Free Math Worksheets

- Recall the period of the parent cosine function. - Identify the coefficient of \(x\) inside the cosine function. - Determine how that coefficient horizontally stretches or compresses the graph.

1. In the form \(a\cos(bx)\), the coefficient of \(x\) is \(b=\frac{\pi}{1.5}\). 2. The period is \(p=\frac{2\pi}{|b|}\). 3. Substitute: \(p=\frac{2\pi}{\pi/1.5}=3\).

c) \(3\)

- Identify the parameter that controls vertical scale. - A shift right changes the input inside cosine. - A downward shift is represented by subtracting from the output. - Amplitude is the maximum distance from the midline.

1. The vertical stretch gives \(3.5\cos(x)\). 2. Shifting right \(\frac{\pi}{4}\) units replaces \(x\) with \(x-\frac{\pi}{4}\). 3. Shifting down \(2\) units gives \(h(x)=3.5\cos\left(x-\frac{\pi}{4}\right)-2\). 4. The amplitude is the absolute value of the outside coefficient, \(3.5\).

Function rule: \(h(x)=3.5\cos\left(x-\frac{\pi}{4}\right)-2\) Amplitude: \(3.5\)

- Identify the new constants inside and outside the sine expression. - A change inside the input affects horizontal position; an outside addition affects vertical position. - The midline is the horizontal line about which the graph oscillates. - Compare the height of the waves with the parent function.

1. The input \(x-\pi\) shifts the graph right by \(\pi\). 2. The outside term \(+4\) shifts the graph up \(4\) units. 3. The outside coefficient is \(1\), so the amplitude is \(1\). 4. The vertical shift gives the midline \(y=4\).

a) Shift right by \(\pi\) and up by \(4\). b) Amplitude: \(1\); midline: \(y=4\)

- A vertical scale factor multiplies the function value. - A horizontal shift appears inside the function input with the opposite sign. - Reflection across the x-axis multiplies the function by \(-1\). - A vertical shift is added outside the sine expression.

1. For part a, multiply the function by \(5\) and add \(2\): \(g(x)=5\sin(x)+2\). 2. For part b, replace \(x\) with \(x-\frac{\pi}{2}\), then multiply by \(-1\): \(g(x)=-\sin\left(x-\frac{\pi}{2}\right)\). 3. For part c, multiply by \(0.4\), replace \(x\) with \(x+3\), and subtract \(1\): \(g(x)=0.4\sin(x+3)-1\).

a) \(g(x)=5\sin(x)+2\) b) \(g(x)=-\sin\left(x-\frac{\pi}{2}\right)\) c) \(g(x)=0.4\sin(x+3)-1\)

- The coefficient outside sine or cosine determines the vertical stretch. - Amplitude is always nonnegative, even when the leading coefficient is negative. - The coefficient of \(x\) inside the function determines the horizontal stretch or compression. - Start with the parent period \(2\pi\).

1. For \(f\), the amplitude is \(|3|=3\). Since \(b=0.5\), the period is \(p=\frac{2\pi}{0.5}=4\pi\). 2. For \(g\), the amplitude is \(|-2|=2\). The vertical shift does not affect amplitude or period. Since \(b=\pi\), the period is \(p=\frac{2\pi}{\pi}=2\). 3. For \(h\), the amplitude is \(\left|\frac{1}{2}\right|=\frac{1}{2}\). Since \(b=\frac{1}{4}\), the period is \(p=\frac{2\pi}{1/4}=8\pi\).

a) \(A=3\); \(p=4\pi\) b) \(A=2\); \(p=2\) c) \(A=\frac{1}{2}\); \(p=8\pi\)

- An outside coefficient changes the height of the waves. - A coefficient multiplying \(x\) changes the horizontal scale. - An outside addition or subtraction changes the vertical position. - A negative outside coefficient reflects the graph across the x-axis.

1. For \(f_1\), the coefficient \(4\) produces a horizontal compression by a factor of \(\frac{1}{4}\). The period changes from \(2\pi\) to \(\frac{\pi}{2}\). 2. For \(f_2\), the coefficient \(0.2\) produces a vertical compression, giving amplitude \(0.2\), and the term \(-5\) shifts the graph down \(5\) units. 3. For \(f_3\), the negative sign reflects the graph across the x-axis, and the factor \(3\) vertically stretches it to amplitude \(3\).

a) Horizontal compression by a factor of \(\frac{1}{4}\); period \(\frac{\pi}{2}\) b) Vertical compression to amplitude \(0.2\), then a shift down \(5\) units c) Reflection across the x-axis and vertical stretch to amplitude \(3\)

- Use the coefficient of \(x\) to calculate the period. - Identify which parameter in \(y=a\sin(bx)\) controls vertical stretching. - Consider how a larger value of \(b\) changes the speed of oscillation.

1. For \(f\), \(p=\frac{2\pi}{2}=\pi\), so statement a is true. 2. The coefficient \(4\) in \(g(x)=4\sin(x)\) produces a vertical stretch by a factor of \(4\), so statement b is true. 3. For \(h\), \(p_h=\frac{2\pi}{4}=\frac{\pi}{2}\). For \(k\), \(p_k=\frac{2\pi}{2}=\pi\). Since \(\frac{\pi}{2}<\pi\), statement c is false; \(h\) has the smaller period.

a) True. b) True. c) False. The period of \(h\) is \(\frac{\pi}{2}\), which is less than the period \(\pi\) of \(k\).

- Recall the minimum and maximum values of the parent sine function. - Consider how an outside coefficient changes the amplitude. - Recall where the sine graph crosses the x-axis. - A vertical stretch does not change the x-intercepts.

1. The parent sine function has range \([-1, 1]\). Multiplying by \(1.5\) gives the range \([-1.5, 1.5]\). 2. The equation \(1.5\sin(x)=0\) is equivalent to \(\sin(x)=0\). 3. In \([0, 2\pi]\), sine is zero at \(x=0\), \(x=\pi\), and \(x=2\pi\).

a) \([-1.5, 1.5]\) b) \(x=0, \pi, 2\pi\)

- Recall where the cosine graph crosses the x-axis. - Subtracting a constant shifts a graph vertically. - Apply the vertical shift to both endpoints of the original range.

1. Cosine is zero where the unit-circle x-coordinate is \(0\). In \([-\pi, \pi]\), this occurs at \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\). 2. The range of \(\cos(x)\) is \([-1, 1]\). Subtracting \(1\) shifts every output down by \(1\), so the range of \(h\) is \([-2, 0]\).

a) \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\) b) \([-2, 0]\)

- Cosine is the x-coordinate on the unit circle. - Use the highest and lowest x-coordinates to identify the range. - Locate the minimum point of one cosine period.

1. The greatest value of cosine is \(1\), and the least value is \(-1\). 2. Therefore, the range is \([-1, 1]\). 3. On \([0, 2\pi]\), cosine equals \(-1\) only at \(x=\pi\).

a) Maximum: \(1\); minimum: \(-1\) b) \([-1, 1]\) c) \(x=\pi\)

- Recall where the sine graph crosses the x-axis. - Use the period of sine to locate repeated minima. - Use the repeating sine pattern over the interval to identify all relevant points. - Check the multiples of \(\pi\) and odd multiples of \(\frac{\pi}{2}\) within the endpoints.

1. The zeros of sine occur at \(x=k\pi\), where \(k \in \mathbb{Z}\). In \([-2\pi, 2\pi]\), these are \(-2\pi\), \(-\pi\), \(0\), \(\pi\), and \(2\pi\). 2. Sine equals \(-1\) at \(x=\frac{3\pi}{2}+2k\pi\), where \(k \in \mathbb{Z}\). 3. The values in the given interval are \(x=-\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).

Zeros: \(x \in \{-2\pi, -\pi, 0, \pi, 2\pi\}\) Values where \(\sin(x)=-1\): \(x \in \left\{-\frac{\pi}{2}, \frac{3\pi}{2}\right\}\)

- What are the highest and lowest values reached by the sine graph? - The sine function has period \(2\pi\). - Identify the maximum and minimum locations in one period, then repeat them as needed.

1. The maximum value of sine is \(1\), attained at \(x=\frac{\pi}{2}+2k\pi\). In \([0, 3\pi]\), this gives \(x=\frac{\pi}{2}\) and \(x=\frac{5\pi}{2}\). 2. The minimum value of sine is \(-1\), attained at \(x=\frac{3\pi}{2}+2k\pi\). In \([0, 3\pi]\), the only such value is \(x=\frac{3\pi}{2}\).

a) \(x=\frac{\pi}{2}\) and \(x=\frac{5\pi}{2}\) b) \(x=\frac{3\pi}{2}\)

- Identify the coefficient that controls the vertical stretch. - Relate the coefficient of \(x\) to the period. - Start with the minimum and maximum values of sine. - Recall the odd-function symmetry of the parent sine function.

1. The amplitude is \(|2.5|=2.5\). 2. The period is \(p=\frac{2\pi}{0.5}=4\pi\). 3. Since sine ranges from \(-1\) to \(1\), the range of \(f\) is \([-2.5, 2.5]\). 4. Because sine is odd, \(f(-x)=2.5\sin(-0.5x)=-f(x)\). Therefore, the graph has rotational symmetry about the origin.

Amplitude: \(2.5\); period: \(4\pi\); range: \([-2.5, 2.5]\); symmetry: rotational symmetry about the origin

- The outside coefficient controls vertical scaling. - A horizontal scale factor changes the input by its reciprocal. - Compare the result with the form \(a\cos(bx)\).

1. The vertical stretch gives an outside coefficient of \(2.5\). 2. A horizontal scale factor of \(0.8\) replaces \(x\) with \(\frac{x}{0.8}=1.25x\). 3. Therefore, \(g(x)=2.5\cos(1.25x)\).

\(g(x)=2.5\cos(1.25x)\)

- Read the amplitude from the maximum and minimum values. - Find the shortest horizontal interval over which the pattern repeats. - Relate the horizontal scale factor to the period. - Use the x-axis labels in multiples of \(\pi\).

1. The graph has a maximum of \(2\) and a minimum of \(-2\), so the amplitude is \(a=2\). 2. One complete cycle has length \(\pi\), so the period is \(T=\pi\). 3. Use \(b=\frac{2\pi}{T}\): \(b=\frac{2\pi}{\pi}=2\). 4. Therefore, \(f(x)=2\sin(2x)\).

\(f(x)=2\sin(2x)\)

- Find the midline first to determine the vertical shift. - The amplitude is the distance from the midline to a maximum or minimum. - Compare the cycle length with the parent sine function.

1. The maximum is \(2\) and the minimum is \(-1\), so the midline is \(\frac{2+(-1)}{2}=0.5\). Thus, \(d=0.5\). 2. The amplitude is the distance from the midline to an extreme value: \(2-0.5=1.5\). Thus, \(a=1.5\). 3. The graph completes one cycle over \(2\pi\), consistent with \(\sin(x)\). 4. Therefore, \(h(x)=1.5\sin(x)+0.5\).

\(a=1.5\); \(d=0.5\); \(h(x)=1.5\sin(x)+0.5\)

- A graph that falls through the origin has a negative sine coefficient. - Find the period from corresponding points on consecutive cycles. - Use \(b=\frac{2\pi}{p}\).

1. Graph a) falls through the origin, so \(a\) is negative. Its amplitude is \(2\), so \(a=-2\). The period is \(2\), so \(b=\frac{2\pi}{2}=\pi\). 2. Graph b) has amplitude \(3\) and period \(5\). Thus, \(a=3\), and \(b=\frac{2\pi}{5}\). 3. Graph c) has amplitude \(1\) and period \(0.5\). Thus, \(a=1\), and \(b=\frac{2\pi}{0.5}=4\pi\).

a) \(a=-2\), \(b=\pi\) b) \(a=3\), \(b=\frac{2\pi}{5}\) c) \(a=1\), \(b=4\pi\)

- Count grid intervals to find each x-coordinate. - Each horizontal grid interval represents \(\frac{\pi}{6}\). - Substitute the x-coordinate into \(f(x)=\cos(x)\). - Recall special cosine values.

1. Read the x-coordinates from the grid: \(x_A=-\frac{\pi}{2}\), \(x_B=0\), \(x_C=\frac{\pi}{3}\), and \(x_D=\pi\). 2. Evaluate cosine at each x-coordinate: \(\cos\left(-\frac{\pi}{2}\right)=0\), \(\cos(0)=1\), \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\), and \(\cos(\pi)=-1\). 3. Therefore, \(A=\left(-\frac{\pi}{2}, 0\right)\), \(B=(0, 1)\), \(C=\left(\frac{\pi}{3}, \frac{1}{2}\right)\), and \(D=(\pi, -1)\).

\(A=\left(-\frac{\pi}{2}, 0\right)\), \(B=(0, 1)\), \(C=\left(\frac{\pi}{3}, \frac{1}{2}\right)\), and \(D=(\pi, -1)\)

- Count grid intervals to find the x-coordinates. - Substitute each x-coordinate into \(f(x)=\cos(x)-1\). - Check the points against the graph.

1. Reading the x-coordinates from the grid gives \(x_A=0\), \(x_B=\frac{\pi}{2}\), \(x_C=\pi\), and \(x_D=2\pi\). 2. Evaluate the function: \(f(0)=0\), \(f\left(\frac{\pi}{2}\right)=-1\), \(f(\pi)=-2\), and \(f(2\pi)=0\). 3. Therefore, the points are \(A(0, 0)\), \(B\left(\frac{\pi}{2}, -1\right)\), \(C(\pi, -2)\), and \(D(2\pi, 0)\).

\(A(0, 0)\), \(B\left(\frac{\pi}{2}, -1\right)\), \(C(\pi, -2)\), \(D(2\pi, 0)\)

- Each horizontal grid interval represents \(\frac{\pi}{6}\). - Count grid intervals from the origin to obtain each x-coordinate. - Use special-angle sine values for the y-coordinates.

1. Read the x-coordinates from the grid: \(x_A=0\), \(x_B=\frac{\pi}{6}\), \(x_C=\frac{\pi}{2}\), and \(x_D=\frac{3\pi}{2}\). 2. Evaluate sine at each x-coordinate: \(\sin(0)=0\), \(\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}\), \(\sin\left(\frac{\pi}{2}\right)=1\), and \(\sin\left(\frac{3\pi}{2}\right)=-1\). 3. Therefore, the points are \(A=(0, 0)\), \(B=\left(\frac{\pi}{6}, \frac{1}{2}\right)\), \(C=\left(\frac{\pi}{2}, 1\right)\), and \(D=\left(\frac{3\pi}{2}, -1\right)\).

\(A=(0, 0)\), \(B=\left(\frac{\pi}{6}, \frac{1}{2}\right)\), \(C=\left(\frac{\pi}{2}, 1\right)\), and \(D=\left(\frac{3\pi}{2}, -1\right)\)

- The parameter \(a\) is the amplitude, the maximum distance from the midline. - Count grid intervals on the x-axis, with each interval representing \(\frac{\pi}{2}\). - After finding \(a\), substitute the read x-values into \(f(x)=a\sin(x)\).

1. The first maximum occurs at \(x=\frac{\pi}{2}\) and has y-coordinate \(2\). Since \(\sin\left(\frac{\pi}{2}\right)=1\), \(a=2\), so \(f(x)=2\sin(x)\). 2. Reading the x-coordinates from the grid gives \(x_A=0\), \(x_B=\frac{\pi}{2}\), \(x_C=\pi\), and \(x_D=\frac{3\pi}{2}\). 3. Evaluating the function gives \(A(0, 0)\), \(B\left(\frac{\pi}{2}, 2\right)\), \(C(\pi, 0)\), and \(D\left(\frac{3\pi}{2}, -2\right)\).

\(a=2\); \(A(0, 0)\), \(B\left(\frac{\pi}{2}, 2\right)\), \(C(\pi, 0)\), \(D\left(\frac{3\pi}{2}, -2\right)\)

- Identify whether each marked point is a maximum, minimum, or zero. - Read each x-coordinate from the grid. - Substitute the x-coordinate into the function to verify the y-coordinate. - Recall sine values at multiples of \(\frac{\pi}{2}\).

1. Point \(A\) is the first maximum. Since \(\sin\left(\frac{\pi}{2}\cdot1\right)=1\), \(A=(1, 1)\). 2. Point \(B\) is the first zero after the origin. Since \(\sin\left(\frac{\pi}{2}\cdot2\right)=\sin(\pi)=0\), \(B=(2, 0)\). 3. Point \(C\) is the first minimum. Since \(\sin\left(\frac{\pi}{2}\cdot3\right)=\sin\left(\frac{3\pi}{2}\right)=-1\), \(C=(3, -1)\). 4. Point \(D\) is the next zero. Since \(\sin\left(\frac{\pi}{2}\cdot4\right)=\sin(2\pi)=0\), \(D=(4, 0)\).

\(A=(1, 1)\), \(B=(2, 0)\), \(C=(3, -1)\), and \(D=(4, 0)\)

- What is \(\cos(0)\)? - A coefficient outside cosine determines the amplitude. - Read the graphs' y-values at \(x=0\).

1. Each graph has the shape and period of a cosine function, so the family has the form \(g_a(x)=a\cos(x)\). 2. Since \(\cos(0)=1\), each graph has \(g_a(0)=a\). 3. Reading the y-values at \(x=0\) gives \(0.5\), \(1\), \(2\), and \(3\). 4. Therefore, \(a \in \{0.5, 1, 2, 3\}\).

\(g_a(x)=a\cos(x)\), with \(a \in \{0.5, 1, 2, 3\}\)

- In \(a\sin(bx)\), identify the roles of \(a\) and \(b\). - Relate the horizontal compression to the length of one complete cycle.

1. Use \(a=0.75\) and \(b=3\): \(h(x)=0.75\sin(3x)\). 2. The amplitude is \(A=0.75\). 3. The period is \(p=\frac{2\pi}{3}\).

a) \(h(x)=0.75\sin(3x)\) b) Amplitude: \(0.75\); period: \(\frac{2\pi}{3}\)

- Compare amplitudes and midlines. - Check the value of each function at \(x=0\) or another simple input. - Identify horizontal and vertical shifts from key points. - Determine whether each graph has the shape of sine or cosine.

1. Graph 1 has amplitude \(2\) and value \(2\) at \(x=0\), so it matches \(p(x)=2\cos x\). 2. Graph 2 is a sine curve shifted right \(1\) unit. It crosses its midline at \(x=1\), so it matches \(q(x)=\sin(x-1)\). 3. Graph 3 is a cosine curve with midline \(y=-2\). It is shifted down \(2\) units, so it matches \(r(x)=\cos x-2\).

Graph 1: \(p(x)\) Graph 2: \(q(x)\) Graph 3: \(r(x)\)

- A horizontal shift changes the input inside the cosine function. - In \(a\cos(b(x-c))+d\), identify which parameter controls amplitude and which controls period. - Use the periodicity of cosine to find equivalent phase shifts.

1. A vertical stretch by a factor of \(2\) and a shift right \(\pi\) units give \(g(x)=2\cos(x-\pi)\). 2. The amplitude is \(|2|=2\). The coefficient of \(x\) is \(1\), so the period remains \(2\pi\). 3. The rule from part a gives \(d=-\pi\). Since cosine has period \(2\pi\), \(\cos(x-\pi)=\cos(x+\pi)\), so \(d=\pi\) also works.

a) \(g(x)=2\cos(x-\pi)\) b) Amplitude: \(2\); period: \(2\pi\) c) \(d=\pi\) and \(d=-\pi\)

- A horizontal stretch changes the coefficient of \(x\) inside the function. - Start with the maximum and minimum values of sine, then apply the vertical shift. - Use a phase-shift identity relating sine and cosine.

1. A horizontal stretch by a factor of \(3\) replaces \(x\) with \(\frac{1}{3}x\). Shifting up \(2\) units gives \(q(x)=\sin\left(\frac{1}{3}x\right)+2\). Since sine has range \([-1, 1]\), the range of \(q\) is \([1, 3]\). 2. For \(\sin(bx)\), the period is \(\frac{2\pi}{|b|}\). Here, \(b=\frac{1}{3}\), so the period is \(\frac{2\pi}{1/3}=6\pi\). 3. Use \(\sin(\alpha)=\cos\left(\alpha-\frac{\pi}{2}\right)\). With \(\alpha=\frac{1}{3}x\), this gives \(q(x)=\cos\left(\frac{1}{3}x-\frac{\pi}{2}\right)+2\), so \(c=\frac{\pi}{2}\).

a) \(q(x)=\sin\left(\frac{1}{3}x\right)+2\); range: \([1, 3]\) b) \(6\pi\) c) \(c=\frac{\pi}{2}\)

- Factor the coefficient of \(x\) from the entire angle expression. - Identify the parameters that control horizontal scale, phase shift, amplitude, and vertical shift. - Distinguish transformations applied to the input from those applied to the output. - Replacing \(x\) with \(3x\) changes the horizontal scale.

1. Factor the argument: \(3x-\pi=3\left(x-\frac{\pi}{3}\right)\), so \(f(x)=2\sin\left(3\left(x-\frac{\pi}{3}\right)\right)+1\). 2. Horizontally compress the graph of \(g\) by a factor of \(\frac{1}{3}\). 3. Shift the result right \(\frac{\pi}{3}\) units. 4. Vertically stretch by a factor of \(2\). 5. Shift up \(1\) unit.

Horizontally compress by a factor of \(\frac{1}{3}\), shift right \(\frac{\pi}{3}\) units, vertically stretch by a factor of \(2\), and shift up \(1\) unit.

- Compare the rule with \(a\sin(x-c)+d\). - Pay attention to the sign of the horizontal shift inside the input. - Use the amplitude and midline to find the highest and lowest values. - The range contains all possible output values.

1. The factor \(1.5\) creates a vertical stretch by a factor of \(1.5\). 2. The input \(x+2\) shifts the graph left \(2\) units. 3. The term \(-3\) shifts the graph down \(3\) units. 4. The amplitude is \(|1.5|=1.5\). 5. Sine has range \([-1, 1]\). After the vertical stretch, the range is \([-1.5, 1.5]\), and after shifting down \(3\), the range is \([-4.5, -1.5]\).

Transformations: vertically stretch by a factor of \(1.5\), shift left \(2\) units, and shift down \(3\) units. Amplitude: \(1.5\) Range: \([-4.5, -1.5]\)

- Track how the vertical stretch and reflection affect the outside coefficient. - A vertical shift is added outside the cosine expression. - Amplitude is a positive distance from the midline. - Use the midline and amplitude to find the minimum and maximum values.

1. The vertical stretch gives \(2.5\cos(x)\). 2. Reflecting across the x-axis gives \(-2.5\cos(x)\). 3. Shifting down \(1.5\) units gives \(g(x)=-2.5\cos(x)-1.5\). 4. The amplitude is \(|-2.5|=2.5\). 5. The midline is \(y=-1.5\), so the range is \([-1.5-2.5, -1.5+2.5]=[-4, 1]\).

\(g(x)=-2.5\cos(x)-1.5\); amplitude: \(2.5\); range: \([-4, 1]\)

- Locate the maximum and minimum points of the sine graph. - Compare the graph immediately to the left and right of an extremum. - Recall where the sine graph crosses the x-axis. - Use the repeating shape of the sine graph to identify line and point symmetries.

1. At \(x=\frac{3\pi}{2}\), sine has a local minimum of \(-1\). The graph has mirror symmetry to the left and right of this minimum, so \(x=\frac{3\pi}{2}\) is an axis of symmetry. 2. In \([0, \pi]\), sine has a maximum at \(x=\frac{\pi}{2}\), so \(x=\frac{\pi}{2}\) is another axis of symmetry. 3. The zeros of sine occur at \(x=k\pi\). In the given interval, possible centers are \((-\pi, 0)\), \((0, 0)\), \((\pi, 0)\), and \((2\pi, 0)\). Any three are valid.

a) The line \(x=\frac{3\pi}{2}\) passes through a minimum, and the graph mirrors across that vertical line. b) \(x=\frac{\pi}{2}\) c) For example, \((-\pi, 0)\), \((0, 0)\), and \((\pi, 0)\)

- Recall the definition of an even function. - Vertical symmetry lines pass through maxima and minima of the cosine graph. - Centers of rotational symmetry occur at x-intercepts. - Check the stated interval endpoints carefully.

1. For a), \(g(-x)=\cos(-x)=\cos(x)=g(x)\). Therefore, cosine is even, and its graph is symmetric about the y-axis. 2. The maxima and minima of cosine occur at \(x=k\pi\), and each such vertical line is an axis of symmetry. In \([0, 3\pi]\), the axes are \(x=0\), \(x=\pi\), \(x=2\pi\), and \(x=3\pi\). 3. The centers of rotational symmetry occur at the x-intercepts \(x=\frac{\pi}{2}+k\pi\). In \([0, 2\pi]\), they are \(\left(\frac{\pi}{2}, 0\right)\) and \(\left(\frac{3\pi}{2}, 0\right)\).

a) \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) \(x=0\), \(x=\pi\), \(x=2\pi\), and \(x=3\pi\) c) \(\left(\frac{\pi}{2}, 0\right)\) and \(\left(\frac{3\pi}{2}, 0\right)\)

- Match the given amplitude and period to the parameters \(a\) and \(b\). - The amplitude is the maximum distance from the midline. - Rearrange \(p=\frac{2\pi}{b}\) to solve for \(b\). - A graph that ranges from \(-2\) to \(2\) has amplitude \(2\).

1. For part a, \(a=1.5\). Using \(p=\frac{2\pi}{b}\), solve \(4\pi=\frac{2\pi}{b}\) to get \(b=0.5\). Thus, \(f(x)=1.5\cos(0.5x)\). 2. For part b, \(a=4\). Solve \(1=\frac{2\pi}{b}\) to get \(b=2\pi\). Thus, \(f(x)=4\cos(2\pi x)\). 3. For part c, the range \([-2, 2]\) gives \(a=2\). Solve \(3=\frac{2\pi}{b}\) to get \(b=\frac{2\pi}{3}\). Thus, \(f(x)=2\cos\left(\frac{2\pi}{3}x\right)\).

a) \(f(x)=1.5\cos(0.5x)\) b) \(f(x)=4\cos(2\pi x)\) c) \(f(x)=2\cos\left(\frac{2\pi}{3}x\right)\)

- Relate the parameter \(a\) to the amplitude. - Recall where the zeros of the parent sine function occur. - Use the relationship between period \(p\) and parameter \(b\). - Determine what fraction of a period lies between the origin and the first positive zero.

1. The amplitude gives \(a=4.5\). 2. For \(\sin(bx)\), the first positive zero occurs after half a period. Thus, \(\frac{p}{2}=3\), so \(p=6\). 3. Use \(b=\frac{2\pi}{p}\): \(b=\frac{2\pi}{6}=\frac{\pi}{3}\). 4. Therefore, \(f(x)=4.5\sin\left(\frac{\pi}{3}x\right)\).

\(a=4.5\); \(b=\frac{\pi}{3}\); \(f(x)=4.5\sin\left(\frac{\pi}{3}x\right)\)

- The distance between consecutive maximum points equals the period. - Use the period to determine \(b\). - Substitute the given point to find the remaining parameter. - Recall the value of \(\sin\left(\frac{\pi}{2}\right)\).

1. The distance between consecutive maximum points is the period, so \(p=\pi\). 2. Use \(b=\frac{2\pi}{p}\): \(b=\frac{2\pi}{\pi}=2\). Thus, \(g(x)=a\sin(2x)\). 3. Substitute the point: \(5=a\sin\left(2\cdot \frac{\pi}{4}\right)=a\sin\left(\frac{\pi}{2}\right)\). 4. Since \(\sin\left(\frac{\pi}{2}\right)=1\), \(a=5\). Therefore, \(g(x)=5\sin(2x)\).

\(a=5\); \(b=2\); \(g(x)=5\sin(2x)\)

- Combine the amplitude and reflection in the outside coefficient. - Use the period formula to determine the coefficient inside sine. - The outside constant controls the vertical shift. - Substitute the parameters into \(a\sin(bx)+d\).

1. The amplitude and reflection give the outside coefficient \(a=-2.5\). 2. The period \(p=4\) gives \(b=\frac{2\pi}{p}=\frac{\pi}{2}\). 3. The upward shift gives \(d=1\). 4. Therefore, \(h(x)=-2.5\sin\left(\frac{\pi}{2}x\right)+1\).

\(h(x)=-2.5\sin\left(\frac{\pi}{2}x\right)+1\)

- Factor the coefficient of \(x\) before reading a horizontal shift. - Find the minimum and maximum values of each function. - Consider the effect of shifting a sine graph by half a period.

1. Factor the input: \(\sin(2x-6)=\sin(2(x-3))\). The shift is \(3\) units right, so statement a is false. 2. The range of \(h\) is \([1, 3]\), while the range of \(k\) is \([-2, 2]\). The ranges differ, so statement b is false. 3. A shift by half a period reverses the sign of sine: \(\sin(x-\pi)=-\sin(x)\). Therefore, statement c is true.

a) False. The graph is shifted \(3\) units right because \(f(x)=\sin(2(x-3))\). b) False. The ranges are \([1, 3]\) and \([-2, 2]\), respectively. c) True.

- Use the midline crossing to determine \(c\) and \(d\). - Find the amplitude from the maximum value and the midline. - The distance from an increasing midline crossing to the next maximum is one-fourth of a period. - Use the period to determine \(b\).

1. Because \((0,2)\) lies on the midline, \(d=2\). 2. The amplitude is the difference between the maximum value and the midline: \(a=5-2=3\). 3. The distance from an increasing midline crossing to the next maximum is one-fourth of a period. Thus, \(\frac{p}{4}=2\), so \(p=8\). 4. Therefore, \(b=\frac{2\pi}{8}=\frac{\pi}{4}\). 5. The increasing midline crossing occurs at \(x=0\), so take \(c=0\). Therefore, \(g(x)=3\sin\left(\frac{\pi}{4}x\right)+2\).

\(g(x)=3\sin\left(\frac{\pi}{4}x\right)+2\)

- Cosine is zero when the unit-circle x-coordinate is \(0\). - Consecutive cosine zeros are \(\pi\) units apart. - Test values of \(\frac{\pi}{2}+k\pi\) against the interval. - Include both positive and negative values of \(k\).

1. In \([0, 2\pi]\), cosine is zero at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). 2. In general, the zeros are \(x=\frac{\pi}{2}+k\pi\), where \(k \in \mathbb{Z}\). 3. Testing integer values of \(k\) for \([-2\pi, 3\pi]\) gives \(-\frac{3\pi}{2}\), \(-\frac{\pi}{2}\), \(\frac{\pi}{2}\), \(\frac{3\pi}{2}\), and \(\frac{5\pi}{2}\). 4. There are \(5\) zeros.

a) \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) b) \(x \in \left\{-\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}\right\}\) c) \(5\) zeros

- Locate the maximum and minimum points of the sine graph. - Think about where the graph rises as \(x\) increases. - Think about where the graph falls as \(x\) increases. - Restrict every interval to the given domain.

1. On \([-\pi, \pi]\), sine reaches its minimum value \(-1\) at \(x=-\frac{\pi}{2}\) and its maximum value \(1\) at \(x=\frac{\pi}{2}\). 2. From \(x=-\frac{\pi}{2}\) to \(x=\frac{\pi}{2}\), the function rises from \(-1\) to \(1\), so it is strictly increasing on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). 3. The function is strictly decreasing on \(\left[-\pi, -\frac{\pi}{2}\right]\) and \(\left[\frac{\pi}{2}, \pi\right]\).

a) Minimum at \(x=-\frac{\pi}{2}\); maximum at \(x=\frac{\pi}{2}\) b) \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) c) \(\left[-\pi, -\frac{\pi}{2}\right]\) and \(\left[\frac{\pi}{2}, \pi\right]\)

- Track the unit-circle x-coordinate as the angle increases. - Locate the maximum and minimum values of cosine. - Recall what strictly decreasing means when comparing two inputs. - Determine which monotonicity interval contains both angles.

1. From \(0^\circ\) to \(180^\circ\), cosine decreases from \(1\) to \(-1\), so it is strictly decreasing on \([0^\circ, 180^\circ]\). 2. From \(180^\circ\) to \(360^\circ\), cosine increases from \(-1\) to \(1\), so it is strictly increasing on \([180^\circ, 360^\circ]\). 3. Both \(110^\circ\) and \(160^\circ\) lie in the decreasing interval. Since \(110^\circ < 160^\circ\), it follows that \(\cos(110^\circ)>\cos(160^\circ)\).

a) Strictly decreasing on \([0^\circ, 180^\circ]\); strictly increasing on \([180^\circ, 360^\circ]\) b) True; \(110^\circ < 160^\circ\) and cosine is decreasing on that interval.

- Identify the coefficient of \(x\) inside the sine function. - Use the period formula for \(\sin(bx)\). - Maximum points repeat after each full period.

1. For \(\sin(bx)\), the period is \(p=\frac{2\pi}{|b|}\). With \(b=0.5\), \(p=\frac{2\pi}{0.5}=4\pi\). 2. Maximum points repeat every period, so all such inputs have the form \(x=\pi+4\pi k\), where \(k\) is an integer. 3. Using \(k=1, 2, -1, -2\) gives \(5\pi, 9\pi, -3\pi, -7\pi\).

a) \(p=4\pi\) b) For example, \(-7\pi, -3\pi, 5\pi, 9\pi\)

- Use the period to determine the coefficient of \(x\). - A zero occurs when the sine value is \(0\). - Recall the spacing between consecutive zeros of the parent sine function. - Check each result against the stated interval.

1. The amplitude gives \(a=5\). The period gives \(b=\frac{360^\circ}{120^\circ}=3\), so \(f(x)=5\sin(3x)\). 2. Set \(5\sin(3x)=0\), which is equivalent to \(\sin(3x)=0\). 3. Sine is zero when \(3x=180^\circ k\), where \(k\) is an integer. Thus, \(x=60^\circ k\). 4. In the given interval, the zeros are \(0^\circ, 60^\circ, 120^\circ, 180^\circ\).

\(x=0^\circ, 60^\circ, 120^\circ, 180^\circ\)

- Recall the minimum and maximum values of sine. - Visualize the graph from \(0\) to \(2\pi\). - Include both endpoints because the interval is closed. - Count intersections with each horizontal line.

1. The range of sine is \([-1, 1]\). 2. For a), the horizontal line \(y=1\) touches the graph once, at \(x=\frac{\pi}{2}\). 3. For b), the line \(y=0.8\) intersects the graph twice. 4. For c), the line \(y=0\) intersects the graph at \(x=0\), \(x=\pi\), and \(x=2\pi\), for three solutions. 5. For d), the line \(y=-1\) touches the graph once, at \(x=\frac{3\pi}{2}\). 6. For e), \(-2\) is outside the range of sine, so there are no solutions. 7. The value \(0\) occurs at both endpoints and at the midpoint of the closed interval, while the maximum value \(1\) occurs at only one point.

a) \(1\) solution b) \(2\) solutions c) \(3\) solutions d) \(1\) solution e) \(0\) solutions There are three solutions for \(c=0\) because the graph meets the x-axis at \(0\), \(\pi\), and \(2\pi\), while it reaches its maximum only once.

- Use the range of cosine. - Consider one complete cosine period from \(0\) to \(2\pi\). - Count intersections with horizontal lines. - Remember that both interval endpoints are included.

1. The range of cosine is \([-1, 1]\), so there are no solutions when \(|k|>1\). 2. The minimum value \(-1\) occurs once on the interval, at \(x=\pi\). 3. Every value strictly between \(-1\) and \(1\) occurs twice in one closed period. 4. The maximum value \(1\) occurs at both endpoints, \(x=0\) and \(x=2\pi\), so there are two solutions.

1. \(0\) 2. \(1\); \(x=\pi\) 3. \(2\) 4. \(2\); \(x_1=0\), \(x_2=2\pi\)

- Relate the outside coefficient to the amplitude and range. - Recall the minimum and maximum values of the parent sine function. - A point on the graph satisfies the function equation. - Use the exact unit-circle value at \(\frac{3\pi}{4}\).

1. The range of \(a\sin(x)\) is \([-|a|, |a|]\). Part a gives \(|a|=2.5\), and the condition \(a>0\) gives \(a=2.5\). 2. For part b, substitute the point: \(3\sqrt{2}=a\sin\left(\frac{3\pi}{4}\right)\). 3. Since \(\sin\left(\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2}\), solve \(3\sqrt{2}=a\frac{\sqrt{2}}{2}\) to obtain \(a=6\).

a) \(a=2.5\) b) \(a=6\)

- The total vertical range is twice the amplitude. - Reflecting across the x-axis does not change the range. - Use the unit circle to evaluate cosine at \(\frac{4\pi}{3}\). - Substitute the given input and output, then solve the resulting linear equation.

1. The minimum and maximum are \(-|k|\) and \(|k|\), so their distance is \(2|k|\). 2. For part a, \(2|k|=24\), so \(|k|=12\) and \(k=12\) or \(k=-12\). 3. For part b, \(k\cos\left(\frac{4\pi}{3}\right)=2\). Since \(\cos\left(\frac{4\pi}{3}\right)=-\frac{1}{2}\), solve \(-\frac{k}{2}=2\) to get \(k=-4\).

a) \(k=12\) or \(k=-12\) b) \(k=-4\)

- Relate the period of cosine to the coefficient of \(x\). - Substitute the coordinates of a point that lies on the graph. - Identify the unit-circle angles with the required cosine value. - Solve the resulting equation for \(k\).

1. The period is \(p=\frac{2\pi}{k}\). 2. For part a, solve \(5\pi=\frac{2\pi}{k}\) to obtain \(k=\frac{2}{5}=0.4\). 3. For part b, substitute the point: \(\cos\left(\frac{k\pi}{4}\right)=-\frac{\sqrt{2}}{2}\). 4. The least positive angle with this cosine value is \(\frac{3\pi}{4}\). Therefore, \(\frac{k\pi}{4}=\frac{3\pi}{4}\), so \(k=3\).

a) \(k=0.4\) b) \(k=3\)

- Find the period first to determine how much of the graph appears in the interval. - Set the sine input equal to integer multiples of \(\pi\) to find zeros. - Recall where the parent sine function reaches its maximum and minimum. - Verify that every input lies in the stated interval.

1. The amplitude is \(0.5\), and the period is \(p=\frac{2\pi}{2}=\pi\). 2. Zeros occur when \(2x=k\pi\). In the interval, this gives \(x=0, \frac{\pi}{2}, \pi\). 3. The maximum occurs when \(2x=\frac{\pi}{2}\), so \(x=\frac{\pi}{4}\) and \(g(x)=0.5\). The maximum point is \(\left(\frac{\pi}{4}, 0.5\right)\). 4. The minimum occurs when \(2x=\frac{3\pi}{2}\), so \(x=\frac{3\pi}{4}\) and \(g(x)=-0.5\). The minimum point is \(\left(\frac{3\pi}{4}, -0.5\right)\).

Amplitude: \(0.5\); period: \(\pi\); zeros: \(0, \frac{\pi}{2}, \pi\); maximum: \(\left(\frac{\pi}{4}, 0.5\right)\); minimum: \(\left(\frac{3\pi}{4}, -0.5\right)\)

- Factor the coefficient of \(x\) from the entire input before reading the horizontal shift. - Interpret the parameters in \(a\sin(b(x-c))+d\). - An inside coefficient greater than \(1\) horizontally compresses the graph.

1. Factor the input: \(f(x)=0.5\sin\left(2\left(x+\frac{\pi}{4}\right)\right)-1\). 2. The outside factor \(0.5\) gives a vertical compression by \(0.5\). 3. The inside factor \(2\) gives a horizontal compression by \(\frac{1}{2}\), so the period is \(\pi\). 4. The term \(x+\frac{\pi}{4}\) shifts the graph left \(\frac{\pi}{4}\). 5. The outside term \(-1\) shifts the graph down \(1\) unit.

Vertical compression by \(0.5\); horizontal compression by \(\frac{1}{2}\); shift left \(\frac{\pi}{4}\); shift down \(1\).

- Use the period to determine the inside coefficient. - Reflection across the x-axis changes the sign of the outside coefficient. - A left shift appears as addition inside the input.

1. Reflection and vertical stretch give \(a=-3\). 2. Doubling the parent period gives \(T=4\pi\), so \(b=\frac{2\pi}{4\pi}=0.5\). 3. A shift left by \(\pi\) gives \(x+\pi\), and a shift down \(2\) gives \(d=-2\). 4. Therefore, \(g(x)=-3\cos\left(0.5(x+\pi)\right)-2\).

\(g(x)=-3\cos\left(0.5(x+\pi)\right)-2\)

- A horizontal stretch changes the function input by the reciprocal factor. - Reflection across the x-axis negates the function value. - Apply the transformations in order. - Substitute the point's x-coordinate to test it.

1. A horizontal stretch by \(4\) replaces \(x\) with \(\frac{x}{4}\). 2. Reflecting across the x-axis gives \(-\cos\left(\frac{x}{4}\right)\). 3. Shifting right by \(\pi\) gives \(h(x)=-\cos\left(\frac{x-\pi}{4}\right)\). 4. Evaluate: \(h(5\pi)=-\cos\left(\frac{5\pi-\pi}{4}\right)=-\cos(\pi)=1\). Therefore, the point lies on the graph.

a) \(h(x)=-\cos\left(\frac{x-\pi}{4}\right)\) b) Yes, because \(h(5\pi)=1\).

- Average the maximum and minimum values to find the midline. - Find the amplitude as the distance from the midline to an extreme value. - Read one full cycle from the graph, and use \(b=\frac{2\pi}{p}\). - Check a clearly visible point in your equation.

1. The maximum value is \(1\), and the minimum value is \(-2\). The midline is \(d=\frac{1+(-2)}{2}=-0.5\). 2. The amplitude is \(a=1-(-0.5)=1.5\). 3. The graph returns to the same phase after \(2\) units, so the period is \(2\). Therefore, \(b=\frac{2\pi}{2}=\pi\). 4. The equation is \(f(x)=1.5\sin(\pi x)-0.5\).

\(a=1.5\), \(b=\pi\), \(d=-0.5\); \(f(x)=1.5\sin(\pi x)-0.5\)

- Find the period from the distance between corresponding points on consecutive cycles. - Use \(b=\frac{2\pi}{p}\). - Find the midline and amplitude from the maximum and minimum values. - Use the behavior at \(x=0\) to confirm whether the required sine or cosine form fits.

1. In graph a), the curve passes through \((0, 0)\) while increasing, so the required sine form is appropriate. 2. Its maximum is \(1.5\), so \(a=1.5\). Its period is \(4\), so \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). Thus, \(f(x)=1.5\sin\left(\frac{\pi}{2}x\right)\). 3. In graph b), the maximum is \(2\), and the minimum is \(0\). Therefore, the midline is \(d=1\), and the amplitude is \(a=1\). 4. The distance between consecutive maxima is \(2\), so \(b=\frac{2\pi}{2}=\pi\). Thus, \(g(x)=\cos(\pi x)+1\).

a) \(f(x)=1.5\sin\left(\frac{\pi}{2}x\right)\) b) \(g(x)=\cos(\pi x)+1\)

- Average the maximum and minimum values to find each midline. - The amplitude is half the distance between the maximum and minimum. - Use the distance between corresponding points on consecutive cycles to find the period. - For the cosine form, decide whether the graph begins at a maximum or a minimum when \(x=0\).

1. For \(f\), the maximum is \(3\), and the minimum is \(-1\). Thus, the midline is \(d=\frac{3+(-1)}{2}=1\), and the amplitude is \(a=2\). 2. The increasing midline crossings at \(x=0\) and \(x=4\) show that the period is \(4\). Therefore, \(b=\frac{2\pi}{4}=\frac{\pi}{2}\), so \(f(x)=2\sin\left(\frac{\pi}{2}x\right)+1\). 3. For \(g\), the maximum is \(1.5\), and the minimum is \(-1.5\). Thus, the midline is \(d=0\), and the amplitude is \(1.5\). 4. The graph has a minimum at \(x=0\), so the cosine coefficient is negative. Consecutive minima are \(2\) units apart, so \(b=\pi\). Therefore, \(g(x)=-1.5\cos(\pi x)\).

\(f(x)=2\sin\left(\frac{\pi}{2}x\right)+1\) \(g(x)=-1.5\cos(\pi x)\)

- Use the midline and extrema to find amplitude and vertical shift. - Find the period from corresponding points on consecutive cycles. - Use \(b=\frac{2\pi}{p}\). - An increasing midline crossing identifies the horizontal shift in a positive sine model.

1. For \(g\), the midline is \(y=0\), and the extrema are \(1.5\) units from the midline, so the amplitude is \(1.5\). 2. One complete cycle has length \(4\), so the period is \(4\), and \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). Thus, \(g(x)=1.5\sin\left(\frac{\pi}{2}x\right)\). 3. For \(h\), the maximum is \(1.5\), and the minimum is \(-0.5\). Thus, the midline is \(d=0.5\), and the amplitude is \(1\). 4. Consecutive minima at \(x=0\) and \(x=4\) show that the period is \(4\), so \(b=\frac{\pi}{2}\). 5. The graph crosses its midline while increasing at \(x=1\), so the smallest positive shift to the right is \(c=1\). Therefore, \(h(x)=\sin\left(\frac{\pi}{2}(x-1)\right)+0.5\).

a) Amplitude \(1.5\), period \(4\), \(b=\frac{\pi}{2}\); \(g(x)=1.5\sin\left(\frac{\pi}{2}x\right)\) b) Amplitude \(1\), period \(4\), \(b=\frac{\pi}{2}\), \(c=1\), \(d=0.5\); \(h(x)=\sin\left(\frac{\pi}{2}(x-1)\right)+0.5\)

- Use the highest and lowest values to find the amplitude. - Find the horizontal distance between consecutive maxima to determine the period. - Use the cosine period formula to find \(b\). - Each x-axis grid interval represents \(\frac{\pi}{2}\). - Substitute each marked x-value into your equation to find its exact y-value.

1. The graph ranges from \(-2\) to \(2\), so the amplitude is \(a=2\). 2. Consecutive maxima occur at \(x=0\) and \(x=6\pi\), so the period is \(6\pi\). 3. Since \(p=\frac{2\pi}{b}\), \(b=\frac{2\pi}{6\pi}=\frac{1}{3}\). Therefore, \(f(x)=2\cos\left(\frac{x}{3}\right)\). 4. At \(x=-\pi\), \(f(-\pi)=2\cos\left(-\frac{\pi}{3}\right)=1\), so \(A=(-\pi,1)\). 5. At the maximum, \(B=(0,2)\). 6. At the next zero shown, \(C=\left(\frac{3\pi}{2}, 0\right)\). 7. At the minimum, \(D=(3\pi,-2)\).

a) Amplitude \(2\), period \(6\pi\); \(f(x)=2\cos\left(\frac{x}{3}\right)\) b) \(A=(-\pi,1)\), \(B=(0,2)\), \(C=\left(\frac{3\pi}{2}, 0\right)\), \(D=(3\pi,-2)\)

- Use the highest and lowest points to find the midline and amplitude. - Find the distance between consecutive maxima to determine the period. - Use \(b=\frac{2\pi}{p}\). - Locate an increasing midline crossing to determine \(c\).

1. The maximum is \(3\), and the minimum is \(-1\). Thus, the midline is \(d=\frac{3+(-1)}{2}=1\), and the amplitude is \(a=2\). 2. Consecutive maxima occur at \(x=2\) and \(x=6\), so the period is \(p=4\). 3. Therefore, \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 4. The graph crosses the midline while increasing at \(x=1\), so choose \(c=1\), which satisfies \(0 \leq c < 4\). 5. The complete equation is \(f(x)=2\sin\left(\frac{\pi}{2}(x-1)\right)+1\).

a) \(a=2\), \(d=1\) b) \(p=4\), \(b=\frac{\pi}{2}\) c) \(c=1\); \(f(x)=2\sin\left(\frac{\pi}{2}(x-1)\right)+1\)

- Read amplitude, period, and vertical shift from each graph. - An outside constant determines the midline. - The outside coefficient determines the amplitude. - Use the period to identify the coefficient of \(x\). - Check whether the graph starts on the midline or at an extreme value when \(x=0\).

1. Graph I has midline \(y=-1\), amplitude \(1\), and period \(2\). It crosses the midline at \(x=0\) with positive slope, so it matches \(C(x)=\sin(\pi x)-1\). 2. Graph II has midline \(y=0\), amplitude \(1.5\), and period \(4\). It passes through the origin with positive slope, so it matches \(B(x)=1.5\sin(0.5\pi x)\). 3. Graph III has midline \(y=0.5\), amplitude \(1\), and period \(4\). It begins at a maximum when \(x=0\), so it matches \(A(x)=\cos(0.5\pi x)+0.5\).

Graph I: \(C(x)\) Graph II: \(B(x)\) Graph III: \(A(x)\)

- Check the scale on both axes before reading the points. - Identify the maximum, minimum, and x-intercepts. - The amplitude is the greatest distance from the x-axis. - Find the horizontal length of one complete cycle. - Use \(b=\frac{2\pi}{p}\).

1. The marked points are \(A=(1.5, 1.5)\), \(B=(3, 0)\), \(C=(4.5, -1.5)\), and \(D=(6, 0)\). 2. The maximum distance from the x-axis is \(1.5\), so the amplitude is \(1.5\). 3. The graph begins an increasing cycle at \(x=0\) and returns to the same phase at \(x=6\), so the period is \(6\). 4. Therefore, \(b=\frac{2\pi}{6}=\frac{\pi}{3}\), and \(f(x)=1.5\sin\left(\frac{\pi}{3}x\right)\).

a) \(A=(1.5, 1.5)\), \(B=(3, 0)\), \(C=(4.5, -1.5)\), \(D=(6, 0)\) b) Amplitude \(1.5\), period \(6\) c) \(b=\frac{\pi}{3}\); \(f(x)=1.5\sin\left(\frac{\pi}{3}x\right)\)

- Use the y-value at \(x=0\) to check the cosine form and sign of the amplitude. - Measure the distance between consecutive maxima to find the period. - Use \(T=\frac{2\pi}{|b|}\).

1. The maximum is \(0.5\) and the minimum is \(-0.5\), so the amplitude is \(|a|=0.5\). Because the graph begins at a maximum, \(a=0.5\). 2. Consecutive maxima occur at \(x=0\) and \(x=\pi\), so the period is \(T=\pi\). 3. Since \(T=\frac{2\pi}{|b|}\), \(|b|=\frac{2\pi}{\pi}=2\). Since \(b>0\), \(b=2\), and the equation is \(g(x)=0.5\cos(2x)\).

\(a=0.5\), \(b=2\), so \(g(x)=0.5\cos(2x)\).

- Use the direction of the graph at the origin to determine the sign of \(a\). - Find the horizontal length of one complete cycle. - Account for the x-axis scale when calculating \(b\).

1. The graph ranges from \(-1\) to \(1\), so \(|a|=1\). 2. The graph falls through the origin, so it is a reflection of a positive sine graph across the x-axis. Therefore, \(a=-1\). 3. Consecutive downward x-axis crossings are \(4\pi\) apart, so the period is \(4\pi\). 4. Thus, \(b=\frac{2\pi}{4\pi}=0.5\), and \(k(x)=-\sin(0.5x)\).

\(k(x)=-\sin(0.5x)\)

- Read the amplitude from the highest point. - The period is the horizontal distance over which the pattern first repeats. - Use \(b=\frac{2\pi}{p}\). - Compare both the height and width of the waves.

1. The maximum value of \(g\) is \(2\), so \(a=2\). 2. The dashed graph completes one cycle in \(6\) units, so \(p_f=6\). The solid graph completes one cycle in \(12\) units, so \(p_g=12\). 3. Use \(b=\frac{2\pi}{p}\): \(b_f=\frac{2\pi}{6}=\frac{\pi}{3}\), and \(b_g=\frac{2\pi}{12}=\frac{\pi}{6}\). 4. The amplitude doubles, giving a vertical stretch by \(2\), and the period doubles, giving a horizontal stretch by \(2\).

1. \(a=2\) 2. \(p_f=6\); \(p_g=12\) 3. \(b_f=\frac{\pi}{3}\); \(b_g=\frac{\pi}{6}\) 4. Vertical stretch by \(2\) and horizontal stretch by \(2\)

- Read the amplitude from the greatest distance between each graph and the x-axis. - Find the horizontal length of one complete cycle. - A larger inside coefficient produces a shorter period.

1. Graph A has amplitude \(2\) and period \(4\). Thus, \(b=\frac{2\pi}{4}=0.5\pi\), so it matches equation (1). 2. Graph B has amplitude \(2\) and period \(2\). Thus, \(b=\pi\), so it matches equation (3). 3. Graph C has amplitude \(1\) and period \(2\). Thus, \(b=\pi\), so it matches equation (2). 4. Equation (4) has amplitude \(2\) and period \(8\), so it is not used.

A: (1) B: (3) C: (2) Unused: (4)

- Compare maximum and minimum values. - Determine the period of each graph. - Check for a vertical shift. - Compare whether the graph initially rises or falls to identify a reflection.

1. The reference function has amplitude \(1\) and period \(4\). 2. Function \(f\) has amplitude \(2\), so it is graph B, a vertical stretch by \(2\). 3. Function \(g\) has the same shape shifted down \(1\), so it is graph A. 4. Function \(h\) has period \(2\), half the reference period, so it is graph D, a horizontal compression by \(\frac{1}{2}\). 5. Function \(i\) is the reflection of the reference graph across the x-axis, so it is graph C.

1. B — vertical stretch by \(2\) 2. A — shift down \(1\) 3. D — horizontal compression by \(\frac{1}{2}\) 4. C — reflection across the x-axis

- Identify the amplitude and vertical shift first. - A horizontal stretch by \(4\) makes the coefficient of \(x\) equal to \(\frac{1}{4}\). - Compare each graph’s midline, amplitude, and period.

1. The vertical compression gives an amplitude of \(0.5\). A horizontal stretch by \(4\) replaces \(x\) with \(\frac{x}{4}\). The upward shift adds \(1\). Thus, \(g(x)=0.5\sin\left(\frac{x}{4}\right)+1\). 2. Graph II has amplitude \(2\), so it does not match. Graph III has amplitude \(0.5\) and midline \(y=1\), but its period is \(2\pi\). Graph I has amplitude \(0.5\), midline \(y=1\), and period \(8\pi\), so Graph I is correct.

\(g(x)=0.5\sin\left(\frac{x}{4}\right)+1\) Matching graph: I

- Reflection across the x-axis gives an outside negative sign. - A horizontal compression makes the graph cycle faster. - Substitute \(x=\frac{\pi}{4}\) into the function. - Use the graph as a visual check.

1. Reflection gives an outside coefficient of \(-1\). 2. A horizontal compression by \(0.5\) gives an inside coefficient of \(2\). 3. Shifting down \(2\) gives \(g(x)=-\cos(2x)-2\). 4. Evaluate: \(g\left(\frac{\pi}{4}\right)=-\cos\left(\frac{\pi}{2}\right)-2=-2\).

\(g(x)=-\cos(2x)-2\); \(g\left(\frac{\pi}{4}\right)=-2\)

- Use the highest and lowest values to find the midline and amplitude. - The distance from a maximum to the next minimum is half a period. - Use \(b=\frac{2\pi}{p}\).

1. The maximum is \(1.5\), and the minimum is \(0.5\). Thus, the midline is \(d=\frac{1.5+0.5}{2}=1\). 2. The amplitude is \(a=1.5-1=0.5\). 3. The distance from the maximum at \(x=2.5\) to the minimum at \(x=7.5\) is \(5\), which is half a period. Therefore, the period is \(10\). 4. Thus, \(b=\frac{2\pi}{10}=\frac{\pi}{5}\).

\(a=0.5\), \(b=\frac{\pi}{5}\), \(d=1\)

- The amplitude is the absolute value of the coefficient outside the sine or cosine. - Use \(T=\frac{2\pi}{|b|}\) for the period. - Find the lowest and highest values by combining the midline and the amplitude. - Check the midline and one full cycle on each graph.

1. For \(f(x)=2\sin(3x)-1\), the amplitude is \(|2|=2\). 2. Its period is \(\frac{2\pi}{3}\), because \(T=\frac{2\pi}{|b|}\) and \(b=3\). 3. Its midline is \(y=-1\), so its range is \([-1-2, -1+2]=[-3, 1]\). 4. For \(g(x)=0.5\cos(0.5x)+2\), the amplitude is \(|0.5|=0.5\). 5. Its period is \(\frac{2\pi}{0.5}=4\pi\). 6. Its midline is \(y=2\), so its range is \([2-0.5, 2+0.5]=[1.5, 2.5]\).

\(f\): amplitude \(2\), period \(\frac{2\pi}{3}\), range \([-3, 1]\) \(g\): amplitude \(0.5\), period \(4\pi\), range \([1.5, 2.5]\)

- Match the amplitude and vertical shift to \(a\) and \(d\). - Use the period formula to determine \(b\). - The range extends one amplitude below and above the midline. - Check the extrema, midline, and distance between corresponding points on consecutive cycles.

1. The amplitude gives \(a=1.5\), and the vertical shift gives \(d=0.5\). 2. Since \(T=\frac{2\pi}{|b|}\), the condition \(T=\pi\) gives \(|b|=2\). Taking \(b=2\), an equation is \(h(x)=1.5\sin(2x)+0.5\). 3. The minimum is \(0.5-1.5=-1\), and the maximum is \(0.5+1.5=2\). Therefore, the range is \([-1, 2]\). 4. The graph oscillates about \(y=0.5\), reaches \(-1\) and \(2\), and repeats every \(\pi\), confirming the equation.

1. \(h(x)=1.5\sin(2x)+0.5\) 2. \([-1, 2]\) 3. The graph has midline \(y=0.5\), extrema \(-1\) and \(2\), and period \(\pi\).

- The amplitude is the greatest distance from the midline. - A graph with a shorter period has a higher frequency. - For \(p\), find the period and check whether the graph crosses the midline at \(x=0\).

1. Graph \(p\) has amplitude \(1.5\). Graph \(q\) oscillates about \(y=2\) from \(1\) to \(3\), so its amplitude is \(1\). Therefore, \(p\) has the greater amplitude. 2. Graph \(p\) has period \(2\pi\), while graph \(q\) has period \(\pi\). Therefore, \(q\) has the higher frequency. 3. For \(p\), the midline is \(y=0\), the amplitude is \(1.5\), and \(b=\frac{2\pi}{2\pi}=1\). Thus, \(p(x)=1.5\sin(x)\).

1. Graph \(p\) 2. Graph \(q\) 3. \(p(x)=1.5\sin(x)\)

- Read the amplitude from the greatest distance from the x-axis. - The period is the horizontal distance between consecutive maximum points. - Substitute the marked x-values to confirm the y-coordinates. - Pay attention to the scale on each axis.

1. The coefficient outside cosine is \(2\), so the amplitude is \(2\). 2. The coefficient of \(x\) is \(b=\frac{\pi}{4}\). Thus, the period is \(p=\frac{2\pi}{\pi/4}=8\). 3. \(g(0)=2\cos(0)=2\), so \(P(0, 2)\). 4. \(g(2)=2\cos\left(\frac{\pi}{2}\right)=0\), so \(Q(2, 0)\). 5. \(g(4)=2\cos(\pi)=-2\), so \(R(4, -2)\).

a) Amplitude: \(2\); period: \(8\) b) \(P(0, 2)\), \(Q(2, 0)\), \(R(4, -2)\)

- Average the maximum and minimum to find the midline. - Find the period from the distance between maxima. - Use an increasing midline crossing for the sine form. - Use a maximum for the cosine form.

1. The maximum is \(2.5\), and the minimum is \(-0.5\). Thus, the midline is \(d=\frac{2.5+(-0.5)}{2}=1\), and the amplitude is \(a=1.5\). 2. Consecutive maxima occur at \(x=1\) and \(x=5\), so the period is \(4\), and \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 3. The graph crosses the midline while increasing at \(x=0\), so a sine equation is \(f(x)=1.5\sin\left(\frac{\pi}{2}x\right)+1\). 4. The graph has a maximum at \(x=1\), so a cosine equation is \(g(x)=1.5\cos\left(\frac{\pi}{2}(x-1)\right)+1\).

Sine form: \(f(x)=1.5\sin\left(\frac{\pi}{2}x\right)+1\) Cosine form: \(g(x)=1.5\cos\left(\frac{\pi}{2}(x-1)\right)+1\)

- Use the grid carefully to identify fractional horizontal shifts. - Each x-axis grid interval represents \(0.25\) unit. - The period is the horizontal length of one complete cycle.

1. The maximum is \(1\), and the minimum is \(-3\). Thus, the midline is \(d=-1\), and the amplitude is \(a=2\). 2. Increasing midline crossings occur at \(x=0.25\) and \(x=2.25\), so the period is \(2\). 3. Therefore, \(b=\frac{2\pi}{2}=\pi\). 4. The increasing midline crossing at \(x=0.25\) gives \(c=0.25\). 5. Thus, \(f(x)=2\sin\left(\pi\left(x-\frac{1}{4}\right)\right)-1\).

\(f(x)=2\sin\left(\pi\left(x-\frac{1}{4}\right)\right)-1\)

- Use the graph's midline to determine the vertical shift. - Use the distance from the midline to a maximum to determine the amplitude. - Compare the location of a maximum with the maximum of \(\cos(x)\) at \(x=0\).

1. The graph has midline \(y=1\), so the cosine graph is shifted up \(1\) unit. 2. The amplitude is \(0.5\), so the graph is vertically compressed by a factor of \(0.5\). 3. A maximum occurs at \(x=\frac{\pi}{2}\) instead of at \(x=0\), so the graph is shifted right \(\frac{\pi}{2}\). 4. Therefore, \(g(x)=0.5\cos\left(x-\frac{\pi}{2}\right)+1\). Equivalently, \(g(x)=0.5\sin(x)+1\).

Vertically compress by a factor of \(0.5\), shift right \(\frac{\pi}{2}\), and shift up \(1\). One equation is \(g(x)=0.5\cos\left(x-\frac{\pi}{2}\right)+1\).

- Examine the outside coefficient, the coefficient inside the sine, the expression in parentheses, and the final constant separately. - Decide whether each parameter causes a stretch, compression, reflection, or shift. - Use the amplitude and period formulas. - Pay close attention to the sign in \(x-1\).

1. The outside coefficient \(2\) vertically stretches the graph by a factor of \(2\), so the amplitude is \(2\). 2. The inside coefficient is \(\frac{\pi}{2}\), so the period is \(\frac{2\pi}{\pi/2}=4\). This is a horizontal compression by a factor of \(\frac{2}{\pi}\). 3. The expression \(x-1\) shifts the graph right \(1\) unit. 4. The final \(+1\) shifts the graph up \(1\) unit.

Vertically stretch by a factor of \(2\), horizontally compress by a factor of \(\frac{2}{\pi}\), shift right \(1\) unit, and shift up \(1\) unit.

- Determine the effects of the sign and magnitude of the outside coefficient separately. - Use the inside coefficient to find the new period. - Compare the new midline with the x-axis. - Check each change against the two graphs.

1. The outside coefficient \(-0.5\) reflects the graph across the x-axis and vertically compresses it by a factor of \(0.5\). The amplitude is \(0.5\). 2. The inside coefficient \(\pi\) changes the period to \(T=\frac{2\pi}{\pi}=2\). This is a horizontal compression by a factor of \(\frac{1}{\pi}\). 3. The final \(-2\) shifts the graph down \(2\) units.

Reflect across the x-axis, vertically compress by a factor of \(0.5\), horizontally compress by a factor of \(\frac{1}{\pi}\), and shift down \(2\) units.

- Find each amplitude from half the distance between the maximum and minimum. - Identify each midline. - Use the period to calculate the inside coefficient. - Locate an increasing midline crossing to determine a horizontal shift.

1. Graph \(g\) has amplitude \(2\), no vertical shift, and period \(4\pi\). Thus, \(b=\frac{2\pi}{4\pi}=0.5\), so \(g(x)=2\sin(0.5x)\). 2. Graph \(h\) has amplitude \(1\), midline \(y=1\), and period \(2\pi\). It crosses the midline while increasing at \(x=\frac{\pi}{2}\), so it is shifted right \(\frac{\pi}{2}\). 3. Therefore, \(h(x)=\sin\left(x-\frac{\pi}{2}\right)+1\).

\(g(x)=2\sin(0.5x)\) \(h(x)=\sin\left(x-\frac{\pi}{2}\right)+1\)

- Compare the amplitudes of the two graphs. - Identify each midline. - Compare the x-coordinates of corresponding maximum points.

1. The amplitude of \(f\) is \(1\), while the amplitude of \(g\) is \(2\). Therefore, the graph is stretched vertically by a factor of \(2\). 2. The midline moves from \(y=0\) to \(y=1\), so the graph shifts up \(1\) unit. 3. A maximum of \(f\) occurs at \((1, 1)\), while the corresponding maximum of \(g\) occurs at \((2, 3)\). Therefore, the graph shifts right \(1\) unit. 4. Thus, \(g(x)=2f(x-1)+1=2\sin\left(\frac{\pi}{2}(x-1)\right)+1\).

\(g(x)=2\sin\left(\frac{\pi}{2}(x-1)\right)+1\)

- Compare the distances between consecutive maximum points. - Determine how the periods of the two graphs are related. - A larger coefficient on \(x\) produces a shorter period.

1. The period of \(f\) is \(2\pi\), while the period of \(g\) is \(\pi\). 2. The period is halved, so the graph is horizontally compressed by a factor of \(\frac{1}{2}\). 3. Replacing \(x\) with \(2x\) produces this compression. Therefore, \(g(x)=f(2x)\).

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

- Use the relationship between \(k\) and the period. - Compare where each graph completes one full cycle. - Zeros can help identify half of a period.

1. The period of \(\sin(kx)\) is \(P=\frac{2\pi}{k}\). 2. Graph p has period \(2\pi\), so \(k=1\). 3. Graph q has period \(\pi\), so \(k=2\). 4. Graph r has period \(4\pi\), so \(k=0.5\).

p) \(k=1\) q) \(k=2\) r) \(k=0.5\)

- Compare how many cycles each graph completes over a fixed interval. - Use the period formula \(\frac{2\pi}{k}\). - The first positive zero of the standard sine function occurs when its input is \(\pi\).

1. Every graph passes through \((0, 0)\) because \(\sin(k\cdot 0)=0\). 2. A larger \(k\) gives a shorter period. Graph I varies most slowly, so \(k=0.5\). Graph II is the standard sine graph, so \(k=1\). Graph III varies most rapidly, so \(k=2\). 3. The period is \(\frac{2\pi}{k}\). Doubling \(k\) halves the period. 4. For the smallest positive zero, \(1.5x=\pi\), so \(x=\frac{2\pi}{3}\).

a) All pass through \((0, 0)\). b) I: \(k=0.5\); II: \(k=1\); III: \(k=2\) c) The period is halved. d) \(x=\frac{2\pi}{3}\)

- Interpret the parameters in \(a\sin(b(x-c))\). - Factor the coefficient of \(x\) before reading the horizontal shift. - Relate the inside coefficient to the period. - Shifts that differ by a whole period produce the same graph.

1. Both functions have amplitude \(1.5\) and period \(T=\frac{2\pi}{3}\). 2. Rewrite \(h(x)=1.5\sin\left(3\left(x-\frac{\pi}{3}\right)\right)\), so one shift is \(\frac{\pi}{3}\) right. The form of \(k\) displays a shift of \(\pi\) right. 3. The displayed shifts differ by \(\pi-\frac{\pi}{3}=\frac{2\pi}{3}\), exactly one period. Also, \(k(x)=1.5\sin(3x-3\pi)=1.5\sin(3x-\pi-2\pi)=h(x)\). Therefore, the graphs are identical.

1. Amplitude: \(1.5\); period: \(\frac{2\pi}{3}\) 2. \(h\): shift right \(\frac{\pi}{3}\); \(k\): shift right \(\pi\) 3. The graphs are identical because the shifts differ by one full period. Neither graph is actually farther right.

- Recall the input where the parent sine function first reaches its maximum. - Set the entire sine input equal to that value. - Solve each equation for \(x\). - Compare the two x-coordinates.

1. A sine function reaches a maximum when its input is \(\frac{\pi}{2}+2\pi k\). 2. For \(f\), solve \(\frac{1}{2}x-\frac{\pi}{4}=\frac{\pi}{2}\). This gives \(x=\frac{3\pi}{2}\). 3. For \(g\), solve \(\frac{1}{2}\left(x-\frac{\pi}{4}\right)=\frac{\pi}{2}\). This gives \(x=\frac{5\pi}{4}\). 4. Since \(\frac{5\pi}{4}<\frac{3\pi}{2}\), \(g\) reaches its first maximum earlier. Both maximum values are \(2\).

\(f\) has its first maximum at \(\left(\frac{3\pi}{2}, 2\right)\). \(g\) has its first maximum at \(\left(\frac{5\pi}{4}, 2\right)\). The graph of \(g\) reaches its first maximum earlier.