- The amplitude is the distance from the midline to a maximum or minimum. - Look at the coefficient multiplying cosine. - Look at the constant added outside the cosine function to locate the midline.

1. The coefficient of cosine is \(3\), so the amplitude is \(|3|=3\). 2. The vertical shift is \(-2\), so the midline is \(y=-2\).

Amplitude: \(3\); midline: \(y=-2\)

- Think about how far the tangent graph moves horizontally before its branch pattern repeats. - Tangent repeats after a smaller rotation than sine or cosine. - State the smallest positive horizontal repeat length.

1. Tangent repeats its values after an input increase of \(\pi\), since \(\tan(x+\pi)=\tan(x)\). Therefore, its period is \(\pi\).

\(\pi\)

- Recall the basic features of the untransformed sine graph. - Think about the horizontal length of one complete cycle.

1. The parent sine function has vertical scale factor \(1\), so its amplitude is \(1\). 2. One complete sine cycle has length \(2\pi\), so its period is \(2\pi\).

Amplitude: \(1\); period: \(2\pi\)

- Compare the domains of the parent reciprocal trigonometric functions. - Use the stated value at \(x=0\) to distinguish between possible reciprocal functions.

1. The undefined inputs are exactly the zeros of cosine. 2. The reciprocal of cosine is secant, and \(\sec(0)=1\). 3. Therefore, the function is \(y=\sec(x)\).

\(y=\sec(x)\)

- Think about which quotient trigonometric function has sine in its denominator. - Compare where the parent tangent and cotangent functions are undefined and where they cross the x-axis.

1. Cotangent is \(\frac{\cos(x)}{\sin(x)}\), so it is undefined where \(\sin(x)=0\), at \(x=k\pi\). 2. It is zero where \(\cos(x)=0\), at \(x=\frac{\pi}{2}+k\pi\). 3. Therefore, the function is \(y=\cot(x)\).

\(y=\cot(x)\)

- Find the highest and lowest y-values of the curve. - Locate the height halfway between them. - Amplitude is the vertical distance from that halfway height to either extreme.

1. The curve's highest point has y-value \(1\), and its lowest point has y-value \(-1\). 2. The midline is \(y=0\), so the distance from the midline to a peak is \(1\). The amplitude is \(1\).

\(1\)

- Look for the highest and lowest parts of the curve. - The midline lies halfway between those levels. - It is a horizontal line, so report its y-value.

1. The maximum value is \(3\), and the minimum is \(-1\). 2. The midline is halfway between these heights: \(\frac{3+(-1)}{2}=1\).

\(y=1\)

- Look for the positive x-value where the curve has two separated branches. - What happens to the y-values on either side as the curve approaches that position? - Use the horizontal scale to locate the position exactly.

1. The middle branch rises without bound as \(x\) approaches the gap from the left, while the next branch falls without bound as \(x\) approaches from the right. 2. This happens at the tick \(x=\frac{\pi}{2}\), so that line is the nearest positive vertical asymptote.

\(x=\frac{\pi}{2}\). The curve rises without bound on one side of the line and falls without bound on the other.

- A complete repeat returns the curve to the same height and direction of motion. - Would a peak-to-valley gap be a full repeat or only part of one? - Check your proposed period against another repeated feature on the curve.

1. The two consecutive peaks occur at \(x=1\) and \(x=5\). 2. Their horizontal distance is \(5-1=4\), so the period is \(4\).

\(4\)

- Shifting a wave sideways does not change the heights of its peaks and valleys. - Could a shift that matches the valleys also match the peaks? - A full-cycle shift leaves a wave unchanged, but the question asks for the smallest positive shift.

1. A peak of \(f\) is at \(x=0\), while the corresponding peak of \(g\) is at \(x=2\). 2. The peaks are \(2\) units apart to the right. Thus the shift is right \(2\) units.

Right \(2\) units

- 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 controlling the vertical stretch. - Relate the coefficient of \(x\) to the period. - Start with the parent sine range and apply the vertical scale. - Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\) before naming the graph symmetry.

1. The amplitude is \(|2.5|=2.5\). 2. The period is \(p=\frac{2\pi}{0.5}=4\pi\). 3. Since the parent sine graph ranges from \(-1\) to \(1\), the range of \(f\) is \([-2.5,2.5]\). 4. Since \(f(-x)=2.5\sin(-0.5x)=-2.5\sin(0.5x)=-f(x)\), the function is odd. 5. Therefore, its graph has rotational symmetry about the origin.

Amplitude: \(2.5\) Period: \(4\pi\) Range: \([-2.5,2.5]\) \(f\) is odd, so the graph has 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, \(f(x)=2\sin(2x)\).

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

- Use the y-value at \(x=0\) to check the sign of the cosine coefficient. - Measure the distance between consecutive maxima to find the period. - Relate the period to the coefficient of \(x\).

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

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

- 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 \(D=(\pi,-1)\)

- The parameter \(a\) controls the vertical distance from the midline to an extreme. - 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 \(D\left(\frac{3\pi}{2},-2\right)\).

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

- Use the highest and lowest values to find the midline and the positive amplitude. - The distance from a maximum to the next minimum is half a period. - Once the period is known, relate it to the positive coefficient \(b\).

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. Since \(a>0\), the amplitude gives \(a=1.5-1=0.5\). 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. Since \(b>0\), \(b=\frac{2\pi}{10}=\frac{\pi}{5}\).

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

- Compare the highest and lowest y-values to locate the common midline. - Measure the horizontal distance between consecutive maxima to determine the common period. - Relate the period of \(\cos(bx)\) to the coefficient \(b\). - Once the common shift is known, examine the curves at \(x=0\) to determine their different vertical scales.

1. All four curves are centered on \(y=0\), so the common vertical shift is \(d=0\). 2. Consecutive maxima are \(2\pi\) units apart, so the common period is \(2\pi\). Since the period of \(\cos(bx)\) is \(\frac{2\pi}{b}\), this gives \(b=1\). 3. With \(b=1\) and \(d=0\), each curve has \(g(0)=a\cos(0)=a\). The four y-intercepts are \(0.5\), \(1\), \(2\), and \(3\), so these are the four values of \(a\). 4. Therefore, the family is \(g_a(x)=a\cos(x)\), with \(a\in\{0.5,1,2,3\}\) for the displayed members.

a) \(b=1\), \(d=0\) b) \(a\in\{0.5,1,2,3\}\) c) \(g_a(x)=a\cos(x)\)

- 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: \(r(x)\)

- Compare one branch with the next branch to find the horizontal repeat length. - Look for x-intercepts that repeat at regular intervals. - A vertical asymptote is an x-value the curve approaches while its y-values grow without bound.

1. The tangent graph repeats after a horizontal shift of \(\pi\), so its period is \(\pi\). 2. Tangent is zero at integer multiples of \(\pi\). In \([-\pi,\pi]\), the zeros are \(-\pi\), \(0\), and \(\pi\). 3. The central branch approaches vertical lines halfway between consecutive zeros, so its bordering vertical asymptotes are \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\).

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

- Compare two consecutive decreasing branches to find the repeat length. - Zeros occur where the graph crosses the x-axis. - Vertical asymptotes separate consecutive branches and occur where the function is undefined.

1. The cotangent pattern repeats every \(\pi\), so its period is \(\pi\). 2. Cotangent is zero where cosine is zero while sine is nonzero. In \((0,2\pi)\), this occurs at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). 3. Cotangent is undefined where sine is zero. In \([0,2\pi]\), the vertical asymptotes are \(x=0\), \(x=\pi\), and \(x=2\pi\).

a) \(\pi\) b) \(x=\frac{\pi}{2}\), \(x=\frac{3\pi}{2}\) c) \(x=0\), \(x=\pi\), \(x=2\pi\)

- Compare what happens at \(x=0\) in the two panels. - Think about where cosine equals \(1\) and where sine equals \(0\). - For the shared range, consider the possible magnitudes of reciprocal sine and cosine values.

1. Panel a has a vertical asymptote at \(x=0\) and a positive vertex at \(x=\frac{\pi}{2}\), matching the reciprocal of sine. Thus panel a is \(y=\csc(x)\). 2. Panel b has a vertex at \((0,1)\), where cosine equals \(1\), so panel b is \(y=\sec(x)\). 3. Both parent functions repeat every \(2\pi\). 4. Since the magnitude of a reciprocal of a sine or cosine value cannot be less than \(1\), both ranges are \(( -\infty,-1]\cup[1,\infty)\).

a) Panel a: \(y=\csc(x)\); panel b: \(y=\sec(x)\) b) For example, cosecant has a vertical asymptote at \(x=0\), while secant has a vertex at \((0,1)\). c) Period: \(2\pi\); range: \(( -\infty,-1]\cup[1,\infty)\)

- First find the angle at the marked point on the graph's horizontal axis. - Recall the exact sine and cosine values for that standard angle. - What does the height of a sine graph represent on the unit circle?

1. The point is above the tick labeled \(\frac{\pi}{3}\), so \(\theta=\frac{\pi}{3}\). 2. At this angle, cosine is \(\frac{1}{2}\) and sine is \(\frac{\sqrt{3}}{2}\). The unit-circle point is \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\). 3. A sine graph plots the unit-circle y-coordinate against the angle, so both points have the same y-coordinate, \(\frac{\sqrt{3}}{2}\).

a) \(\theta=\frac{\pi}{3}\) b) \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\) c) Both y-coordinates equal \(\sin(\theta)=\frac{\sqrt{3}}{2}\).

- Count complete waves between the left and right edges. - A shorter period means more cycles over the same horizontal distance. - For each curve, estimate the halfway height between the peaks and valleys.

1. Curve \(g\) completes two cycles while \(f\) completes one over the same interval, so \(g\) has the shorter period. 2. Curve \(f\) oscillates around \(y=0\); curve \(g\) oscillates around \(y=-1\). Therefore, \(f\) has the higher midline.

a) \(g\), because it completes more cycles over the same interval. b) \(f\), whose midline is \(y=0\), rather than \(y=-1\) for \(g\).

- Identify the curve's highest and lowest heights before checking the marked inputs. - At the middle input, is the curve at an extreme or halfway between its extremes? - Compare the heights at the first and last inputs before writing the ordered pairs.

1. At \(x=0\), the graph's y-value is \(3\). 2. At \(x=\frac{\pi}{2}\), it crosses its midline at \(y=1\). 3. At \(x=\pi\), it reaches a minimum at \(y=-1\). 4. The points are \((0,3)\), \(\left(\frac{\pi}{2},1\right)\), and \((\pi,-1)\).

\((0,3)\), \(\left(\frac{\pi}{2},1\right)\), \((\pi,-1)\)

- Read the x-positions where the graph's branches rise or fall sharply. - What makes a fraction undefined? - Use the unit circle to locate where cosine has that value.

1. The secant branches grow without bound near \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\), identifying the vertical asymptotes. 2. Cosine is zero at both angles. Since \(\sec(x)=\frac{1}{\cos(x)}\), division by zero makes secant undefined there.

\(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\); cosine is zero at each angle, so its reciprocal is undefined.

- 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 \((\pi, 0)\)

- On the unit circle, compare the x-coordinates of the endpoints for angles \(x\) and \(-x\). - 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 angles \(x\) and \(-x\), the corresponding unit-circle points are reflections of each other across the x-axis, so they have the same x-coordinate. Because cosine is that x-coordinate, the graph has the same height at opposite inputs and is symmetric about the y-axis. 2. The maxima and minima of the cosine graph occur at \(x=k\pi\), and each vertical line through one of these extrema 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) Opposite input angles have unit-circle endpoints with the same x-coordinate, so the cosine 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, \(-7\pi, -3\pi, 5\pi, 9\pi\)

- Compare one complete turn in degrees with the given period. - At a zero, which factor of the function must equal zero? - Think about what multiplication by a nonzero number does to a zero. - Include both interval endpoints when checking possible zeros.

1. The period in degrees is \(\frac{360^\circ}{|b|}\). Therefore \(|b|=3\), so \(b=3\) or \(b=-3\). 2. Both signs give the same zeros because \(\sin(-3x)=-\sin(3x)\). We can solve \(5\sin(3x)=0\), equivalent to \(\sin(3x)=0\). 3. Sine is zero when \(3x=180^\circ k\), where \(k\) is an integer, so \(x=60^\circ k\). 4. In \([0^\circ,180^\circ]\), the zeros are \(0^\circ,60^\circ,120^\circ,180^\circ\). 5. Multiplication by a nonzero number changes the heights of nonzero values, but preserves all zeros.

a) \(x=0^\circ,60^\circ,120^\circ,180^\circ\) b) Multiplying by the nonzero factor \(5\) does not change where sine equals \(0\).

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

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 sine graph meets the x-axis at \(0\), \(\pi\), and \(2\pi\), while it reaches its maximum only once.

- First recall which output values cosine can take. - Look at one full cosine cycle, including its two endpoints. - Count how often a horizontal line at height \(k\) meets that cycle. - Treat the maximum and minimum values separately from interior heights.

1. The range of cosine is \([-1,1]\), so there are no solutions when \(|k|>1\). 2. The minimum value \(-1\) occurs once, at \(x=\pi\). 3. Every value strictly between \(-1\) and \(1\) occurs twice over one complete period. 4. The maximum value \(1\) occurs at both included 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 \(a\sin(b(x-c))+d\). - A coefficient of \(x\) greater than \(1\) inside the sine makes the graph cycle faster.

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 coefficient \(2\) of \(x\) 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 new period to find the coefficient of \(x\) in the cosine input. - Reflection across the x-axis changes the sign of the outside coefficient. - A left shift appears as addition within the cosine 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 has a maximum or a minimum at \(x=0\).

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, \(g(x)=-1.5\cos(\pi x)\).

\(f\): midline \(y=1\), amplitude \(2\), period \(4\); \(f(x)=2\sin\left(\frac{\pi}{2}x\right)+1\). \(g\): midline \(y=0\), amplitude \(1.5\), period \(2\); \(g(x)=-1.5\cos(\pi x)\).

- Use the midline and extrema to find amplitude and vertical shift. - Find the period from corresponding points on consecutive cycles. - Relate the coefficient of \(x\) to the period. - 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 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 \(2\)

- Compare corresponding zeros or maximum points of the two graphs. - A shift right appears as subtraction inside the function input. - Shift the location of the parent sine function's first maximum by the same amount.

1. The graph of \(g\) crosses the x-axis with positive slope at \(x=\frac{\pi}{2}\), so the parent sine graph has been shifted right by \(\frac{\pi}{2}\). 2. Therefore, \(g(x)=\sin\left(x-\frac{\pi}{2}\right)\), which is also \(-\cos(x)\). 3. The first maximum of \(\sin(x)\) occurs at \(x=\frac{\pi}{2}\). After the shift, it occurs at \(x=\pi\), with y-coordinate \(1\).

a) \(g(x)=\sin\left(x-\frac{\pi}{2}\right)\), equivalently \(g(x)=-\cos(x)\) b) \((\pi, 1)\)

- Read the amplitude from the greatest distance between each graph and the x-axis. - Find the horizontal length of one complete cycle. - A larger coefficient of \(x\) in the sine input 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 curve'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. Curve q has amplitude \(2\), so it does not match. Curve r has amplitude \(0.5\) and midline \(y=1\), but its period is \(2\pi\). Curve p has amplitude \(0.5\), midline \(y=1\), and period \(8\pi\), so curve p is correct.

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

- 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 a coefficient of \(2\) multiplying \(x\) inside the cosine. 3. Shifting down \(2\) gives \(g(x)=-\cos(2x)-2\). 4. Evaluate: \(g\left(\frac{\pi}{4}\right)=-\cos\left(\frac{\pi}{2}\right)-2=-2\). The plotted point at this x-coordinate has the same y-value.

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

- 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. \(p(x)=1.5\sin(x)\)

- Find each graph's midline, amplitude, and period. - Check for horizontal shifts and reflections. - Use identities that relate sine and cosine after a phase shift.

1. Graph (1) has midline \(y=1\), amplitude \(2\), period \(4\), and an increasing midline crossing at \(x=0\). Thus, equation A matches. 2. In equation B, \(0.5\pi(x-1)=0.5\pi x-\frac{\pi}{2}\), and \(\cos\left(\theta-\frac{\pi}{2}\right)=\sin(\theta)\). Therefore, B is equivalent to A and also matches graph (1). 3. Graph (2) has the same midline, amplitude, and period but is shifted right \(1\) unit. Thus, equation C matches. 4. Since \(-\cos(\theta)=\sin\left(\theta-\frac{\pi}{2}\right)\), equation D is equivalent to C and also matches graph (2).

(1) A and B; (2) C and D

- 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 maximum and minimum to find the midline and amplitude. - Use the grid carefully to identify the period and the increasing midline crossing. - The restriction on \(c\) selects one representative from shifts differing by a full period.

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 \(c=0.25=\frac{1}{4}\), within \(0\le c<2\). 5. Thus, \(f(x)=2\sin\left(\pi\left(x-\frac{1}{4}\right)\right)-1\).

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

- 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 coefficient outside sine, the coefficient of \(x\), 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 \(2\) stretches the graph vertically by a factor of \(2\), so the amplitude is \(2\). 2. The coefficient \(\frac{\pi}{2}\) of \(x\) changes the period to \(\frac{2\pi}{\pi/2}=4\). 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.

- What are the amplitude and period of the parent cosine wave? - Does a negative factor outside cosine change its height, its direction, or both? - Compare the new period with the parent period to describe the horizontal change. - Which part of the formula moves the middle height of the wave?

1. The outside coefficient \(-0.5\) reflects the graph across the x-axis and compresses it vertically by \(0.5\). The amplitude is \(0.5\). 2. The coefficient \(\pi\) of \(x\) 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.

a) 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. b) Amplitude: \(0.5\); period: \(2\); midline: \(y=-2\).

- Find each amplitude from half the distance between the maximum and minimum. - Identify each midline. - Use the period to find the coefficient of \(x\) in each sine input. - 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, \(h(x)=\sin\left(x-\frac{\pi}{2}\right)+1\).

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

- An odd function must pass through the origin. - Compare the graph with the parent sine function and its reflection. - Recall the phase-shift relationship between sine and cosine. - Express the minimum location near \(1.57\) as a multiple of \(\pi\).

1. A graph symmetric about the origin must be an odd function. The graph passes through the origin and decreases for small positive \(x\), so it matches \(-\sin(x)\). 2. Use \(\cos\left(x+\frac{\pi}{2}\right)=-\sin(x)\). 3. The first minimum of \(-\sin(x)\) occurs at \(x=\frac{\pi}{2}\), matching the graph. 4. Therefore, the least positive value is \(c=\frac{\pi}{2}\).

\(c=\frac{\pi}{2}\)

- 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 \(\frac{1}{2}\); \(g(x)=f(2x)\)

- Read the maximum distance of \(h\) from the x-axis. - Measure the horizontal length of one complete cycle of \(h\). - Compare the amplitude and period of \(h\) with those of the dashed parent graph. - Relate a shorter period to the coefficient multiplying \(x\).

1. The graph of \(h\) reaches \(0.75\) and \(-0.75\), so its amplitude is \(0.75\). 2. Consecutive corresponding points on \(h\) are \(\frac{2\pi}{3}\) units apart, so its period is \(\frac{2\pi}{3}\). 3. The parent sine graph has amplitude \(1\), so the vertical compression factor is \(0.75\). Its period is \(2\pi\), so the horizontal compression factor is \(\frac{1}{3}\). 4. A horizontal compression by \(\frac{1}{3}\) gives a coefficient of \(3\) multiplying \(x\) in sine. Therefore, \(h(x)=0.75\sin(3x)\).

a) Amplitude: \(0.75\); period: \(\frac{2\pi}{3}\) b) Vertical compression factor: \(0.75\); horizontal compression factor: \(\frac{1}{3}\) c) \(h(x)=0.75\sin(3x)\)

- Compare how many cycles each curve 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 \(\sin(k\cdot0)=0\). 2. A larger \(k\) gives a shorter period. Curve p varies most slowly, so \(k=0.5\). Curve q is the standard sine graph, so \(k=1\). Curve r 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) p: \(k=0.5\); q: \(k=1\); r: \(k=2\) c) The period is halved. d) \(x=\frac{2\pi}{3}\)

- Separate the effects of the outside coefficient, the expression inside tangent, and the final constant. - Track how the central tangent branch shifts horizontally. - The neighboring vertical asymptotes stay equally spaced from the central midline crossing. - Decide whether an unbounded graph can have a greatest distance from its midline.

1. The parent tangent period is \(\pi\). The coefficient of \(x\) is \(1\), so the period remains \(\pi\). The graph is shifted right \(\frac{\pi}{4}\), shifted down \(1\), and vertically stretched by a factor of \(2\). 2. The parent tangent asymptotes are \(\frac{\pi}{2}\) units to either side of its central zero. After shifting right \(\frac{\pi}{4}\), the asymptotes around the central branch are \(x=-\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\). 3. The parent tangent graph crosses its midline at its central zero. After the shifts, the crossing is \(\left(\frac{\pi}{4},-1\right)\). Tangent has no amplitude because its branches are unbounded rather than having a maximum and minimum.

a) Period: \(\pi\); shift right \(\frac{\pi}{4}\); shift down \(1\); vertical stretch factor \(2\) b) \(x=-\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\) c) \(\left(\frac{\pi}{4},-1\right)\); tangent has no amplitude because it is unbounded.

- Compare each solid reciprocal curve with the dashed base curve at the same x-values. - Look for x-values where the dashed curve crosses the x-axis and the solid curve breaks into separate branches. - For the range, compare each reciprocal branch with the horizontal levels \(y=1\) and \(y=-1\).

1. For \(s(x)=\sec(x)=\frac{1}{\cos(x)}\), vertical asymptotes occur where \(\cos(x)=0\). In the shown interval \([-\pi,\pi]\), these are \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\). Since \(-1\le\cos(x)\le1\), secant cannot have magnitude less than \(1\), so its range is \(( -\infty,-1]\cup[1,\infty)\). 2. For \(q(x)=\csc(x)=\frac{1}{\sin(x)}\), vertical asymptotes occur where \(\sin(x)=0\). In the shown interval \([0,2\pi]\), these are \(x=0\), \(x=\pi\), and \(x=2\pi\). Its range is also \(( -\infty,-1]\cup[1,\infty)\). 3. At a zero of the dashed base function, the denominator of the reciprocal function is zero. The reciprocal is therefore undefined there, and its graph approaches a vertical asymptote.

a) Secant asymptotes: \(x=-\frac{\pi}{2}\), \(x=\frac{\pi}{2}\); range: \(( -\infty,-1]\cup[1,\infty)\) b) Cosecant asymptotes: \(x=0\), \(x=\pi\), \(x=2\pi\); range: \(( -\infty,-1]\cup[1,\infty)\) c) Each zero of the base sine or cosine function makes the reciprocal denominator zero, producing a vertical asymptote.

- Compare the y-values of a positive vertex and a negative vertex to recover the vertical shift and vertical scale. - The restriction \(a>0\) tells you how to use the positive vertex to locate the phase shift. - Read the repeating asymptote pattern to confirm the parent secant period. - The two vertex levels are also the boundary values of the two pieces of the range.

1. The positive and negative vertex levels are \(3\) and \(-1\). Their midpoint is \(d=\frac{3+(-1)}{2}=1\), and their distance from the midline is \(a=2\). 2. With \(a>0\), a positive secant vertex occurs when the secant input is \(0\). The positive vertex is at \(x=\frac{\pi}{2}\), so \(c=\frac{\pi}{2}\). Therefore, \(g(x)=2\sec\left(x-\frac{\pi}{2}\right)+1\). 3. The period is \(2\pi\). The displayed vertical asymptotes are \(x=0\), \(x=\pi\), and \(x=2\pi\). 4. The vertex levels give the range \(( -\infty,-1]\cup[3,\infty)\). The two interior vertices are \(\left(\frac{\pi}{2},3\right)\) and \(\left(\frac{3\pi}{2},-1\right)\).

a) \(a=2\), \(c=\frac{\pi}{2}\), \(d=1\); \(g(x)=2\sec\left(x-\frac{\pi}{2}\right)+1\) b) Period: \(2\pi\); asymptotes: \(x=0\), \(x=\pi\), \(x=2\pi\) c) Range: \(( -\infty,-1]\cup[3,\infty)\); vertices: \(\left(\frac{\pi}{2},3\right)\) and \(\left(\frac{3\pi}{2},-1\right)\)

- Measure the spacing between neighboring vertical asymptotes first. - Use the stated interval for \(c\) to select one representative from the repeating asymptote locations. - The y-value halfway between consecutive asymptotes identifies the vertical shift. - Check that your equation reproduces both an asymptote and a midline crossing from the graph.

1. Consecutive asymptotes are \(\frac{\pi}{2}\) apart, so the period is \(\frac{\pi}{2}\). Since cotangent has period \(\frac{\pi}{b}\), \(b=2\). 2. An asymptote occurs at \(x=c\). The restriction \(-\frac{\pi}{2}<c\le0\) selects \(c=-\frac{\pi}{4}\) from the displayed asymptote pattern. 3. Each branch crosses its midline halfway between consecutive asymptotes. The graph crosses at \((0,-1)\), so \(d=-1\). 4. Therefore, \(h(x)=\cot\left(2\left(x+\frac{\pi}{4}\right)\right)-1\), equivalently \(h(x)=\cot\left(2x+\frac{\pi}{2}\right)-1\). 5. In the requested interval, the asymptotes are \(x=-\frac{\pi}{4}\), \(x=\frac{\pi}{4}\), and \(x=\frac{3\pi}{4}\); the midline crossings are \((0,-1)\) and \(\left(\frac{\pi}{2},-1\right)\).

a) Period: \(\frac{\pi}{2}\); \(b=2\) b) \(c=-\frac{\pi}{4}\), \(d=-1\); \(h(x)=\cot\left(2\left(x+\frac{\pi}{4}\right)\right)-1\) c) Asymptotes: \(x=-\frac{\pi}{4}\), \(x=\frac{\pi}{4}\), \(x=\frac{3\pi}{4}\); midline crossings: \((0,-1)\), \(\left(\frac{\pi}{2},-1\right)\)

- For tangent, the spacing between consecutive vertical asymptotes equals one period. - The middle of a branch between its asymptotes locates the horizontal shift and midline crossing. - Use another clearly readable point on the central branch to determine the vertical scale. - Compare whether the branch rises or falls as \(x\) increases.

1. Consecutive vertical asymptotes around the central branch are \(x=-\frac{\pi}{8}\) and \(x=\frac{3\pi}{8}\). Their spacing is \(\frac{\pi}{2}\), so the period is \(\frac{\pi}{2}\). Therefore, \(b=\frac{\pi}{\pi/2}=2\). 2. The central branch crosses its midline at \(\left(\frac{\pi}{8},1\right)\), so \(c=\frac{\pi}{8}\) and \(d=1\). 3. At \(x=\frac{\pi}{4}\), the graph has \(y=-1\). Then \(-1=a\tan\left(2\left(\frac{\pi}{4}-\frac{\pi}{8}\right)\right)+1=a\tan\left(\frac{\pi}{4}\right)+1=a+1\), so \(a=-2\). 4. The central branch decreases from left to right, while the parent tangent branch increases. This confirms that \(a\) is negative. 5. Thus, \(g(x)=-2\tan\left(2\left(x-\frac{\pi}{8}\right)\right)+1\).

Period: \(\frac{\pi}{2}\); midline: \(y=1\) \(a=-2\), \(b=2\), \(c=\frac{\pi}{8}\), \(d=1\) \(g(x)=-2\tan\left(2\left(x-\frac{\pi}{8}\right)\right)+1\)

- Use the spacing between consecutive asymptotes to determine the horizontal scale. - Use the two visible vertex levels to recover the vertical shift and the magnitude of the vertical scale. - Check one vertex against the sign of the parent cosecant value there to determine the sign of \(a\). - The visible vertex levels form the two boundary values of the range.

1. Consecutive asymptotes are \(\frac{\pi}{2}\) apart. For \(\csc(bx)\), consecutive asymptotes are \(\frac{\pi}{b}\) apart, so \(b=2\) and the period is \(\pi\). 2. The two vertex levels are \(-1\) and \(5\). Their midpoint is \(d=2\), and the vertical scale has magnitude \(3\). 3. At \(x=\frac{\pi}{4}\), \(\sin(2x)=1\), while the graph has value \(-1\). Thus, \(a+2=-1\), so \(a=-3\). Therefore, \(g(x)=-3\csc(2x)+2\). 4. The asymptotes in \([0,\pi]\) are \(x=0\), \(x=\frac{\pi}{2}\), and \(x=\pi\). The range is \((-\infty,-1]\cup[5,\infty)\). 5. Since \(a<0\), the parent cosecant branches are reflected vertically before the shift: the branch where sine is positive opens downward, and the branch where sine is negative opens upward.

a) \(a=-3\), \(b=2\), \(d=2\); \(g(x)=-3\csc(2x)+2\) b) Period: \(\pi\); asymptotes: \(x=0\), \(x=\frac{\pi}{2}\), \(x=\pi\) c) Range: \((-\infty,-1]\cup[5,\infty)\); because \(a<0\), the cosecant branches are vertically reflected before the upward shift.

- Interpret the parameters in \(a\sin(b(x-c))\). - Factor the coefficient of \(x\) before reading the horizontal shift. - Relate the coefficient of \(x\) to the period. - Try simplifying one function using sine's repeating pattern.

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.

- Use the two y-values to find the midline and amplitude. - Compare the stretch of the graph from a minimum to the next maximum with one full cycle. - The positive coefficient of \(x\) relates to the period. - Think about where the upward midline crossing occurs relative to the given maximum.

1. The midline is halfway between the minimum and maximum values: \(d=\frac{1+5}{2}=3\). 2. The amplitude is the distance from the midline to either extreme: \(a=5-3=2\). 3. The horizontal distance from a minimum to the next maximum is half a period. Thus, \(\frac{p}{2}=4-1=3\), so \(p=6\). 4. Since \(b>0\), \(b=\frac{2\pi}{p}=\frac{\pi}{3}\). 5. The maximum occurs one-fourth of a period after the upward midline crossing at \(x=c\). Hence, \(4=c+\frac{6}{4}\), so \(c=\frac{5}{2}\), within \([0,6)\).

\(a=2\), \(b=\frac{\pi}{3}\), \(c=\frac{5}{2}\), \(d=3\)

- Compare how far you must shift each entire graph before every branch has the same shape and orientation again. - Locate the x-values where both functions become undefined. - Distinguish the spacing between neighboring asymptotes from the horizontal length required for the complete graph pattern to repeat.

1. The parent tangent period is \(\pi\), so replacing \(x\) by \(2x\) gives \(u\) period \(\frac{\pi}{2}\). The parent secant period is \(2\pi\), so replacing \(x\) by \(2x\) gives \(v\) period \(\pi\). 2. Both functions are undefined where \(\cos(2x)=0\). Thus \(2x=\frac{\pi}{2}+k\pi\), so \(x=\frac{\pi}{4}+k\frac{\pi}{2}\). In \((-\pi,\pi)\), the asymptotes are \(x=-\frac{3\pi}{4}\), \(x=-\frac{\pi}{4}\), \(x=\frac{\pi}{4}\), and \(x=\frac{3\pi}{4}\). 3. The claim is false. Although the tangent and secant graphs have the same vertical-asymptote locations here, the tangent branches repeat after \(\frac{\pi}{2}\), while the secant graph needs a horizontal shift of \(\pi\) to repeat its full pattern.

a) \(u\): \(\frac{\pi}{2}\); \(v\): \(\pi\) b) \(x=-\frac{3\pi}{4}\), \(x=-\frac{\pi}{4}\), \(x=\frac{\pi}{4}\), \(x=\frac{3\pi}{4}\) c) The claim is false. The graphs share asymptotes, but their full branch patterns repeat over different horizontal lengths.

- First separate the unbounded tangent/cotangent branches from the U-shaped reciprocal sine/cosine branches. - At \(x=0\), check which graphs cross an axis, have a vertex, or are undefined. - For tangent versus cotangent, compare whether the graph has a central crossing or an asymptote at the origin. - For secant versus cosecant, compare whether the reciprocal graph is defined at the origin.

1. Panel 4 is \(p(x)=\tan(2x)\): it crosses the origin and has increasing tangent branches with period \(\frac{\pi}{2}\). 2. Panel 2 is \(q(x)=-\cot(2x)\): it has an asymptote at \(x=0\), and the negative sign makes its branches increase. 3. Panel 1 is \(r(x)=2\sec(x)\): it is defined at \(x=0\) with a positive vertex at \((0,2)\) and has asymptotes where cosine is zero. 4. Panel 3 is \(s(x)=2\csc(x)\): it has an asymptote at \(x=0\) and reciprocal-sine branches with vertices halfway between consecutive sine zeros. 5. At \(x=0\), \(q\) and \(s\) are undefined because both depend on a sine denominator that is zero there. 6. Although \(p\) and \(q\) both have period \(\frac{\pi}{2}\), \(p(0)=0\) while \(q\) is undefined at \(0\). A midline crossing versus a vertical asymptote distinguishes the graphs.

a) \(p\to4\), \(q\to2\), \(r\to1\), \(s\to3\) b) \(q\) and \(s\) c) \(p\) crosses the x-axis at \(x=0\), while \(q\) has a vertical asymptote there.