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suter [353]
2 years ago
10

The graph of which function passes through (0,4) and has a minimum value at (StartFraction 3 pi Over 2 EndFraction, 3)? f (x) =

sine (x) + 4 f (x) = cosine (x) + 3 f (x) = negative 3 sine (x) f (x) = 4 cosine (x)

Mathematics
2 answers:
Nikitich [7]2 years ago
8 0

Answer:

<h2>f(x) = sin(x) + 4</h2>

Step-by-step explanation:

The right choice is the first one.

You can rapidly get the answer by identifying the interception with the vertical axis.

The problem specifies that the functions passes through (0,4), which is a vertical interception. So, vertical interceptions appears in the equations as an independent term, and the only function among options that has an independent term of 4 is the first choice.

In addition, the graph of the function is attached, there you could observe the interception point with the vertical axis.

solniwko [45]2 years ago
7 0

Answer:

f(x) = sin(x) + 4  Let me know if you don't see why.

Step-by-step explanation:

As much as I hate it I think it might be best to check each one.  

sin(x) has a point at (0,0) so that's not right.

sin(x)+4 has (0,4) as a point and a minimum at (3*pi/2+2*pi*n,3) where n is some integer.  if we have n = 0 then it becomes (3*pi/2, 3) So that looks like our answer.

cos(x) + 3 has a point at (0,4) then minimums at (pi+2*pi*n, 2)  the y coordinate is wrong so this won't work

-3sin(x) has a point at (0,0) so that's wrong

4cos(x) has a point at (0,4) then minimums at (pi+2*pi*n, -4) which again has the wrong y value so this is wrong.

Let me know if you don't understand how I got the results I did, I would be happy to explain.  

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