What is the answer for question 3) ?
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(1) y=sinx+c through `(225^@,3)`
Since `sin225^@=-1/sqrt(2)` , and the only transformation is a vertical shift, `c=3-(-1/sqrt(2))=3+1/sqrt(2)`
(2) y=cosx+c through `(30^@,-2)`
Since `cos30^@=sqrt(3)/2` , and the only transformation is a vertical translation, `c=-2-sqrt(3)/2`
(3) y=atanx through `(pi/6,-sqrt(3))`
Since `tan(pi/6)=1/sqrt(3)` , and the only transformation is a vertical stretch, we have `a*1/sqrt(3)=-sqrt(3)==>a=-3` .
(4) y=sin(x-d) through `(270^@,1/2)`
The sine is 1/2 at `30^@,150^@` . There are an infinite number of possible phase shifts. `-120^@+n360` or ` ` `120^@+n360` .(for integer n.)
For example `y=sin(x-120)` will work.
(5) y=cos(x-d) through `(pi/4,-1/sqrt(2))`
Again there are an infinite number of possible phase shifts. `d=pi/2+2npi,d=pi+2npi` will work for integer n.
For example `y=cos(x-pi)` will work.
(6) y=tankx through `(290^@,-1)`
The k affects the period. The tangent is -1 at `135^@` and `315^@` .
Let `290k=135==>k=135/290=27/58` . (There are other values for k.)
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