Free Math Worksheets

- A negative coefficient outside the function creates a reflection across the x-axis. - Compare the rule with the form \(a q(x-c)+d\). - A coefficient with absolute value between \(0\) and \(1\) creates a vertical compression. - Determine the direction of the shift from the expression \(x+2\).

1. Replacing \(x\) with \(x+2\) shifts the graph left \(2\) units. 2. Multiplying the output by \(0.5\) vertically compresses the graph by a factor of \(0.5\). 3. The negative sign reflects the graph across the x-axis. 4. Subtracting \(3\) shifts the graph down \(3\) units.

Shift left \(2\) units, vertically compress by a factor of \(0.5\), reflect across the x-axis, and shift down \(3\) units.

- 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: \(y=3\)

- The exponential factor is always positive. - Set the linear factor equal to zero. - You can also compare the y-intercepts, since \(f_a(0)=-a\).

1. Since \(e^{-0.5x}>0\), the only zero occurs when \(x-a=0\). Thus, each graph has its x-intercept at \(x=a\). 2. Graph p crosses at \(x=-2\), so \(a=-2\). Graph q crosses at \(x=-1\), so \(a=-1\). Graph r crosses at \(x=1\), so \(a=1\). Graph s crosses at \(x=2\), so \(a=2\). 3. The y-intercept \(f_a(0)=-a\) provides the same matches.

p: \(a=-2\) q: \(a=-1\) r: \(a=1\) s: \(a=2\)

- Compare the horizontal asymptotes to determine the vertical shift. - Track a recognizable point from the parent graph to the transformed graph. - Test your equation with an easy input such as \(x=1\) or \(x=2\).

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 both shifts 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)

- Identify the graph with a vertex first. - Look for the graph that repeats and determine its amplitude. - For the exponential graph, track the reflection and vertical shift of the parent function.

1. Graph \(f\) is an upward-opening parabola with vertex \((1, -3)\). This matches \(t_1(x)=(x-1)^2-3\). 2. Graph \(g\) is periodic, passes through the origin, and has amplitude \(2\). This matches \(t_2(x)=2\sin(0.5x)\). 3. Graph \(h\) approaches \(y=3\) as \(x\to\infty\). Starting with \((0.5)^x\), reflecting across the x-axis gives \(-(0.5)^x\), and shifting up \(3\) units gives \(3-(0.5)^x\). Therefore, graph \(h\) matches \(t_3\), and its horizontal asymptote is \(y=3\).

\(f\rightarrow t_1\) \(g\rightarrow t_2\) \(h\rightarrow t_3\); the horizontal asymptote of \(t_3\) is \(y=3\).

- 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)

- 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) \((0, 1.5)\)

- 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 \(x\to\infty\).

- A number added outside an exponential expression produces a vertical shift. - Set the function equal to zero to find an x-intercept. - Consider whether a horizontal shift changes the possible output values.

1. Subtracting \(9\) shifts the graph of \(f\) down \(9\) units, so the horizontal asymptote of \(g\) is \(y=-9\). 2. For the x-intercept, solve \(3^x-9=0\). Then \(3^x=9=3^2\), so \(x=2\). The x-intercept is \((2, 0)\). 3. Replacing \(x\) with \(x+1\) shifts the graph left \(1\) unit. 4. A horizontal shift does not change the set of output values. Both \(f\) and \(h\) have range \((0, \infty)\).

a) Horizontal asymptote: \(y=-9\); x-intercept: \((2, 0)\) b) Shift left \(1\) unit c) The range does not change; it is \((0, \infty)\).

- 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.

- 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.

- 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.

- 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 \((0, 4)\).

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

- 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. \(r(x)=-4^x\)

- 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.

- Identify the horizontal line approached on the far left. - Compare whether the graph increases or decreases relative to \(2^x\). - A reflection across the x-axis changes the sign of the function. - A vertical translation changes the horizontal asymptote. - Check your equation using visible points such as the y-intercept.

1. The graph approaches the dashed horizontal line \(y=3\) as \(x\to-\infty\), so the horizontal asymptote is \(y=3\). 2. The graph is decreasing and lies below its asymptote, indicating that \(2^x\) was reflected across the x-axis to give \(-2^x\). 3. Moving that graph up \(3\) units gives \(g(x)=-2^x+3=3-2^x\). 4. A check gives \(g(0)=2\), \(g(1)=1\), and \(g(2)=-1\), which agree with the graph.

a) \(y=3\) b) \(g(x)=3-2^x\); 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 the rule for a left shift and the rule for a vertical stretch. - Use an exponent rule to separate \(2^{x+2}\). - A horizontal shift changes only the x-coordinate of a point. - A vertical stretch changes only the y-coordinate.

1. A shift left \(2\) units gives \(g(x)=f(x+2)=2^{x+2}\). 2. A vertical stretch by a factor of \(4\) gives \(g(x)=4f(x)=4\cdot2^x\). 3. By the exponent rule, \(2^{x+2}=2^x\cdot2^2=4\cdot2^x\). Thus, both descriptions define the same function. 4. A left shift changes \(P(0, 1)\) to \(Q(-2, 1)\). Since \(g(-2)=2^0=1\), \(Q\) lies on \(g\). 5. A vertical stretch changes \(P(0, 1)\) to \(R(0, 4)\). Since \(g(0)=2^2=4\), \(R\) also lies on \(g\).

a) \(2^{x+2}=2^2\cdot2^x=4\cdot2^x\), so both transformations produce the same function. b) The left shift maps \(P(0, 1)\) to \(Q(-2, 1)\), and the vertical stretch maps it to \(R(0, 4)\). Both \(Q\) and \(R\) lie on \(g\).

- 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)

- Use the horizontal asymptote to find the vertical translation. - Reflection across the x-axis places a negative sign in front of the exponential term. - Check your equation with the y-intercept.

1. The horizontal asymptote is \(y=3\), so the vertical translation is up \(3\) units. 2. Reflection across the x-axis changes the sign of the exponential term, giving \(g(x)=-\left(\frac{1}{2}\right)^x+3\). 3. The y-intercept checks the result: \(g(0)=-1+3=2\), matching the graph.

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

- 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 horizontal asymptotes to find the vertical translation. - Track a convenient point such as the y-intercept of \(f\). - A shift right changes the input to \(x-c\). - A vertical shift adds a constant outside the exponential expression.

1. The horizontal asymptote moves from \(y=0\) to \(y=1\), so the graph shifts up \(1\) unit. 2. The point \((0, 4)\) on \(f\) corresponds to \((2, 5)\) on \(g\), so the graph also shifts right \(2\) units. 3. Therefore, \(g(x)=f(x-2)+1\). 4. Substituting the rule for \(f\) gives \(g(x)=4\cdot0.5^{x-2}+1\).

1) Shift 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)

- 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)

- 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 \(a<0\).