hey guys! I need some help with this:
Find the function of the form y = sin x + c that passes through the given
points:
- (225°, 3)
- (30°, -2)
i'm lost. all i know is 225° is 5pi/4 and 30° is pi/6 in radians
hey guys! I need some help with this:
Find the function of the form y = sin x + c that passes through the given
points:
- (225°, 3)
- (30°, -2)
i'm lost. all i know is 225° is 5pi/4 and 30° is pi/6 in radians
1. Function that passes through ( ,3)
Answer:
Explaination:
If you look at the normal sin graph:
You'll notice it reaches it peak at ( ,1)
The amplitude is basically the height = 1/2(max-min) so here its 1/2(1-(-1)) = 1.
You want it to reach to your y value 3, so if we look at a function, e.g. y = 3sin(x) the amplitude is always the absolute value of the # in front of the sin. That means it max is going to be 3, while its min is going to be -3.
y=3sin(x):
My x-scale is set to . is half-way between (where it crosses the x-axis) and We need to SHIFT this line to the right to get it to where we need.
So you answer will be
2. You will need to put a (-) in front of sin to flip the graph. y=-2sin(x+ )
Here is the graph, divided into 6ths.
thanks, but i don't think that's right. the equation needs to be y = sin x + c, so the c would be a vertical shift of the graph y = sin x
if you do y = sin(x) on your graphing calc, and you set the X-Scale to 5pi/4, you'll see a sin(x) graph, and the 5pi/4 point. The graph needs to pass through the point (5pi/4, 3). the mark on the x-axis will show you where 5pi/4 is, but the y = sin(x) graph needs to have a vertical shift (upwards) to pass through that point.
EDIT: okay, i solved the first one. the equation must pass through (5pi/4, 3) but the regular y=sin(x) graph at x = 5pi/4 is (5pi/4, -1/sqrt[2]). So if you add a vertical shift of 3 units upwards + 1/sqrt[2], the graph will pass through the point, (5pi/4, 3).
no no no lol you don't understand. the equation must be in the form of y = sin x + c. that means the coefficint of sin is 1, it cannot be anything else.
but i actually figured out the second one, logically.
- for the graph y = sin(x): when x = pi/6, y = 0.5
- this means there is a displacement of 0.5
- the equation needs to pass through (pi/6, -2)
- therefore, you need to do a vertical shift of (2 + displacement) * (-1 since its a downward shift)
so sin(x) - 2.5 will pass through the point (pi/6, -2)
it works because if you do sin(pi/6) you get 0.5
I'm going to thank you anyway because you took the trouble of actually graphing it on a ticalc emulator and posted it up. thanks for your help!!!