For a given input value q, the function f outputs a value r to satisfy the following equation.
1 answer:
To solve this, you re-arrange the equation to get r by itself:
3r - 6 = 11q - 4
3r = 11q + 2 <span>(add 6 to both sides to cancel out the -6)
</span>r =
<span>(divide both sides by 3 to get what r is)
</span>
This can also be written as: r =
<span>-------------------------------------------------------------------------------------------------------------
</span>ANSWER:
f(q) =
<span>OR
</span><span>
</span> f(q) =
OR
f(q) = (11q + 2) / 3 (if you can't put in fractions)
3r - 6 = 11q - 4
3r = 11q + 2 <span>(add 6 to both sides to cancel out the -6)
</span>r =
</span>
This can also be written as: r =
<span>-------------------------------------------------------------------------------------------------------------
</span>ANSWER:
f(q) =
<span>OR
</span><span>
</span> f(q) =
OR
f(q) = (11q + 2) / 3 (if you can't put in fractions)
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Short Answer: y = 3*cos(2x) + 1
Remark
This is not part of the answer, but it will help you to see what is going on. Begin by shifting the cos graph 1 unit down. That will that the cos has a minimum of -3 and a max of + 3
What that tells you is the part of the equation is
y = 3*cos(x)
Step One
Show that the graph is of something that resembles y = 3*cos(x)
the test way to check this is to put in 0 for x
3*cos(0) = 3*1 = 3. But why is it 4 and -2 instead of 3 and -3.
Step Two
Show how the graph is shifted up one space.
y movement is always recorded behind how the variable is determined.
So if the graph is shifting up one, you should do this.
y = 3cos(x) + 1
Step Three
The graph seems to be starting over at n*pi rather than n*2pi. How do we adjust for that?
There are 2 choices. Either there is a 4 in front of the x or there is a 1/2 in front of the x. Before just telling you, consider the graph below.
Violet is 3*cos(1/2 x)
Blue is 3*cos (x)
orange is 3*cos(2*x)
You want the graph that starts over again at x = 3.14 rather than at x = 6.28
The one that starts over at y = 3*cos(2x)
The rule is that if you want to compress a trigonometric function, use a constant such that a > 1 y = 3*cos(a*x). It is the a I'm trying to explain.
Step 4
Is there a phase shift?
No. If there was cos(x) would not have a maximum at x = 0
Answer y = 3*cos(2x) + 1
Remark
This is not part of the answer, but it will help you to see what is going on. Begin by shifting the cos graph 1 unit down. That will that the cos has a minimum of -3 and a max of + 3
What that tells you is the part of the equation is
y = 3*cos(x)
Step One
Show that the graph is of something that resembles y = 3*cos(x)
the test way to check this is to put in 0 for x
3*cos(0) = 3*1 = 3. But why is it 4 and -2 instead of 3 and -3.
Step Two
Show how the graph is shifted up one space.
y movement is always recorded behind how the variable is determined.
So if the graph is shifting up one, you should do this.
y = 3cos(x) + 1
Step Three
The graph seems to be starting over at n*pi rather than n*2pi. How do we adjust for that?
There are 2 choices. Either there is a 4 in front of the x or there is a 1/2 in front of the x. Before just telling you, consider the graph below.
Violet is 3*cos(1/2 x)
Blue is 3*cos (x)
orange is 3*cos(2*x)
You want the graph that starts over again at x = 3.14 rather than at x = 6.28
The one that starts over at y = 3*cos(2x)
The rule is that if you want to compress a trigonometric function, use a constant such that a > 1 y = 3*cos(a*x). It is the a I'm trying to explain.
Step 4
Is there a phase shift?
No. If there was cos(x) would not have a maximum at x = 0
Answer y = 3*cos(2x) + 1