We have dealt with three ways to change a trig function.
1) Amplitude
f(x) = asinx or g(x) = acosx
This changes the height of the graph. Instead of oscillating between -1 and 1, the high and low points will be |a| and -|a|. (It could be that a is negative, so we need the absolute value signs.)
2) Frequency (or period)
f(x) = sinpx or g(x) = cospx
This changes the distance between the high points. For sine and cosine, the period is 2pi. This changes to |2pi/p|. If |p| > 1, it takes a shorter distance to repeat. If |p| < 1, the wave is stretched and takes a longer distance to repeat.
3) Adding a constant
f(x) = c + sinx or g(x) = c + cosx
This makes a graph fluctuates between c - 1 and c + 1. If there is an amplitude multiplier a, the fluctuation is between c - |a| and c + |a|.
The last way to change a trig function is called phase shift.
f(x) = sin(px+h) or g(x) = cos(px+h)
The "important" points in a trig function happen at multiples of ½pi.
sinx = 0 when x = ...-2pi, -pi, 0, pi, 2pi...
sinx = 1 when x = ... -3½pi, -1½pi, ½pi, 2½pi, 4½pi....
sinx = -1 when x = ... -2½pi, -½pi, 1½pi, 3½pi, 5½pi....
cosx = 0 when x = ...-1½pi, -½pi, ½pi,
1½pi...
cosx = 1 when x = ... -2pi, 0, 2pi, 4pi....
cosx = -1 when x = ... -3pi, -pi, pi,
3pi, 5pi....
If we want to look at multiples of ½pi, we let k be any integer and consider kpi/2.
px+h = kpi/2
px = kpi/2 - h
x = (kpi/2 - h)/p
These will be the values of x where the graph will be either in the middle or at the maximum or minimum. The easiest thing to do is plug in k = 0, k = 1 and k = 2 in a row to see what you get. That means
x = - h/p
x = (pi/2 - h)/p
x = (pi - h)/p
There are four possible situations you will get plugging these numbers in.
Middle, high, middle
High, middle, low
Middle, low, middle
Low, middle, high
Once you have three such points, you can draw the entire graph.
For the following functions, find three points in a row where the function either reaches an extreme value or a median value.
f(x) = -3 + 2cos(pi(x + ¾))
g(x) = -4sin(2x - 5)
k(x) = cos(2 - 3x)
Answers in the notes.
f(x) = -3 + 2cos(pi(x + ¾))
ReplyDeletePoint # 1: pi(x + ¾)= 0
x = -¾
f(-¾) = -1
Point # 2: pi(x + ¾)= pi/2
x = -¼
f(-¼) = -3
Point # 3: pi(x + ¾)= pi
x = ¼
f(¼) = -5
These three points are high, middle and low, so we know the pattern.
g(x) = -4sin(2x - 5)
Point # 1: 2x - 5 = 0
x = 2½
g(2½) = 0
Point # 2: 2x - 5 = pi/2
x = 5/2 + pi/4
g(5/2 + pi/4) = -4
Point # 3: 2x - 5 = pi
x = (5 + pi)/2
g((5 + pi)/2) = 0
These three points are middle, low and middle, so we know the pattern.
k(x) = cos(2 - 3x)
Point # 1: 2 - 3x = 0
x = 2/3
k(2/3) = 1
Point # 2: 2 - 3x = pi/2
x = 2/3 - pi/6
k(2/3 - pi/6) = 0
Point # 3: 2 - 3x = pi
x = (2 - pi)/3
g((2 - pi)/3) = -1
Counting from point #1 to point #3, we get high, middle, low, but notice that the points are positioned from right to left instead of left to right.
I'm not getting it. When do we use the formula kpi/2 ???
ReplyDeleteI am not understanding how the answers: -4sin(2x-5).......g((5/2+pi/4)) = -4
How is g((5+pi/2))=0 ??.?.? And the last problems three points as well.....
Instead of looking at ALL values of k, we look at k=0, k=1 and k=2.
ReplyDeleteg(x) = -4sin(2x-5)
If 2x-5 = 0, then sin(2x-5) = 0,
If 2x-5 = pi/2, then sin(2x-5) = 1 and -4sin(2x-5) = -4
If 2x-5 = pi, then sin(2x-5) = 0.
Solve
2x - 5 = 0 => x = 5/2
2x - 5 = pi/2 => 5/2 + pi/4
2x - 5 = pi = 5/2 + pi/2, which I wrote at (5+pi)/2
Be careful with parentheses.