The graph of a sinusoidal function intersects its midline at (0,5) and then has a maximum point at (\pi,6)
1 answer:
First, let's use the given information to determine the function's amplitude, midline, and period.
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
<span>= Solution </span>
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
You might be interested in
Step-by-step explanation:
hi im not really sure if this is right
The limit as a definite integral on the interval on [2π , 4π] is
.
<h3>What is meant by definite integral?</h3>
A definite integral uses infinitesimal slivers or stripes of the region to calculate the area beneath a function. Integrals can be used to represent a region's (signed) area, the cumulative value of a function changing over time, or the amount of a substance given its density.
Definite integral, a term used in mathematics. is the region in the xy plane defined by the graph of f, the x-axis, and the lines x = a and x = b, where the area above the x-axis adds to the total and the area below the x-axis subtracts from the total.
If an antiderivative F exists for the interval [a, b], the definite integral of the function is the difference of the values at points a and b. The definite integral of any function can also be expressed as the limit of a sum.
Let the equation be
substitute the values in the above equation, we get
= on [2π, 4π],
simplifying the above equation
To learn more about definite integral refer to:
#SPJ4