# Thread: Where the Graphs of two different equations intersect

1. ## Where the Graphs of two different equations intersect

The Problem: I am given the graph $y = sin x$ and $5 *\pi *y = 2x$

I am ask to find how many different points they intersect. I am literally blank when it comes to this question. I know solving for the system would give me one point where they intersect, but aside from graphing I have no idea as to how I would solve this.

Thank you.

2. solution by graphing is all you can do.

$\sin{x} - \frac{2x}{5\pi} = 0$

look for the roots.

3. Originally Posted by D. Martin
The Problem: I am given the graph $y = sin x$ and $5 *\pi *y = 2x$

I am ask to find how many different points they intersect. I am literally blank when it comes to this question. I know solving for the system would give me one point where they intersect, but aside from graphing I have no idea as to how I would solve this.

Thank you.
Another way is by iteration.
For people like me who doesn't know how to use any graphing calculator, in graphing, and who doesn't know either how to graph in computers, the iteration method is a way.
Search the Web for the Newton's Method, or the Newton-Raphson method for iteration, if you know some Calculus.

Iteration is repetition. Trial and error. But there is a method whereby you know or can guess what value to try as you do repeated trials.

Using my own iteration method,
y = sin(x)
y = 2x / 5pi
----------------
sin(x) = 2x /5pi
sin(x) -[2x /(5pi)] = 0 ----------try some x-values to get to zero.

sin(x) is from -1.0 to 1.0 only, so try x's that would give (2x / 5pi) within (-1.0 to 1.0) only.

I was getting x = 7.6 when I stopped iterating.

4. Originally Posted by ticbol
Another way is by iteration.
For people like me who doesn't know how to use any graphing calculator, in graphing, and who doesn't know either how to graph in computers, the iteration method is a way.
Search the Web for the Newton's Method, or the Newton-Raphson method for iteration, if you know some Calculus.

Iteration is repetition. Trial and error. But there is a method whereby you know or can guess what value to try as you do repeated trials.

Using my own iteration method,
y = sin(x)
y = 2x / 5pi
----------------
sin(x) = 2x /5pi
sin(x) -[2x /(5pi)] = 0 ----------try some x-values to get to zero.

sin(x) is from -1.0 to 1.0 only, so try x's that would give (2x / 5pi) within (-1.0 to 1.0) only.

I was getting x = 7.6 when I stopped iterating.
indeed. but that is going to be REALLY painful here. there are 7 zeros (see graph below, the zeros that look like double roots are actually 2 roots), so to do an iteration for each is overkill. plus we have two pairs of roots where the zeros are very close to each other. that will make the guessing confusing

5. Originally Posted by Jhevon
indeed. but that is going to be REALLY painful here. there are 7 zeros (see graph below, the zeros that look like double roots are actually 2 roots), so to do an iteration for each is overkill. plus we have two pairs of roots where the zeros are very close to each other. that will make the guessing confusing
You only know those because you know how to graph on the computers and/or the graphing calculators. :-)

6. Originally Posted by ticbol
You only know those because you know how to graph on the computers and/or the graphing calculators. :-)
exactly my point. it would take me forever to find them without this. this problem was probably made to be solved with some kind of CAS

7. Originally Posted by Jhevon
exactly my point. it would take me forever to find them without this. this problem was probably made to be solved with some kind of CAS
Okay, let's prolong this a bit.

I did not say everything in my first reply to you, but the two curves,
y = sin(x) ---------------a sine curve of amplitude 1.0
and y = (2 / 5pi)x --------a straight line with positive slope,
may not have seven intersection points.

In fact, I guess there could only be one intersection point.

Is that the graph of the y = sin(x) shown in your post? It's neutral axis is sloping downwards?
If so, then, it is not okay.
A sin curve, the basic one, has horizontal neutral axis.

8. Originally Posted by ticbol
Okay, let's prolong this a bit.

I did not say everything in my first reply to you, but the two curves,
y = sin(x) ---------------a sine curve of amplitude 1.0
and y = (2 / 5pi)x --------a straight line with positive slope,
may not have seven intersection points.

In fact, I guess there could only be one intersection point.

Is that the graph of the y = sin(x) shown in your post? It's neutral axis is sloping downwards?
If so, then, it is not okay.
A sin curve, the basic one, has horizontal neutral axis.
no, i graphed the function $\sin x - \frac {2x}{5 \pi}$. the zeros of that graph are the x-values we seek

the graph of both functions separately is below

9. Originally Posted by Jhevon
no, i graphed the function $\sin x - \frac {2x}{5 \pi}$. the zeros of that graph are the x-values we seek

the graph of both functions separately is below
Umm, now those graphs look familiar.
So there really are 7 intersection points, according to exact graphings?

Then, yes, to get all of the possible intersection points, computer assisted graphings is necessary.