Q:

Which functions has the graph shown?

Accepted Solution

A:
Answer:C.Step-by-step explanation:Let's identify some points here that are on the graph:(0,0), (pi/2,-1), (pi,0).Let's see if this is enough.We want to see which equation holds for these points.Let's try A.(0,0)?y=cos(x-pi/2)0=cos(0-pi/2)0=cos(-pi/2)0=0 is true so (0,0) is on A.(pi/2,-1)?y=cos(x-pi/2)-1=cos(pi/2-pi/2)-1=cos(0)-1=1 is false so (pi/2,-1) is not on A.The answer is not A.Let's try B.(0,0)?y=cos(x)0=cos(0)0=1 is false so (0,0) is not on B.The answer is not B.Let's try C.(0,0)?y=sin(-x)0=sin(-0)0=sin(0)0=0 is true so (0,0) is on C.(pi/2,-1)?y=sin(-x)-1=sin(-pi/2)-1=-1 is true so (pi/2,-1) is on C.(pi,0)?y=sin(-x)0=sin(-pi)0=0 is true so (pi,0) is on C.So far C is winning!Let's try D.(0,0)?y=-cos(x)0=-cos(0)0=-(1)0=-1 is not true so (0,0) is not on D.So D is wrong.Okay if you do look at the curve it does appear to be a reflection of the sine function.