I've got a diagram full of graphs that make upa picture. I'm trying to get all of the equations. There are linear graphs, quadratics and sine graphs. I need some help with the sine graph.
It starts at the (10,5) point, then does a ~ shape to end on (22,5). It only does the ~ process the one time.
I am unsure of which rule to use and what process to do to get the equation from this graph, but any help will be appreciated, and the first correct answer that helps will get the 10 points.
It starts at the (10,5) point, then does a ~ shape to end on (22,5). It only does the ~ process the one time.
I am unsure of which rule to use and what process to do to get the equation from this graph, but any help will be appreciated, and the first correct answer that helps will get the 10 points.
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This is one period of the sine graph, with p= 22-10=12
So b = 2pi/12 = pi/6
The horizontal axis is y= 5
The horizontal shift is right 10, since it starts at 10.
The only thing you haven't mentioned is how tall it is. The amplitude is 1 for the sine curve, so that from the horizontal axis, it goes up 1 unit and down 1 unit .
So if your sine curve is 2 units up and 2units down, for example, then the amplitude is 2.
Then the highest point would be (13, 7) and the lowest would be (19, 3). At (16,5) it is in line with the beginning and end.
also, since it is a piece of the sine curve, I would specify the domain.
The equation would be y= 2sin[(pi/6)(x-10)] +5, (10<=x<=22)
But set the asymptote to whatever height you need.( I saw your change :)
Hoping this helps!
***** set the coefficient to pi/6
You will get about 24 cycles if you set the coefficient to 12.
:)
So b = 2pi/12 = pi/6
The horizontal axis is y= 5
The horizontal shift is right 10, since it starts at 10.
The only thing you haven't mentioned is how tall it is. The amplitude is 1 for the sine curve, so that from the horizontal axis, it goes up 1 unit and down 1 unit .
So if your sine curve is 2 units up and 2units down, for example, then the amplitude is 2.
Then the highest point would be (13, 7) and the lowest would be (19, 3). At (16,5) it is in line with the beginning and end.
also, since it is a piece of the sine curve, I would specify the domain.
The equation would be y= 2sin[(pi/6)(x-10)] +5, (10<=x<=22)
But set the asymptote to whatever height you need.( I saw your change :)
Hoping this helps!
***** set the coefficient to pi/6
You will get about 24 cycles if you set the coefficient to 12.
:)
-
Period = 12
y = A sin((pi/6)(x-10)) + 5 , missing info about max or min value---> can not find A
A = 2
y = A sin((pi/6)(x-10)) + 5 , missing info about max or min value---> can not find A
A = 2