Question

graph y=5/2cos(x-pi/4) and plot x-intercepts, maxima, minima for one period

Answer 1

To find the x-intercepts: When does the cosine function become zero?

Answer 2

set y=0?

Answer 3

Well, in one period, y = cos(x) becomes zero when x = pi/2 and 3pi/2
Therefore, y=5/2cos(x-pi/4) becomes zero when x - pi/4 = pi/2 and x - pi/4 = 3pi/2 or x = pi/2 + pi/4 or x = 3pi/2 + pi/4.
x = 3pi/4 or x = 7pi/4. These are the x-intercepts.

At what x does y = cos(x) attain the maximum and minimum?

Answer 4

the maximum between 3pi/4 and 7pi/4. and the minimum before 3pi/4?

Answer 5

y = cos(x) attains its maximum at x = 0 and minimum at x = pi.
Therefore, y = 5/2cos(x-pi/4) attains its maximum at x-pi/4 = 0 and minimum at x - pi/4 = pi. Solve for x.
The max value (that is, the y value) will be +5/2. The min value will be -5/2.

Answer 6

so if i plotted the min and max on a graph would it be (0, 5/2) for the max and (pi, -5/2) for the min? thank you so much by the way :)

Answer 7

No. y = cos(x) attains it max at x = 0 and min at x = pi.
y = 5/2cos(x-pi/4) attains its maximum when x-pi/4 = 0 and
attains minimum when x - pi/4 = pi. You need to solve for x.
max at x = pi/4 and min at x = pi + pi/4 = 5pi/4.
Max: (pi/4, 5/2)
Min: (5pi/4, -5/2)

Answer 8

and this would all be within the period [0, 2pi]?

Answer 9

Yes. pi/4 = 0.25pi which is in [0, 2pi]
5pi/4 = 1.25pi which is in [0, 2pi]

Answer 10

0 < 0.25pi < 2pi
0 < 1.25pi < 2pi

Answer 11

perfect. would you mind helping me graph it? i'm still a little confused on how it should be graphed?

Answer 12

If you know the graph of y = cos(x), you can easily graph y = 5/2cos(x-pi/4).
The first graph goes from -1 to +1. The second graph goes from -5/2 to +5/2.
Since the second graph has x - pi/4, it would be like shifting the first graph to the RIGHT by pi/4.

Answer 13

thank you. where would pi and 2pi be on the graph?