Thread: Graphing trig function

Hey guys!
So for an assignment, we're given some trig functions and asked to graph it, from it's base graph to it's final state: but i'm having problems with these two:

b) y = 3cos2x + 1 for 0 < x < 360°
d) y = -sin(x/3) - 2 for -pi/2 < x < pi/2

Solution for b):
So I know the base graph is y=cos(x), so the first graph I draw is y=cos(x). The second graph would be y=cos(x) but all the x-values are multiplied by 3. I don't know what to do with the 2x part. The last graph would be the whole thing, shifted up one unit.

Solution for d):
The base graph is y=sin(x), so the first graph to draw is sin(x). The second graph to draw is -sin(x), which is basically sin(x) flipped on the x-axis. I don't know what to do for the (x/3) part. The last graph would be the whole thing shifted downwards.

the parts i'm not sure what to do is the ones italicized above. Any help would be appreciated !

Originally Posted by snypeshow

Hey guys!
So for an assignment, we're given some trig functions and asked to graph it, from it's base graph to it's final state: but i'm having problems with these two:

b) y = 3cos2x + 1 for 0 < x < 360°
d) y = -sin(x/3) - 2 for -pi/2 < x < pi/2

Solution for b):
So I know the base graph is y=cos(x), so the first graph I draw is y=cos(x). The second graph would be y=cos(x) but all the x-values are multiplied by 3. I don't know what to do with the 2x part. The last graph would be the whole thing, shifted up one unit.

Solution for d):
The base graph is y=sin(x), so the first graph to draw is sin(x). The second graph to draw is -sin(x), which is basically sin(x) flipped on the x-axis. I don't know what to do for the (x/3) part. The last graph would be the whole thing shifted downwards.

the parts i'm not sure what to do is the ones italicized above. Any help would be appreciated !

HI

the 2x here implies that the sin graph will go 2 rounds in one cycle .

1/3 x implies that the sin graph will go 1/3 round (stop at x=120 degree) in one cycle .

take a look at the graphs i attached .

thanks alot for those graphs!!!

so i'm guessing all the y-values will be multiplied by the coefficient of x?

for b) the y-values would be multiplied by 2
for d) the y-values would be multiplied by 1/3

right?

Originally Posted by snypeshow

Hey guys!
So for an assignment, we're given some trig functions and asked to graph it, from it's base graph to it's final state: but i'm having problems with these two:

b) y = 3cos2x + 1 for 0 < x < 360°
d) y = -sin(x/3) - 2 for -pi/2 < x < pi/2

Solution for b):
So I know the base graph is y=cos(x), so the first graph I draw is y=cos(x). The second graph would be y=cos(x) but all the x-values are multiplied by 3. I don't know what to do with the 2x part. The last graph would be the whole thing, shifted up one unit.

Solution for d):
The base graph is y=sin(x), so the first graph to draw is sin(x). The second graph to draw is -sin(x), which is basically sin(x) flipped on the x-axis. I don't know what to do for the (x/3) part. The last graph would be the whole thing shifted downwards.

the parts i'm not sure what to do is the ones italicized above. Any help would be appreciated !

hey , for (d) i made a mistake , i din notice the range given . It should look sth like this .