2\), but only graph a) approaches \(y=3\) as \(x\to-\infty\). 4. Therefore, graph a) is correct.

Graph a)

- Identify the horizontal asymptote before using any plotted point. - In the form \(b^{c(x-h)}+d\), which parameter gives the asymptote directly? - After finding the vertical shift, substitute one clear non-asymptotic point from the graph.

1. The horizontal asymptote is \(y=-2\), so \(d=-2\). 2. The plotted point \((1.5,1)\) lies on the graph. Substitute it: \(1=3^{c(1.5-1)}-2\). 3. Then \(3=3^{0.5c}\), so \(0.5c=1\) and \(c=2\). 4. Thus, \(g(x)=3^{2(x-1)}-2\).

\(c=2\) and \(d=-2\)

- What output does the plotted graph give when the input is zero? - Use that output in the rule containing the unknown constant. - Which exponent rule lets you write the product as one power of \(2\)?

1. The graph crosses the y-axis at \((0,4)\). Since \(g(0)=4\cdot2^{-1}+c=2+c\), we get \(c=2\). 2. Since \(4=2^2\), \(4\cdot2^{x-1}=2^{x+1}\). Thus \(s=1\), giving \(g(x)=2^{x+1}+2\). 3. Relative to \(f(x)=2^x\), the graph shifts left \(1\) unit and up \(2\) units.

a) \(g(0)=4\), so \(c=2\). b) \(s=1\); \(g(x)=2^{x+1}+2\) c) Left \(1\) unit and up \(2\) units.

- Rewrite \(10^{x+1}\) as a product. - Express \(0.01\) as a power of \(10\). - Determine the sign of \(b^c\) when \(b>0\). - Consider whether a horizontal shift can move an always-positive graph onto or below the x-axis.

1. \(h(x)=10^{x+1}=10\cdot10^x\), so the graph is vertically stretched by a factor of \(10\). 2. Since \(0.01=10^{-2}\), \(k(x)=10^{-2}\cdot10^x=10^{x-2}=f(x-2)\). Therefore, the graph is shifted right \(2\) units. 3. A horizontal shift produces a factor of the form \(b^c\), which is always positive. Every positive factor can be represented by some horizontal shift, but \(a=0\) or \(a<0\) cannot. Therefore, the claim is false.

a) Stretch factor \(10\) b) Shift right \(2\) units c) False; horizontal translations can produce only positive multiplicative factors, not \(a=0\) or \(a<0\).