Part 1The function graphed is of the form y equals a sine bxy=asinbx or y equals a cosine bxy=acosbx, where b 0. Determine the equation of the graph. Part 1
-44xy00piπ2 pi2π Edit coordinates (0,0) A coordinate system has a horizontal x-axis labeled from 0 to 2 pi in increments of StartFraction pi Over 2 EndFraction and a vertical y-axis labeled from negative 4 to 4 in increments of 1. A curve rises from a minimum at (0, negative 2) to a point at (2 pi, 0). Part 1yequals=[input]wrong wrong (Use integers or fractions for any numbers in the expression.)

Question

Part 1The function graphed is of the form y equals a sine bxy=asinbx or y equals a cosine bxy=acosbx​, where b​ > 0. Determine the equation of the graph. Part 1
-44xy00piπ2 pi2π Edit coordinates (0,0) A coordinate system has a horizontal x-axis labeled from 0 to 2 pi in increments of StartFraction pi Over 2 EndFraction and a vertical y-axis labeled from negative 4 to 4 in increments of 1. A curve rises from a  minimum at (0, negative 2) to a point at (2 pi, 0). Part 1yequals=[input]wrong wrong ​(Use integers or fractions for any numbers in the​ expression.)

BlackTom AI · Accepted Answer

We are told the graph is of the form y = a sin(bx) or y = a cos(bx) with b > 0, and we have specific features: at x = 0 the graph has a minimum value of -2, and at x = 2π the graph crosses the y-value 0 (i.e., y = 0).

First, examine the possibility using a cosine form. If we take y = a cos(bx), then at x = 0 we have y = a cos(0) = a. The graph shows a minimum value of -2 at x = 0, so a must be -2 to satisfy y(0) = -2.

Thus a cosine model consistent with the starting point is y = -2 cos(bx).

Next, use the information at x = 2π. Substituting, y(2π) = -2 cos(b · 2π) = 0. This gives cos(2πb) = 0.

The cosine function is zero at angles π/2 + kπ, where k is an integer. Therefore 2πb = π/2 + kπ. Solving for b, we get b = (1/4) + k/2. The smallest positive value that makes sense for a frequency parameter is b = 1/4 (taking k = 0).

With b = 1/4, the model becomes y = -2 cos((1/4)x) = -2 cos(x/4).

Now consider the sine possibility briefly. If the graph were y = a sin(bx), the value at x = 0 would be y(0) = a sin(0) = 0, which contradicts the observed minimum of -2 at x = 0. Therefore a sine form does not fit the given starting point, reinforcing the cosine form as the appropriate choice.

In summary, using the given starting minimum at x = 0 and the zero crossing at x = 2π leads to the cosine model with a = -2 and b = 1/4, yielding the equation y = -2 cos(x/4).

Homework:Section 6.3 Homework

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