SOLUTION: Can someone show me a graph that represents y = 3 sec x?

Question 86087: Can someone show me a graph that represents y = 3 sec x?
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!

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This is the graph you are looking for. Some explanation is needed. The nearly vertical
lines on the graph are not part of the graph of 3*sec(x). They are asymptotes that the graph
approaches but never reaches.
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The actual graph of 3*sec(x) are the curved parts that look somewhat like parabolas.
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The x-axis needs to be translated to units of pi. [I could not get the graph maker to label
the x-axis as angles.] To calibrate the x-axis as angles, let's start at the origin and
work to the right. 0 is the first point, and it represents 0 degrees or 0 radians.
Recall that 90 degrees is and . So you can label
the point on the x-axis where x = 1.5708 as 90 degrees or radians. Similarly,
.

.
which means that where x = 3.1416 you can label that point as 180 degrees or radians.
.
And furthermore:
.

.
which means that where x = 4.7124 you can label that point on the x-axis as 270 degrees or
radians.
.
Finally,
.

.
so you can label the point 6.2832 on the x-axis as or 360 degrees. Continue
this pattern as far as you want to the right.
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Then you need to do it in the negative x direction. The spacing of the critical values is
the same except for the fact that the values are negative. And you can fill in intermediate
values if you would like. For example 45 degrees is and it occurs where
x is equal to 3.1416 divided by 4 which is at x = 0.7854.
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A way you can create the secant graph is to draw the cosine graph. Then recognize
that the secant function is the inverse of the cosine function. Pick an angle on the
horizontal axis of the cosine graph. Go up to the corresponding value on the cosine curve.
Read that value and divide it into 1 to get the secant. For example, at 45 degrees,
the value of the cosine is 0.7071. Divide that into 1 and you get that the secant at
45 degrees is . Note that this will not be correct on the graph pictured
above because it is the graph of 3*sec(x), not of sec(x). It will show on the pictured
graph as a value of 3*1.4142 or 4.2426 as read off the y-axis. Another thing to check is
that the cosine has a value of zero at 90 degrees and at 270 degrees. At those two angles
the secant is, therefore . But division by zero is not allowed. Therefore
at 90 degrees and at 270 degrees there is a vertical asymptote.
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You can reconstruct the secant graph by following this process. For various angles
find the corresponding value of the cosine and divide that value into 1 to find the secant
at that angle. [Since your problem asks for 3*secant you will have to multiply each of
your secant values by 3.]
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Hope this helps you to understand the problem and to visualize the graph. Sorry for the
complex discussion, but hang with it. You'll get the idea.

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