Free math worksheets

- Look for a quantity that returns to the same pattern after a fixed amount of time. - Distinguish repeating behavior from quantities that mainly keep increasing.

1. A periodic quantity repeats the same pattern after a fixed interval. 2. A Ferris-wheel rider returns to the same heights in the same order once every revolution. 3. The savings balance and total distance generally increase rather than repeat.

b) The rider's height is periodic because the same height pattern repeats every Ferris-wheel revolution.

- The midline lies halfway between the maximum and minimum values. - The amplitude measures the vertical distance from the midline to either extreme. - Check that midline plus and minus amplitude reproduce the given extremes.

1. The amplitude is half the distance from the minimum to the maximum: \(\frac{8.0-2.4}{2}=2.8\). 2. The midline is the average of the maximum and minimum: \(\frac{8.0+2.4}{2}=5.2\).

Amplitude: \(2.8\,\text{ft}\) Midline: \(5.2\,\text{ft}\)

- The period is the time required for one complete repeating cycle. - Frequency tells how many cycles occur in one unit of time. - Use the reciprocal relationship between period and frequency.

1. The height pattern repeats after one complete revolution, so the period is \(48\) seconds. 2. Frequency is the reciprocal of the period: \(\frac{1}{48}\) cycle per second.

a) \(48\,\text{s}\) b) \(\frac{1}{48}\) cycle per second

- Compare the starting behavior of the basic sine and cosine graphs. - Focus on the value and direction of change at \(t=0\).

1. A positive cosine function starts at its maximum when its input is \(0\). 2. The described quantity is also at its maximum at \(t=0\) and then decreases. 3. Therefore, a cosine-based model is the natural choice.

Cosine, because a positive cosine model starts at a maximum and then decreases.

- Think about where the basic sine graph begins relative to its midline. - Compare the initial direction of motion with the context.

1. A positive sine function starts at its midline when its input is \(0\). 2. It initially increases from that midline. 3. Therefore, a sine-based model matches the stated starting behavior naturally.

Sine, because a positive sine model starts on the midline and increases.

- Read the maximum and minimum y-values from the graph. - The midline is halfway between those values. - Measure the horizontal distance between matching points on consecutive cycles.

1. The graph has maximum \(2\) and minimum \(-2\), so its midline is \(y=0\) and its amplitude is \(2\). 2. One full wave repeats every \(2\pi\) units along the x-axis. 3. Therefore, the period is \(2\pi\).

Amplitude: \(2\) Midline: \(y=0\) Period: \(2\pi\)

- Identify the highest and lowest y-values first. - Use the average and half-difference of the extremes for midline and amplitude. - Find two matching locations in consecutive cycles to measure the period.

1. The graph ranges from \(-2\) to \(4\), so the maximum is \(4\) and the minimum is \(-2\). 2. The midline is \(\frac{4+(-2)}{2}=1\), and the amplitude is \(\frac{4-(-2)}{2}=3\). 3. Consecutive maxima are \(4\pi\) units apart, so the period is \(4\pi\).

Maximum: \(4\) Minimum: \(-2\) Midline: \(y=1\) Amplitude: \(3\) Period: \(4\pi\)

- Compare vertical distances from each graph's midline to its peaks. - Check whether the peaks occur at the same x-values. - Describe the change in outputs without changing the x-values.

1. Both functions have midline \(y=0\) and period \(2\pi\). 2. The amplitude of \(f\) is \(1\), while the amplitude of \(g\) is \(2\). 3. Multiplying every output of \(f\) by \(2\) produces \(g\), so \(g\) is a vertical stretch of \(f\) by a factor of \(2\).

\(f\): amplitude \(1\), midline \(y=0\), period \(2\pi\) \(g\): amplitude \(2\), midline \(y=0\), period \(2\pi\) The transformation is a vertical stretch by a factor of \(2\).

- First compare the vertical heights of the two graphs. - Then measure how far each graph travels horizontally before its pattern repeats. - Count complete repetitions on the stated interval.

1. Both graphs range from \(-1\) to \(1\), so both amplitudes are \(1\). 2. The period of \(f\) is \(2\pi\). 3. The graph of \(g\) repeats after \(\pi\), so its period is \(\pi\). 4. Therefore, \(g\) completes two cycles on \([0,2\pi]\), while \(f\) completes one.

Both amplitudes are \(1\). The period of \(f\) is \(2\pi\), and the period of \(g\) is \(\pi\). Function \(g\) completes more cycles.

- Read the highest and lowest y-values from the graph. - Average the extremes for the midline and take half their difference for amplitude. - Measure the horizontal distance between consecutive peaks.

1. The graph has maximum \(3\) and minimum \(1\). 2. The midline is \(\frac{3+1}{2}=2\), and the amplitude is \(\frac{3-1}{2}=1\). 3. The pattern repeats every \(2\pi\), so the period is \(2\pi\).

Maximum: \(3\) Minimum: \(1\) Midline: \(y=2\) Amplitude: \(1\) Period: \(2\pi\)

- Substitute the given time into the model before simplifying. - Recognize the resulting special-angle sine value. - Keep the exact form until the final decimal approximation.

1. Substitute \(t=2\): \(y(2)=10+4\sin\left(\frac{\pi}{3}\right)\). 2. Since \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), \(y(2)=10+2\sqrt{3}\). 3. Numerically, \(10+2\sqrt{3}\approx13.46\).

\(y(2)=10+2\sqrt{3}\approx13.46\). The height is about \(13.46\,\text{ft}\) at \(t=2\,\text{s}\).

- Use the extremes to locate the center and vertical distance to an extreme. - Compare the starting minimum with the basic cosine starting point. - Leave the horizontal coefficient unspecified because no period is given.

1. The amplitude is \(\frac{17-5}{2}=6\). 2. The midline is \(\frac{17+5}{2}=11\). 3. A positive cosine starts at a maximum, so starting at a minimum requires a negative amplitude coefficient. 4. Thus a suitable form is \(y=11-6\cos(Bt)\), where \(B\) depends on the period.

a) Amplitude \(6\); midline \(y=11\) b) \(y=11-6\cos(Bt)\)

- Focus on the coefficient multiplying \(t\) inside the sine function. - Relate that coefficient to the horizontal length of one full cycle. - Divide the total time span by the period only after finding the period.

1. For \(\sin(Bt)\), the period is \(\frac{2\pi}{|B|}\). 2. Here \(B=\frac{\pi}{4}\), so the period is \(\frac{2\pi}{\pi/4}=8\). 3. The interval from \(0\) to \(24\) has length \(24\), so it contains \(\frac{24}{8}=3\) complete cycles.

Period: \(8\) Complete cycles from \(0\) to \(24\): \(3\)

- Measure the horizontal distance between two matching peaks. - Read the highest and lowest y-values from the graph. - Use the extremes to find amplitude and midline.

1. Consecutive peaks occur \(6\) units apart, so the period is \(6\). 2. The maximum is \(3\) and the minimum is \(-1\). 3. The amplitude is \(\frac{3-(-1)}{2}=2\), and the midline is \(\frac{3+(-1)}{2}=1\).

Period: \(6\) Amplitude: \(2\) Midline: \(y=1\)

- Distinguish the total peak-to-trough distance from the distance from the midline to a peak. - The midline must be equally far from the maximum and minimum. - Check that midline plus and minus amplitude reproduce both extremes.

1. The difference \(14-6=8\) is the full vertical distance from minimum to maximum. 2. Amplitude is half that distance, so the amplitude is \(\frac{8}{2}=4\). 3. The midline is the average of the extremes: \(\frac{14+6}{2}=10\).

The student used the full range instead of half the range. The amplitude is \(4\), and the midline is \(10\).

- Compare vertical peak distances before comparing horizontal repetition. - Measure how far each graph travels horizontally before repeating. - A longer period means slower progress through a cycle for the same change in x.

1. Both functions have amplitude \(2\). 2. The period of \(f\) is \(2\pi\). 3. For \(g\), the inside coefficient is \(\frac{1}{2}\), so the period is \(\frac{2\pi}{1/2}=4\pi\). 4. Therefore, \(g\) takes longer to complete a cycle and changes through its cycle more slowly.

a) Both amplitudes are \(2\). The periods are \(2\pi\) for \(f\) and \(4\pi\) for \(g\). b) \(g\) changes through its cycle more slowly.

- Read the vertical shift and outside coefficient first. - Use the inside coefficient to determine the period. - Combine midline and amplitude to find the extremes.

1. The vertical shift gives midline \(y=12\), and the cosine coefficient gives amplitude \(3\). 2. The period is \(\frac{2\pi}{\pi/6}=12\). 3. The maximum is \(12+3=15\), and the minimum is \(12-3=9\). 4. A period of \(12\) hours means the full pattern repeats every \(12\) hours.

Midline: \(y=12\) Amplitude: \(3\) Period: \(12\) hours Maximum: \(15\) Minimum: \(9\)

- Use the starting position and direction to select a basic sine or cosine form. - Relate the coefficient inside the trigonometric function to the desired period. - Place the amplitude outside the function and the midline as a vertical shift.

1. Midline-and-increasing behavior at \(t=0\) suggests a positive sine model. 2. For period \(6\), the angular frequency is \(B=\frac{2\pi}{6}=\frac{\pi}{3}\). 3. Using amplitude \(4\) and vertical shift \(10\) gives \(y(t)=10+4\sin\left(\frac{\pi}{3}t\right)\).

\(y(t)=10+4\sin\left(\frac{\pi}{3}t\right)\)

- Match the starting maximum with a basic trigonometric graph. - Convert the desired period into the coefficient multiplying \(t\). - Use the amplitude and midline as the outside coefficient and vertical shift.

1. Starting at a maximum suggests a positive cosine model. 2. For period \(8\), the angular frequency is \(B=\frac{2\pi}{8}=\frac{\pi}{4}\). 3. Using amplitude \(3\) and vertical shift \(5\) gives \(y(t)=5+3\cos\left(\frac{\pi}{4}t\right)\).

\(y(t)=5+3\cos\left(\frac{\pi}{4}t\right)\)

- Find amplitude and midline from the maximum and minimum depths. - Relate the period to the coefficient inside the trigonometric function. - Use the fact that the cycle starts at a maximum to select the phase.

1. The amplitude is \(\frac{8-2}{2}=3\), and the midline is \(\frac{8+2}{2}=5\). 2. The period is \(12\), so \(B=\frac{2\pi}{12}=\frac{\pi}{6}\). 3. Since the depth starts at a maximum, use cosine: \(d(t)=5+3\cos\left(\frac{\pi}{6}t\right)\).

\(d(t)=5+3\cos\left(\frac{\pi}{6}t\right)\)

- Connect the wheel's radius with amplitude and the center height with midline. - Use the revolution time as the period. - Decide how to make the model start at a minimum rather than a maximum.

1. The radius gives amplitude \(20\), and the center height gives midline \(25\). 2. The period is \(30\), so \(B=\frac{2\pi}{30}=\frac{\pi}{15}\). 3. Starting at the lowest point is modeled by subtracting cosine: \(h(t)=25-20\cos\left(\frac{\pi}{15}t\right)\).

\(h(t)=25-20\cos\left(\frac{\pi}{15}t\right)\)

- Use the maximum and minimum to determine amplitude and midline. - Convert the twelve-month period to the inside coefficient. - A cosine maximum occurs when its inside angle is \(0\), so place that at the stated month.

1. The amplitude is \(\frac{80-50}{2}=15\), and the midline is \(\frac{80+50}{2}=65\). 2. A period of \(12\) gives \(B=\frac{2\pi}{12}=\frac{\pi}{6}\). 3. To place a cosine maximum at \(m=7\), shift the input by \(7\): \(T(m)=65+15\cos\left(\frac{\pi}{6}(m-7)\right)\).

\(T(m)=65+15\cos\left(\frac{\pi}{6}(m-7)\right)\)

- Identify the model's midline from the vertical shift. - Ask when the cosine contribution is zero. - Restrict the solutions to the stated single-period interval.

1. The midline is \(y=5\), so set \(5+3\cos\left(\frac{\pi}{4}t\right)=5\). 2. This requires \(\cos\left(\frac{\pi}{4}t\right)=0\). 3. In one cycle, cosine is zero at inside angles \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). 4. Solving gives \(t=2\) and \(t=6\).

\(t=2\) and \(t=6\)

- Distinguish the period from the coefficient multiplying the input. - Use the relationship between \(B\) and one full cycle of \(2\pi\). - Check what period the student's coefficient would actually produce.

1. For \(\sin(Bt)\), the period is \(\frac{2\pi}{|B|}\), so the inside coefficient is not the period itself. 2. Setting \(\frac{2\pi}{B}=10\) gives \(B=\frac{\pi}{5}\). 3. The student's coefficient \(10\) would instead give period \(\frac{2\pi}{10}=\frac{\pi}{5}\).

The coefficient should be \(\frac{\pi}{5}\), not \(10\). A suitable model is \(y=2+3\sin\left(\frac{\pi}{5}t\right)\).

- Read the midline and amplitude from the maximum and minimum. - Check the graph's value at \(x=0\). - Use the spacing between repeated minima or maxima to verify the period.

1. The graph has midline \(y=1\), amplitude \(3\), and period \(4\). 2. At \(x=0\), the graph is at its minimum value \(-2\). 3. A negative cosine term with vertical shift \(1\) starts at that minimum, so choice b) matches.

b) \(q(x)=1-3\cos\left(\frac{\pi}{2}x\right)\)

- Substitute the shift point into the inside expression. - Determine the period before reasoning about quarter-cycle timing. - Locate where a positive sine cycle reaches its maximum relative to an upward midline crossing.

1. At \(t=1\), the sine input is \(0\), so \(y=5\), the midline, and the positive sine pattern is increasing there. 2. The period is \(\frac{2\pi}{\pi/2}=4\). 3. A positive sine function reaches its first maximum one quarter-period after the upward midline crossing. 4. One quarter of \(4\) is \(1\), so the first maximum after \(t=1\) occurs at \(t=2\).

a) At the midline \(y=5\) and increasing b) Period \(4\) c) \(t=2\)

- Use the highest and lowest table values to determine vertical parameters. - Track the sequence midline, maximum, midline, minimum, midline to identify one full cycle. - Match the starting pattern with a basic sine or cosine cycle before writing the equation.

1. The maximum is \(14\) and the minimum is \(6\), so the amplitude is \(4\) and the midline is \(10\). 2. The values return to the same midline position and direction after \(8\) time units, so the period is \(8\). 3. The table starts on the midline, rises to a maximum, returns to the midline, falls to a minimum, and returns again, matching a positive sine cycle. 4. With period \(8\), \(B=\frac{2\pi}{8}=\frac{\pi}{4}\), so a model is \(y=10+4\sin\left(\frac{\pi}{4}t\right)\).

Amplitude: \(4\) Midline: \(y=10\) Period: \(8\) Model: \(y=10+4\sin\left(\frac{\pi}{4}t\right)\)

- Use the highest and lowest points to find the midline and amplitude. - Measure the horizontal distance between consecutive maxima to find the period. - A positive cosine model reaches a maximum when its inside angle is \(0\). - Place the horizontal shift at a visible maximum.

1. The graph has maximum \(3\) and minimum \(-1\), so the amplitude is \(2\) and the midline is \(D=1\). 2. Consecutive maxima are \(4\) units apart, so the period is \(4\). Thus \(B=\frac{2\pi}{4}=\frac{\pi}{2}\). 3. A maximum occurs at \(x=1\), so choose a cosine shift of \(C=1\). 4. One matching equation is \(r(x)=1+2\cos\left(\frac{\pi}{2}(x-1)\right)\).

One matching equation is \(r(x)=1+2\cos\left(\frac{\pi}{2}(x-1)\right)\).