Examkin

Graphs of Trigonometric Functions

16 soru

Tüm türler16Çoktan seçmeli10Sayısal1Eşleştirme5
Soru 1Soru →

Match each of the trigonometric functions listed on the left with the correct description of its amplitude and period listed on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

y=3sin⁡(2x)y = 3\sin(2x)
y=2cos⁡(3x)y = 2\cos(3x)
y=4sin⁡(πx)y = 4\sin(\pi x)

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

The function y=3sin⁡(2x)y = 3\sin(2x) matches the description stating 'Amplitude is 3 and period is π\pi'. The function y=2cos⁡(3x)y = 2\cos(3x) matches the description stating 'Amplitude is 2 and period is 2π3\frac{2\pi}{3}'. The function y=4sin⁡(πx)y = 4\sin(\pi x) matches the description stating 'Amplitude is 4 and period is 2'.
Each trigonometric function of the form y=asin⁡(bx)y = a\sin(bx) or y=acos⁡(bx)y = a\cos(bx) has an amplitude equal to the absolute value of the coefficient of the trigonometric term (∣a∣|a|) and a period equal to 2π2\pi divided by the absolute value of the coefficient of the angle variable (∣b∣|b|). Applying these formulas gives the correct properties for each function.

Adım Adım Çözüm

1
Identify the general form of the trigonometric functions
The functions are in the form y=asin⁡(bx)y = a\sin(bx) or y=acos⁡(bx)y = a\cos(bx), where the amplitude is given by the absolute value of the vertical stretch coefficient (∣a∣|a|), and the period is calculated as 2π∣b∣\frac{2\pi}{|b|}.
This establishes the formulas needed to determine the amplitude and period for each equation.
2
Calculate the properties for y=3sin⁡(2x)y = 3\sin(2x)
The vertical stretch coefficient is 3, so the amplitude is 3. The frequency coefficient is 2, so the period is 2π2=π\frac{2\pi}{2} = \pi.
To find the amplitude and period for the first function.
3
Calculate the properties for y=2cos⁡(3x)y = 2\cos(3x)
The vertical stretch coefficient is 2, so the amplitude is 2. The frequency coefficient is 3, so the period is 2π3\frac{2\pi}{3}.
To find the amplitude and period for the second function.
4
Calculate the properties for y=4sin⁡(πx)y = 4\sin(\pi x)
The vertical stretch coefficient is 4, so the amplitude is 4. The frequency coefficient is π\pi, so the period is 2ππ=2\frac{2\pi}{\pi} = 2.
To find the amplitude and period for the third function.

Anahtar Kavram

Identifying the amplitude and calculating the period of trigonometric functions from their equations
Tahmini Süre:1m 30s
Soru 2Soru →

What is the period of the function f(x)=4cos⁡(13x)f(x) = 4\cos\left(\frac{1}{3}x\right)?

Cevabı ve açıklamayı göster

Cevap: 6π6\pi

Cevap

The period of the function is 6π6\pi.
The standard period of the cosine function y=cos⁡(x)y = \cos(x) is 2π2\pi. For a transformed trigonometric function of the form y=Acos⁡(Bx)y = A\cos(Bx), the period is given by the formula 2π∣B∣\frac{2\pi}{|B|}. In the function f(x)=4cos⁡(13x)f(x) = 4\cos\left(\frac{1}{3}x\right), the coefficient of xx is B=13B = \frac{1}{3}. Dividing the standard period 2π2\pi by 13\frac{1}{3} yields a period of 2π×3=6π2\pi \times 3 = 6\pi.

Adım Adım Çözüm

1
Identify the standard period of the parent cosine function and the coefficient BB of the variable xx in the given function.
The parent function is y=cos⁡(x)y = \cos(x), which has a standard period of 2π2\pi. In the function f(x)=4cos⁡(13x)f(x) = 4\cos\left(\frac{1}{3}x\right), the coefficient of xx is B=13B = \frac{1}{3}.
This sets up the parameters needed for the period formula Period=2π∣B∣\text{Period} = \frac{2\pi}{|B|}.
2
Substitute the value of BB into the period formula and simplify.
The period is 2π13=2π×3=6π\frac{2\pi}{\frac{1}{3}} = 2\pi \times 3 = 6\pi.
Dividing by a fraction is equivalent to multiplying by its reciprocal, which gives the final horizontal distance for one complete cycle.

Anahtar Kavram

Graphs of Trigonometric Functions
Soru 3Soru →

What is the period of the function f(x)=tan⁡(3x)f(x) = \tan(3x)?

Cevabı ve açıklamayı göster

Cevap: π3\frac{\pi}{3}

Cevap

The period of the function is π3\frac{\pi}{3}.
The parent function y=tan⁡(x)y = \tan(x) has a standard period of π\pi. To find the period of a transformed tangent function of the form y=tan⁡(Bx)y = \tan(Bx), the standard period must be divided by the coefficient of xx, yielding π∣B∣\frac{\pi}{|B|}. Substituting B=3B = 3 gives π3\frac{\pi}{3}.

Adım Adım Çözüm

1
Determine the standard period of the parent function.
The parent function is the tangent function, y=tan⁡(x)y = \tan(x), which has a standard period of π\pi.
The tangent function completes one full cycle of its graph between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}.
2
Identify the horizontal compression/stretch coefficient from the given equation.
In f(x)=tan⁡(3x)f(x) = \tan(3x), the coefficient of xx is B=3B = 3.
This coefficient determines how many cycles occur in a standard interval.
3
Calculate the period using the formula for the tangent function.
Period = π∣B∣=π3\frac{\pi}{|B|} = \frac{\pi}{3}.
Dividing the standard period of π\pi by the absolute value of the coefficient BB gives the compressed period of the transformed function.

Anahtar Kavram

The period of a transformed tangent function y=tan⁡(Bx)y = \tan(Bx) is given by π∣B∣\frac{\pi}{|B|}.
Soru 4Soru →

The graph of the function f(x)=acos⁡(b(x−c))+df(x) = a \cos(b(x - c)) + d is shown below for constants a>0a > 0, b>0b > 0, c∈[0,π]c \in [0, \pi], and dd. The graph has a local maximum at (π3,5)(\frac{\pi}{3}, 5) and the nearest local minimum to its right is at (5π6,−1)(\frac{5\pi}{6}, -1). What is the y-intercept of the graph of f(x)f(x)?

Cevabı ve açıklamayı göster

Cevap: 12\frac{1}{2}

Cevap

The y-intercept of the graph is 12\frac{1}{2}
The correct answer is 12\frac{1}{2}. The amplitude of the function is a=5−(−1)2=3a = \frac{5 - (-1)}{2} = 3, and the midline is d=5+(−1)2=2d = \frac{5 + (-1)}{2} = 2. The distance between the consecutive local maximum and local minimum represents half of a period: T2=5π6−π3=π2\frac{T}{2} = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}, meaning the period is T=πT = \pi, so b=2ππ=2b = \frac{2\pi}{\pi} = 2. A local maximum occurs when the argument of the cosine function is 00, so 2(π3−c)=0  ⟹  c=π32(\frac{\pi}{3} - c) = 0 \implies c = \frac{\pi}{3}. This gives the equation f(x)=3cos⁡(2(x−π3))+2f(x) = 3 \cos(2(x - \frac{\pi}{3})) + 2. Finding the y-intercept requires evaluating the function at x=0x = 0: f(0)=3cos⁡(2(0−π3))+2=3cos⁡(−2π3)+2=3(−12)+2=12f(0) = 3 \cos(2(0 - \frac{\pi}{3})) + 2 = 3 \cos(-\frac{2\pi}{3}) + 2 = 3(-\frac{1}{2}) + 2 = \frac{1}{2}.

Adım Adım Çözüm

1
Find the amplitude aa and midline dd of the trigonometric function.
a=3a = 3 and d=2d = 2
The amplitude is half the distance between the maximum and minimum values: a=5−(−1)2=3a = \frac{5 - (-1)}{2} = 3. The midline is the average of these values: d=5+(−1)2=2d = \frac{5 + (-1)}{2} = 2.
2
Determine the period TT and the frequency coefficient bb.
T=πT = \pi and b=2b = 2
The horizontal distance between a consecutive maximum and minimum is half of the period: 5π6−π3=π2\frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}. Thus, the full period is T=πT = \pi. Since T=2πbT = \frac{2\pi}{b}, we find b=2b = 2.
3
Determine the horizontal phase shift cc.
c=π3c = \frac{\pi}{3}
A cosine function achieves its maximum when its argument is a multiple of 2π2\pi. Since the maximum is at x=π3x = \frac{\pi}{3}, we set 2(π3−c)=02(\frac{\pi}{3} - c) = 0, giving c=π3c = \frac{\pi}{3}.
4
Evaluate the function at x=0x = 0 to find the y-intercept.
f(0)=12f(0) = \frac{1}{2}
Substitute the parameters into the function: f(x)=3cos⁡(2(x−π3))+2f(x) = 3 \cos(2(x - \frac{\pi}{3})) + 2. Substituting x=0x = 0 gives f(0)=3cos⁡(−2π3)+2=3(−12)+2=12f(0) = 3 \cos(-\frac{2\pi}{3}) + 2 = 3(-\frac{1}{2}) + 2 = \frac{1}{2}.

Anahtar Kavram

Determining the equation of a transformed trigonometric function from key features (maximum and minimum points) and evaluating it.
Tahmini Süre:3m 0s
Soru 5Soru →

For the trigonometric equations on the left, which description on the right correctly matches the graphical features of each equation?

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

y=−3sin⁡(2x−π3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1
y=2cos⁡(12x+π4)−1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1
y=−tan⁡(3x−π2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2
y=3sin⁡(23x+π)−2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

The correct matches are: the equation y=−3sin⁡(2x−π3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1 matches the description with a period of π\pi, a range of [−2,4][-2, 4], and a positive yy-intercept; the equation y=2cos⁡(12x+π4)−1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1 matches the description with a period of 4π4\pi, a range of [−3,1][-3, 1], and a phase shift of π2\frac{\pi}{2} units to the left; the equation y=−tan⁡(3x−π2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2 matches the description with a period of π3\frac{\pi}{3} that is undefined at the yy-axis; and the equation y=3sin⁡(23x+π)−2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2 matches the description with a period of 3π3\pi, a range of [−5,1][-5, 1], and a yy-intercept of (0,−2)(0, -2).
Each equation is correctly paired with the unique graphical properties determined by calculating its amplitude, period, range, phase shift, and yy-intercept. The first equation features a period of π\pi, a range of [−2,4][-2, 4], and a positive yy-intercept. The second equation has a period of 4π4\pi, a range of [−3,1][-3, 1], and a phase shift of π2\frac{\pi}{2} units left. The third equation has a period of π3\frac{\pi}{3} and is undefined at the yy-axis since x=0x=0 creates a vertical asymptote. The fourth equation has a period of 3π3\pi, a range of [−5,1][-5, 1], and a yy-intercept of (0,−2)(0, -2).

Adım Adım Çözüm

1
Determine the period, range, and y-intercept for the function y=−3sin⁡(2x−π3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1.
The period is π\pi, range is [−2,4][-2, 4], and yy-intercept is 1+332>01 + \frac{3\sqrt{3}}{2} > 0.
The period is 2π2=π\frac{2\pi}{2} = \pi, the range is [1−3,1+3]=[−2,4][1-3, 1+3] = [-2, 4], and evaluating at x=0x=0 yields y=−3sin⁡(−π/3)+1>0y = -3\sin(-\pi/3) + 1 > 0.
2
Determine the period, range, and phase shift for the function y=2cos⁡(12x+π4)−1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1.
The period is 4π4\pi, range is [−3,1][-3, 1], and phase shift is π2\frac{\pi}{2} units left.
The period is 2π1/2=4π\frac{2\pi}{1/2} = 4\pi, the range is [−1−2,−1+2]=[−3,1][-1-2, -1+2] = [-3, 1], and factoring the argument yields 12(x+π2)\frac{1}{2}(x + \frac{\pi}{2}), giving a shift of π2\frac{\pi}{2} units left.
3
Determine the period and domain restriction for the function y=−tan⁡(3x−π2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2.
The period is π3\frac{\pi}{3} and the function is undefined at x=0x = 0 (the yy-axis).
For tangent, the period is π3\frac{\pi}{3}. At x=0x=0, the argument is −π2-\frac{\pi}{2}, where tangent is undefined, meaning the function has a vertical asymptote at the yy-axis.
4
Determine the period, range, and y-intercept for the function y=3sin⁡(23x+π)−2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2.
The period is 3π3\pi, range is [−5,1][-5, 1], and yy-intercept is (0,−2)(0, -2).
The period is 2π2/3=3π\frac{2\pi}{2/3} = 3\pi, the range is [−2−3,−2+3]=[−5,1][-2-3, -2+3] = [-5, 1], and at x=0x=0, y=3sin⁡(π)−2=−2y = 3\sin(\pi) - 2 = -2.

Anahtar Kavram

Identifying graphs of trigonometric functions from their equations by determining amplitude, period, phase shift, midline, range, and asymptotes.
Soru 6Soru →

The graph of the function f(x)=Asin⁡(Bx+C)+Df(x) = A \sin(Bx + C) + D has a minimum point at (π4,−2)\left(\frac{\pi}{4}, -2\right) and its consecutive maximum point at (5π4,8)\left(\frac{5\pi}{4}, 8\right), where A>0A > 0 and B>0B > 0. What is the value of f(7π4)f\left(\frac{7\pi}{4}\right)?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3
The midline of a sinusoidal curve is the average of its maximum and minimum values, which is 8+(−2)2=3\frac{8 + (-2)}{2} = 3. The horizontal distance between consecutive minimum and maximum points is half of a period: 5π4−π4=π\frac{5\pi}{4} - \frac{\pi}{4} = \pi. Thus, the full period is 2π2\pi, and the next minimum occurs at 5π4+π=9π4\frac{5\pi}{4} + \pi = \frac{9\pi}{4}. The value x=7π4x = \frac{7\pi}{4} lies halfway between the maximum at 5π4\frac{5\pi}{4} and the minimum at 9π4\frac{9\pi}{4}. Midway between a maximum and a minimum, the function crosses its midline, so f(7π4)=3f\left(\frac{7\pi}{4}\right) = 3.

Adım Adım Çözüm

1
Determine the midline (vertical shift DD) of the trigonometric function.
The midline is D=Maximum+Minimum2=8+(−2)2=3D = \frac{\text{Maximum} + \text{Minimum}}{2} = \frac{8 + (-2)}{2} = 3.
The midline lies exactly halfway between the vertical peak and trough.
2
Find the horizontal distance for a half-period and determine the full period.
Half-period = 5π4−π4=π\frac{5\pi}{4} - \frac{\pi}{4} = \pi, so the full period T=2πT = 2\pi.
The horizontal distance between consecutive minimum and maximum points equals half of one full period.
3
Identify the behavior of the graph at x=7π4x = \frac{7\pi}{4}.
Since x=7π4x = \frac{7\pi}{4} is halfway between the maximum at x=5π4x = \frac{5\pi}{4} and the next minimum at x=9π4x = \frac{9\pi}{4}, the function value equals the midline height D=3D = 3.
A sinusoidal wave crosses its midline exactly midway between a peak and a trough.

Anahtar Kavram

Key features of transformed trigonometric functions (amplitude, midline, period, and symmetry)
Tahmini Süre:2m 0s
Soru 7Soru →

Match each transformed trigonometric function listed on the left with the correct description of its key graphical features listed on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

f(x)=−3cos⁡(2x−π2)+1f(x) = -3\cos\left(2x - \frac{\pi}{2}\right) + 1
g(x)=2sin⁡(12x+π)−1g(x) = 2\sin\left(\frac{1}{2}x + \pi\right) - 1
h(x)=−tan⁡(3x+3π4)+2h(x) = -\tan\left(3x + \frac{3\pi}{4}\right) + 2
k(x)=4cos⁡(πx−π2)−3k(x) = 4\cos\left(\pi x - \frac{\pi}{2}\right) - 3

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

The trigonometric functions correctly match their graphical features as follows: f(x)f(x) matches the description with period π\pi and midline y=1y = 1; g(x)g(x) matches the description with period 4π4\pi and phase shift 2π2\pi left; h(x)h(x) matches the description with period π/3\pi/3 and y-intercept (0,3)(0, 3); k(x)k(x) matches the description with period 22 and midline y=−3y = -3.
Each trigonometric equation is mapped to its unique set of graphical properties by evaluating its period, midline, phase shift, and specific points like y-intercepts or extrema using standard trigonometric transformation formulas.

Adım Adım Çözüm

1
Analyze f(x)=−3cos⁡(2x−π2)+1f(x) = -3\cos\left(2x - \frac{\pi}{2}\right) + 1.
Factor out the coefficient of xx: f(x)=−3cos⁡(2(x−π4))+1f(x) = -3\cos\left(2\left(x - \frac{\pi}{4}\right)\right) + 1. The period is 2πB=2π2=π\frac{2\pi}{B} = \frac{2\pi}{2} = \pi. The midline is y=D=1y = D = 1. The maximum value occurs when the cosine term equals −1-1 (due to the −3-3 coefficient): −3(−1)+1=4-3(-1) + 1 = 4, which happens at 2x−π2=π  ⟹  x=3π42x - \frac{\pi}{2} = \pi \implies x = \frac{3\pi}{4}.
Identify period, midline, phase shift, and extrema from standard form y=Acos⁡(B(x−C))+Dy = A\cos(B(x-C)) + D.
2
Analyze g(x)=2sin⁡(12x+π)−1g(x) = 2\sin\left(\frac{1}{2}x + \pi\right) - 1.
Rewrite as g(x)=2sin⁡(12(x−(−2π)))−1g(x) = 2\sin\left(\frac{1}{2}(x - (-2\pi))\right) - 1. The period is 2π1/2=4π\frac{2\pi}{1/2} = 4\pi, and the phase shift is 2π2\pi units to the left. The minimum value is 2(−1)−1=−32(-1) - 1 = -3, which occurs when 12x+π=3π2  ⟹  x=π\frac{1}{2}x + \pi = \frac{3\pi}{2} \implies x = \pi.
Determine horizontal shift, period, and minimum location.
3
Analyze h(x)=−tan⁡(3x+3π4)+2h(x) = -\tan\left(3x + \frac{3\pi}{4}\right) + 2.
The period for tangent is πB=π3\frac{\pi}{B} = \frac{\pi}{3}. Consecutive vertical asymptotes occur every period π3\frac{\pi}{3}. Evaluating at x=0x = 0 gives h(0)=−tan⁡(3π4)+2=−(−1)+2=3h(0) = -\tan\left(\frac{3\pi}{4}\right) + 2 = -(-1) + 2 = 3, giving a y-intercept of (0,3)(0, 3).
Apply tangent period formula π∣B∣\frac{\pi}{|B|} and evaluate y-intercept.
4
Analyze k(x)=4cos⁡(πx−π2)−3k(x) = 4\cos\left(\pi x - \frac{\pi}{2}\right) - 3.
Rewrite as k(x)=4cos⁡(π(x−12))−3k(x) = 4\cos\left(\pi\left(x - \frac{1}{2}\right)\right) - 3. The period is 2ππ=2\frac{2\pi}{\pi} = 2. The phase shift is 12\frac{1}{2} unit to the right, and the midline is y=−3y = -3.
Extract parameters from cosine function with π\pi in argument.

Anahtar Kavram

Graphical transformations of trigonometric functions (amplitude, period T=2π∣B∣T = \frac{2\pi}{|B|} or π∣B∣\frac{\pi}{|B|}, phase shift CC, and midline DD).
Tahmini Süre:2m 0s
Soru 8Soru →

The vertical displacement, d(t)d(t) in centimeters, of a particle executing simple harmonic motion is modeled by the trigonometric function d(t)=Asin⁡(B(t−C))+Dd(t) = A \sin(B(t - C)) + D, where A>0A > 0, B>0B > 0, and CC represents the smallest non-negative phase shift in seconds. The graph of d(t)d(t) completes one full cycle every 2π3\frac{2\pi}{3} seconds, has a maximum value of 7 cm7\text{ cm} at t=5π18 secondst = \frac{5\pi}{18}\text{ seconds}, and has a minimum value of −3 cm-3\text{ cm}. What is the value of CC?

Cevabı ve açıklamayı göster

Cevap: π9\frac{\pi}{9}

Cevap

π9\frac{\pi}{9}
To find the phase shift CC, first determine BB using the period formula Period=2πB\text{Period} = \frac{2\pi}{B}. Since the period is 2π3\frac{2\pi}{3}, B=3B = 3. The function achieves a maximum when its sine argument equals π2+2kπ\frac{\pi}{2} + 2k\pi. Setting 3(5π18−C)=π23\left(\frac{5\pi}{18} - C\right) = \frac{\pi}{2} yields 5π6−3C=π2\frac{5\pi}{6} - 3C = \frac{\pi}{2}, which simplifies to 3C=π33C = \frac{\pi}{3}, giving C=π9C = \frac{\pi}{9}.

Adım Adım Çözüm

1
Determine the value of BB from the period of the function.
B=3B = 3
The standard period for a sine function is 2π2\pi. Given that the period is 2π3\frac{2\pi}{3}, we set 2πB=2π3\frac{2\pi}{B} = \frac{2\pi}{3}, which yields B=3B = 3.
2
Determine the maximum value equation for the parent sine function.
Argument equals π2\frac{\pi}{2}
The standard sine function sin⁡(θ)\sin(\theta) achieves its first positive maximum at θ=π2\theta = \frac{\pi}{2}. Thus, for d(t)d(t), the maximum occurs when B(t−C)=π2B(t - C) = \frac{\pi}{2}.
3
Substitute B=3B = 3 and t=5π18t = \frac{5\pi}{18} into the argument equation and solve for CC.
C=π9C = \frac{\pi}{9}
Substitute the given values: 3(5π18−C)=π2  ⟹  5π6−3C=π23\left(\frac{5\pi}{18} - C\right) = \frac{\pi}{2} \implies \frac{5\pi}{6} - 3C = \frac{\pi}{2}. Subtract 5π6\frac{5\pi}{6} from both sides to get −3C=3π6−5π6=−2π6=−π3-3C = \frac{3\pi}{6} - \frac{5\pi}{6} = -\frac{2\pi}{6} = -\frac{\pi}{3}. Dividing by −3-3 gives C=π9C = \frac{\pi}{9}.

Anahtar Kavram

Phase shift and parameter identification from graphs of transformed sine functions
Soru 9Soru →

In the standard (x,y)(x,y) coordinate plane, what is the distance along the xx-axis between any two consecutive maximum points on the graph of the function f(x)=5cos⁡(3x−π2)+4f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4?

Cevabı ve açıklamayı göster

Cevap: 2π3\frac{2\pi}{3}

Cevap

The distance between any two consecutive maximum points is 2π3\frac{2\pi}{3}.
The distance along the xx-axis between consecutive peaks of a cosine wave represents one full period. For the function f(x)=5cos⁡(3x−π2)+4f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4, the coefficient of xx is 3. The period formula for cosine is 2π∣B∣\frac{2\pi}{|B|}, which evaluates to 2π3\frac{2\pi}{3}.

Adım Adım Çözüm

1
Relate consecutive maximum points to the function's period
The distance along the xx-axis between consecutive maximum points of a periodic trigonometric function equals one period length, TT.
Cosine graphs repeat their peak values once per complete wave cycle.
2
Identify the coefficient BB of the variable xx
In f(x)=5cos⁡(3x−π2)+4f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4, the coefficient of xx is B=3B = 3.
The standard transformation model is y=Acos⁡(Bx−C)+Dy = A \cos(Bx - C) + D.
3
Calculate the period using T=2π∣B∣T = \frac{2\pi}{|B|}
T=2π3T = \frac{2\pi}{3}.
Dividing the base cosine period of 2π2\pi by ∣B∣=3|B| = 3 gives the period of the transformed function.

Anahtar Kavram

The period of a transformed cosine function y=Acos⁡(Bx−C)+Dy = A \cos(Bx - C) + D is 2π∣B∣\frac{2\pi}{|B|}, which measures the horizontal distance between consecutive peak values.
Soru 10Soru →

What is the period of the trigonometric function f(x)=3cos⁡(π5x+π2)−4f(x) = 3 \cos\left(\frac{\pi}{5}x + \frac{\pi}{2}\right) - 4?

Cevabı ve açıklamayı göster

Cevap: 1010

Cevap

10
The period of a function of the form f(x)=Acos⁡(Bx+C)+Df(x) = A \cos(Bx + C) + D is given by T=2π∣B∣T = \frac{2\pi}{|B|}. Substituting B=π5B = \frac{\pi}{5} yields T=2ππ5=2π⋅5π=10T = \frac{2\pi}{\frac{\pi}{5}} = 2\pi \cdot \frac{5}{\pi} = 10.

Adım Adım Çözüm

1
Identify the coefficient of xx (BB) in the given trigonometric function.
B=π5B = \frac{\pi}{5}
The standard form for a transformed cosine function is f(x)=Acos⁡(Bx+C)+Df(x) = A \cos(Bx + C) + D, where BB controls the horizontal stretch or compression.
2
Apply the period formula for cosine, T=2π∣B∣T = \frac{2\pi}{|B|}.
T=2ππ5T = \frac{2\pi}{\frac{\pi}{5}}
The fundamental period of cos⁡(x)\cos(x) is 2π2\pi, which is scaled inversely by ∣B∣|B|.
3
Simplify the fraction by multiplying by the reciprocal.
T=2π⋅5π=10T = 2\pi \cdot \frac{5}{\pi} = 10
Dividing by a fraction is equivalent to multiplying by its reciprocal, and the factor of π\pi cancels out.

Anahtar Kavram

Period of Transformed Cosine Functions
Soru 11Soru →

Match each transformed trigonometric function on the left with its set of defining graphical properties on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

f(x)=4sin⁡(3x)+2f(x) = 4\sin(3x) + 2
g(x)=−2cos⁡(2x−π)g(x) = -2\cos\left(2x - \pi\right)
h(x)=3tan⁡(12x)−1h(x) = 3\tan\left(\frac{1}{2}x\right) - 1
k(x)=cos⁡(x+π3)−4k(x) = \cos\left(x + \frac{\pi}{3}\right) - 4

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

The correct pairings are: f(x)=4sin⁡(3x)+2f(x) = 4\sin(3x) + 2 matches with 'Amplitude of 4, period of 2π/3, and a midline at y = 2'; g(x)=−2cos⁡(2x−π)g(x) = -2\cos(2x - π) matches with 'Amplitude of 2, period of π, and a phase shift of π/2 units to the right'; h(x)=3tan⁡(x/2)−1h(x) = 3\tan(x/2) - 1 matches with 'Period of 2π, vertical shift of 1 unit down, with vertical asymptotes at x = π + 2kπ'; and k(x)=cos⁡(x+π/3)−4k(x) = \cos(x + π/3) - 4 matches with 'Midline at y = -4, amplitude of 1, and a phase shift of π/3 units to the left'.
Each trigonometric function is matched according to its standard parameter transformation rules: y=Asin⁡(B(x−C))+Dy = A\sin(B(x - C)) + D or y=Acos⁡(B(x−C))+Dy = A\cos(B(x - C)) + D, where ∣A∣|A| is amplitude, period is 2π∣B∣\frac{2\pi}{|B|} for sine/cosine and π∣B∣\frac{\pi}{|B|} for tangent, CC is horizontal phase shift, and y=Dy = D is the midline.

Adım Adım Çözüm

1
Analyze f(x)=4sin⁡(3x)+2f(x) = 4\sin(3x) + 2 using the standard form y=Asin⁡(B(x−C))+Dy = A\sin(B(x - C)) + D
Amplitude =∣A∣=4= |A| = 4, period =2πB=2π3= \frac{2\pi}{B} = \frac{2\pi}{3}, midline =D  ⟹  y=2= D \implies y = 2.
Direct extraction of parameters for sine graphs.
2
Factor out B=2B = 2 from g(x)=−2cos⁡(2x−π)g(x) = -2\cos(2x - \pi)
g(x)=−2cos⁡(2(x−π2))g(x) = -2\cos\left(2\left(x - \frac{\pi}{2}\right)\right), so amplitude =∣−2∣=2= |-2| = 2, period =2π2=π= \frac{2\pi}{2} = \pi, phase shift =π2= \frac{\pi}{2} to the right.
Factoring BB is necessary to correctly identify the horizontal phase shift.
3
Analyze tangent function parameters for h(x)=3tan⁡(12x)−1h(x) = 3\tan\left(\frac{1}{2}x\right) - 1
Period =πB=π1/2=2π= \frac{\pi}{B} = \frac{\pi}{1/2} = 2\pi, shifted down 11 unit (y=−1y = -1). Asymptotes occur when 12x=π2+kπ  ⟹  x=π+2kπ\frac{1}{2}x = \frac{\pi}{2} + k\pi \implies x = \pi + 2k\pi.
Tangent period uses πB\frac{\pi}{B} instead of 2πB\frac{2\pi}{B}, and asymptotes occur where tangent arguments equal odd multiples of π2\frac{\pi}{2}.
4
Analyze transformation parameters for k(x)=cos⁡(x+π3)−4k(x) = \cos\left(x + \frac{\pi}{3}\right) - 4
Amplitude =1= 1, midline =y=−4= y = -4, phase shift =π3= \frac{\pi}{3} to the left.
Addition inside the function argument (x+C)(x + C) corresponds to a horizontal shift to the left.

Anahtar Kavram

Identifying amplitude, period, midline, phase shift, and asymptotes from transformed trigonometric equations
Soru 12Soru →

In the standard (x,y)(x,y) coordinate plane, a cosine function of the form y=acos⁡(b(x−c))+dy = a \cos(b(x - c)) + d, where a>0a > 0 and b>0b > 0, has a local maximum at (2,9)(2, 9) and the very next local minimum at (6,1)(6, 1). What is the value of bb?

Cevabı ve açıklamayı göster

Cevap: π4\frac{\pi}{4}

Cevap

π4\frac{\pi}{4}
The distance along the xx-axis from a maximum to the consecutive minimum is half of one full period of the cosine wave. Here, that distance is 6−2=46 - 2 = 4, which means the full period is 2×4=82 \times 4 = 8. Using the relationship Period=2πb\text{Period} = \frac{2\pi}{b}, we solve for bb to get b=2π8=π4b = \frac{2\pi}{8} = \frac{\pi}{4}.

Adım Adım Çözüm

1
Determine the horizontal distance between the consecutive maximum and minimum points.
The horizontal distance is 6−2=46 - 2 = 4 units.
The xx-coordinates of the maximum and minimum points are 22 and 66, respectively.
2
Calculate the period of the cosine function.
Period=2×4=8\text{Period} = 2 \times 4 = 8 units.
The horizontal distance between a peak and the immediately following trough of a cosine wave represents exactly half of one full period.
3
Solve for the coefficient bb using the period formula b=2πPeriodb = \frac{2\pi}{\text{Period}}.
b=2π8=π4b = \frac{2\pi}{8} = \frac{\pi}{4}.
For a trigonometric function of the form y=acos⁡(b(x−c))+dy = a \cos(b(x - c)) + d, the relationship between the period and bb is Period=2πb\text{Period} = \frac{2\pi}{b}.

Anahtar Kavram

Determining Period and Frequency Coefficient of a Cosine Function
Soru 13Soru →

In the standard (x,y)(x,y) coordinate plane, a sinusoidal function f(x)=Asin⁡(Bx−C)+Df(x) = A \sin(Bx - C) + D reaches a maximum value of 99 at x=1x = 1 and its immediate next minimum value of −1-1 at x=4x = 4. What is the period of f(x)f(x)?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The period of the sinusoidal function is 6.
For any sinusoidal graph, the horizontal distance between a maximum point and the immediate next minimum point corresponds to one-half of the period. Given that the maximum occurs at x=1x = 1 and the next minimum occurs at x=4x = 4, the half-period is 4−1=34 - 1 = 3. Multiplying this half-period by 22 gives the full period of 66.

Adım Adım Çözüm

1
Identify the horizontal distance between the consecutive maximum and minimum points.
The horizontal distance is 4−1=34 - 1 = 3.
The maximum occurs at x=1x = 1 and the consecutive minimum occurs at x=4x = 4.
2
Relate the horizontal distance between consecutive extrema to the period of the function.
Half of the period is equal to 33.
In any sinusoidal function, the horizontal distance between a peak (maximum) and the adjacent trough (minimum) represents exactly one-half of a complete cycle.
3
Calculate the full period of the function.
Period = 2×3=62 \times 3 = 6.
Multiplying the half-period by 2 yields the full period of the function.

Anahtar Kavram

The horizontal distance between consecutive maximum and minimum points of a sinusoidal graph is half of the function's period.
Tahmini Süre:1m 15s
Soru 14Soru →

Which of the following values represents the period, in radians, of the trigonometric function f(x)=3tan⁡(23x−π6)+5f(x) = 3 \tan\left(\frac{2}{3}x - \frac{\pi}{6}\right) + 5?

Cevabı ve açıklamayı göster

Cevap: 3π2\frac{3\pi}{2}

Cevap

The period of the given tangent function is 3π2\frac{3\pi}{2} radians.
The parent function y=tan⁡(x)y = \tan(x) repeats every π\pi radians. For a function in the form f(x)=Atan⁡(Bx−C)+Df(x) = A \tan(Bx - C) + D, the horizontal scale factor BB alters the period according to Period=π∣B∣\text{Period} = \frac{\pi}{|B|}. With B=23B = \frac{2}{3}, dividing π\pi by 23\frac{2}{3} gives 3π2\frac{3\pi}{2}.

Adım Adım Çözüm

1
Identify the standard form of the transformed tangent function and its parameters.
For f(x)=Atan⁡(Bx−C)+Df(x) = A \tan(Bx - C) + D, the coefficient of xx is B=23B = \frac{2}{3}.
The horizontal stretch/compression factor BB determines the period of the function.
2
Apply the period formula for the tangent function.
\text{Period} = \frac{\pi}{|B|} = \frac{\pi}{\frac{2}{3}} = \frac{3\pi}{2}
Unlike sine and cosine functions which have a fundamental period of 2π2\pi, the parent tangent function y=tan⁡(x)y = \tan(x) has a period of π\pi radians.

Anahtar Kavram

Period of Transformed Tangent Functions
Tahmini Süre:1m 0s
Soru 15Soru →

Match each trigonometric function on the left with its correct combination of amplitude and period on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

f(x)=4sin⁡(3x)f(x) = 4 \sin(3x)
g(x)=2cos⁡(12x)g(x) = 2 \cos\left(\frac{1}{2}x\right)
h(x)=−3sin⁡(2x)h(x) = -3 \sin(2x)

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

f(x)=4sin⁡(3x)f(x) = 4 \sin(3x) matches Amplitude = 4, Period = 2π3\frac{2\pi}{3}; g(x)=2cos⁡(12x)g(x) = 2 \cos\left(\frac{1}{2}x\right) matches Amplitude = 2, Period = 4π4\pi; h(x)=−3sin⁡(2x)h(x) = -3 \sin(2x) matches Amplitude = 3, Period = π\pi.
Each trigonometric function is correctly evaluated using the general properties: Amplitude equals ∣A∣|A| and Period equals 2π∣B∣\frac{2\pi}{|B|}.

Adım Adım Çözüm

1
Identify the standard trigonometric form parameters.
For equations of the form y=Asin⁡(Bx)y = A \sin(Bx) or y=Acos⁡(Bx)y = A \cos(Bx), Amplitude =∣A∣= |A| and Period =2π∣B∣= \frac{2\pi}{|B|}.
Applying the definitions of amplitude and period for sine and cosine functions.
2
Calculate amplitude and period for f(x)=4sin⁡(3x)f(x) = 4 \sin(3x).
Amplitude =∣4∣=4= |4| = 4, Period =2π3= \frac{2\pi}{3}.
Here A=4A = 4 and B=3B = 3.
3
Calculate amplitude and period for g(x)=2cos⁡(12x)g(x) = 2 \cos\left(\frac{1}{2}x\right).
Amplitude =∣2∣=2= |2| = 2, Period =2π1/2=4π= \frac{2\pi}{1/2} = 4\pi.
Here A=2A = 2 and B=12B = \frac{1}{2}.
4
Calculate amplitude and period for h(x)=−3sin⁡(2x)h(x) = -3 \sin(2x).
Amplitude =∣−3∣=3= |-3| = 3, Period =2π2=π= \frac{2\pi}{2} = \pi.
Here A=−3A = -3 (so ∣A∣=3|A| = 3) and B=2B = 2.

Anahtar Kavram

Amplitude and Period of Sine and Cosine Graphs
Soru 16Soru →

In the standard (x,y)(x,y) coordinate plane, the graph of a trigonometric function is given by f(x)=4sin⁡(3x−π2)+2f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2. What is the horizontal distance between any two consecutive points where the graph intersects its midline y=2y = 2?

Cevabı ve açıklamayı göster

Cevap: π3\frac{\pi}{3}

Cevap

The horizontal distance between consecutive midline intersections is π3\frac{\pi}{3}.
The midline of f(x)=4sin⁡(3x−π2)+2f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2 is the horizontal line y=2y = 2. Intersections with this line occur when sin⁡(3x−π2)=0\sin\left(3x - \frac{\pi}{2}\right) = 0. Since the sine function equals zero at integer multiples of π\pi, the difference in the argument between consecutive zero points is π\pi. Setting 3Δx=π3\Delta x = \pi gives Δx=π3\Delta x = \frac{\pi}{3}, which is half the period of the function.

Adım Adım Çözüm

1
Identify the frequency parameter BB from the function f(x)=4sin⁡(3x−π2)+2f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2.
The frequency coefficient inside the sine expression is B=3B = 3.
The standard form of a transformed sine function is f(x)=Asin⁡(Bx−C)+Df(x) = A \sin(Bx - C) + D.
2
Calculate the full period TT of the function.
T=2π∣B∣=2π3T = \frac{2\pi}{|B|} = \frac{2\pi}{3}.
The standard period 2π2\pi of a sine function is scaled horizontally by a factor of 1B\frac{1}{B}.
3
Determine the horizontal distance between consecutive intersections with the midline y=2y = 2.
\text{Distance} = \frac{T}{2} = \frac{\frac{2\pi}{3}}{2} = \frac{\pi}{3}.
A sinusoidal wave completes one full period over length TT and intersects its midline twice during each full cycle, making consecutive midline crossings separated by half of the period.

Anahtar Kavram

Distance between consecutive midline intersections of a sinusoidal function
Tahmini Süre:1m 0s