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On the graph of f(x)=3sin(2x), points P and Q are at consecutive lowest and highest points with P occuring before Q. Find the slope of the line which passes through P and Q.

User Icon for MathMate MathMate answered
13 years ago

We will start with the general case of the function sin(x).

The minima (lowest points) of the graph of sin(x) occurs at
x=xmin=3Ο€/2+2kΟ€ where kβˆˆβ„€ (i.e. k=integer)

The maxima (highest points) of sin(x) occurs at
x=xmax=Ο€/2+2kΟ€ where kβˆˆβ„€.

Two consecutive minimum/maximum could therefore occur at xmin=3Ο€/2 and xmax=5Ο€/2.

The given function is 3sin(2x), so
2x=3Ο€/2, or x1= 3Ο€/4 for minimum.
The ordinate at this point is
f(x1)=3sin(2*3Ο€/4)=-3
Therefore x1(3Ο€/4,-3).
and
2x=5Ο€/2, or x2= 5Ο€/4 for maximum.
The ordinate at this point is
f(x2)=3sin(2*5Ο€/2)=3
Therefore x2(5Ο€/4,3)

The slope is therefore
m=(y2-y1)/(x2-x1)
=(3-(-3))/(5Ο€/4-3Ο€4)
=3.82

Check my calculations.

User Icon for Explain Bot Explain Bot answered
11 months ago

To find the slope of the line passing through points P and Q on the graph of f(x) = 3sin(2x), the first step is to determine the x-coordinates of points P and Q.

Points P and Q occur at consecutive lowest and highest points of the function f(x) = 3sin(2x). The general formula for the x-coordinate of the highest and lowest points of the sine function is given by: x = (2n + 1)(Ο€/2), where n is an integer.

Since P occurs before Q, we can take n = 0 for P and n = 1 for Q.

For P: xP = (2*0 + 1)(Ο€/2) = Ο€/2
For Q: xQ = (2*1 + 1)(Ο€/2) = (3Ο€/2)

Next, we need to find the corresponding y-coordinates for P and Q. We can substitute the x-coordinates into the function f(x) = 3sin(2x) to find the y-values.

For P: f(Ο€/2) = 3sin(2(Ο€/2)) = 3sin(Ο€) = 0
For Q: f(3Ο€/2) = 3sin(2(3Ο€/2)) = 3sin(3Ο€) = 0

Since both P and Q have y-coordinates of 0, the line passing through points P and Q is a horizontal line. The slope of a horizontal line is always 0.

Therefore, the slope of the line passing through P and Q is 0.