What is the answer for question 3) ?

http://postimg.org/image/grjobmz2f/

(Reminder): this is 1 question.

### 1 Answer | Add Yours

(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.)

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes