Vertical Shift Effect Quiz: Effect of Vertical Shift on Maxima, Minima, and Midline

The cosine function oscillates between -1 and 1. Multiplying by A scales that range to between negative absolute value of A and positive absolute value of A. Adding D shifts the entire range up by D, giving a final range from D minus absolute value of A to D plus absolute value of A. The minimum is the lower bound of this range, which is D minus absolute value of A. Option B gives the maximum, not the minimum. Option C reverses the subtraction incorrectly. Option D multiplies instead of adding, which has no connection to how vertical shifts work.

For y = -4cos(x) + 1, the amplitude is 4 and the vertical shift is D = 1. The maximum value equals D + amplitude = 1 + 4 = 5, and the minimum value equals D - amplitude = 1 - 4 = -3. Option B swaps the maximum and minimum incorrectly. Option C gives the range of -4cos(x) before the vertical shift is applied. Option D gives the range of plain cos(x) with neither the amplitude scaling nor the vertical shift taken into account.

The midline equals the vertical shift D in the standard form. In option A, D = -3. In option B, D = -3. In option C, the term (x - 1) is a horizontal shift inside the function and does not affect the vertical shift, so D = -3 there as well. Option D has the form 3 + 2sin(x), meaning D = +3, so its midline is y = 3, not y = -3. Only options A, B, and C have a vertical shift of -3.

For y = 2cos(x) + D, the amplitude is always 2 because it equals the absolute value of the leading coefficient, which does not change with D. The maximum always equals D + 2 and the minimum always equals D - 2, following the standard formulas D plus or minus the amplitude. The period is always 2pi because the coefficient of x inside the cosine is 1. All four statements hold true for any value of D, and the midline is always y = D.

The answer is False. The amplitude of a trigonometric function is determined by the absolute value of the coefficient A that multiplies the sine or cosine. The parameter D controls the vertical shift, which moves the entire graph up or down by a fixed amount. Changing D raises or lowers the midline, maximum, and minimum equally, but does not change the distance from the midline to the peak or trough, which is what amplitude measures.

The midline of a sinusoidal function in the form y = A times sin(Bx) + D is the horizontal line y = D. For y = -6sin(3x) - 4, the vertical shift D equals -4, so the midline is y = -4. Option B confuses the amplitude coefficient -6 with the midline. Options C and D give the positive versions of the amplitude and shift respectively, both of which are incorrect. The negative sign in front of the constant is essential and must be included.

A vertical shift of -6 subtracts 6 from every output value of the function. Applying this to the original range gives -4 - 6 = -10 as the new minimum and 4 - 6 = -2 as the new maximum, so the range becomes -10 to -2. Option B shows the original range with no shift applied. Option C incorrectly doubles the effect of the shift. Option D applies a shift of +6 instead of -6, moving the range in the wrong direction.

The amplitude of y = 5cos(2x) + D is 5, so the maximum value equals D + 5. Setting D + 5 = 12 and solving gives D = 7. Option B incorrectly sets D equal to the maximum value itself, ignoring the contribution of the amplitude. Option C adds 5 instead of subtracting, giving D = 17. Option D sets D equal to the amplitude, which would make the maximum 10, not 12. Only D = 7 satisfies the equation.

The midline is the horizontal line that runs exactly halfway between the maximum and minimum values of the function. The maximum is D plus absolute value of A and the minimum is D minus absolute value of A, so the average of the two is D. The vertical shift D therefore defines the midline as y = D. Option A gives the midline only when D = 0. Option B confuses the amplitude coefficient with the midline. Option D confuses the frequency coefficient with the midline.

The sine function outputs values between -1 and 1. Multiplying by A scales the output to the interval from negative absolute value of A to positive absolute value of A. Adding D then shifts this interval upward by D, producing a final range from D minus absolute value of A to D plus absolute value of A. Option B shows the range before the vertical shift is applied. Option C uses A without the absolute value, which gives incorrect bounds when A is negative. Option D swaps the roles of A and D entirely.

Adding D = 3 shifts every y-value upward by 3. The midline rises from y = 0 to y = 3, the maximum rises from 1 to 4, and the minimum rises from -1 to 2. Amplitude is determined by the coefficient of sin(x), which remains 1 and is unaffected by a vertical shift. The period is also unchanged because it depends only on the coefficient of x inside the function.

For y = -2sin(x) + 5, the vertical shift is D = 5, so the midline is y = 5. The amplitude is the absolute value of the coefficient, giving 2. The maximum equals D + amplitude = 5 + 2 = 7, and the minimum equals D - amplitude = 5 - 2 = 3. The period is 2pi because the coefficient of x inside the function is 1. All four statements listed are correct and consistent with the standard form of the function.

The answer is True. The x-axis crossings of y = f(x) + D occur where f(x) = -D. As D changes, the value that f(x) must reach to cross the x-axis changes. If D is large enough that -D falls outside the range of f(x), no crossings exist in any interval. If -D is within the range, crossings appear. Shifting the graph vertically therefore directly determines whether and where x-axis crossings occur within any specific interval.

For y = -2cos(2x) + D, the amplitude is 2. The maximum value equals D + 2. Setting D + 2 = 1 and solving gives D = -1. Option B gives D = 3, which would produce a maximum of 5, not 1. Option C gives D = 1, making the maximum 3, not 1. Option D gives D = -3, producing a maximum of -1, not 1. Only D = -1 satisfies the condition that the maximum is 1.

The answer is False. The spacing between consecutive maxima along the x-axis is the period of the function. The period is determined by the coefficient B inside the argument of the trigonometric function, calculated as 2pi divided by B. Adding a vertical shift D moves every point up or down by the same amount but does not alter the horizontal positions of the maxima at all. The period therefore stays exactly the same after any vertical shift.

The minimum of y = 3sin(x) + D occurs when sin(x) = -1, giving a minimum value of D - 3. Setting D - 3 = -1 and solving gives D = 2. Option B gives D = -2, which would make the minimum -5, not -1. Option C gives D = 4, which would make the minimum 1, not -1. Option D gives D = -4, producing a minimum of -7. Only D = 2 satisfies the given condition.

The answer is True. A vertical shift adds the constant D to every output of the function without exception. This applies equally to the maximum and minimum values. If the original range is from y_min to y_max, the shifted range becomes y_min + D to y_max + D. The gap between maximum and minimum, which equals twice the amplitude, is preserved. Neither the amplitude nor the period is changed by adding D.

Adding 2 shifts the entire graph of y = sin(x) upward by 2 units. The midline moves from y = 0 to y = 2, the maximum rises from 1 to 3, and the minimum rises from -1 to 1. The amplitude, which is the distance from the midline to a peak, stays at 1 because the shift does not stretch or compress the graph. The period also remains 2pi and is not affected by adding a constant.

The maximum of y = 3cos(x) occurs when cos(x) = 1, giving a maximum value of 3. Adding D to the function shifts every output up by D, so the new maximum becomes 3 + D. Option B incorrectly multiplies D by 3, which would represent a change in amplitude, not a shift. Option C subtracts instead of adds. Option D ignores the shift entirely and leaves the maximum unchanged.

The answer is True. A vertical shift of D adds the same constant D to every output value of the function. Since the maximum is the largest output and the minimum is the smallest output, both increase by exactly D. The distance between them, which equals twice the amplitude, stays the same. This confirms that vertical shift translates the entire graph up or down without stretching or compressing it.