finding equation of curve from simple graph

hurliehoo

Hi can anyone please tell me how to find an equation from a graph.

It's a fairly simple graph. A symmetrical curve with a maximum at y=0.5, and y=0 at x=0 and x=7.5 ... any suggestions appreciated!

arildno

Generally, it is impossible to find a unique equation describing some particular graph.

What you might do, is to construct equations that show SOME of the properties of the graph.

You say you have a symmetric curve with maximum at (0,0.5) and a zero at (7.5,0).

Well, is it a parabola, perhaps? I don't know. It might as well be the graph of a fourth-order polynomial, or some other function.

If it IS the graph of a parabola, i.e, that it is a graph of a second-order polynomial, we might make some headway:

1. We have that:
[tex]y=ax^{2}+bx+c[/tex], and we are to determine a,b,c.

From your info, we have that the maximum occurs at x=0.
Since, in general, we have that the maximum will occur at [tex]x=-\frac{b}{2a}[/tex], we see that b=0 in your case.

2. The zero:
We know this occurs at x=7.5, therefore we have, using the info in (1):
[tex]0=a*7.5^{2}+c\to{c}=-a*7.5^{2}[/tex]

3. Preliminary expression:
We now have that:
[tex]y=a*(x^{2}-7.5^{2})[/tex]

4. Value at maximum:
We have that at x=0, y=0.5, we therefore get:
[tex]0.5=-a*7.5^{2}\to{a}=-\frac{0.5}{7.5^{2}}[/tex]

5. Final expression
We have now strictly determined how a second-order polynomial that fullfills all the given info looks like, namely, for example:
[tex]y=-\frac{1}{112.5}(x-7.5)(x+7.5)[/tex]

Mentallic

arildno you've misread the info. Assuming it's a parabola, it's given that the zeroes are x=0 and x=7.5 and the maximum is at y=0.5
So far we can conclude it will be of the form y=kx(x-7.5) for some constant k. Since y=0.5 is a maximum and this occurs when x is in between the roots 0 and 7.5 - because parabola's are symmetric about their turning point, we can substitute in x=3.75 and y=0.5 to find k.

hurliehoo

Sorry I didn't make it clear the midpoint maximum for this symmetrical curve is at (3.75, 0.5).

Yes it's a parabola though, and your answer is very helpful so thank you very much indeed!

arildno

Oh, you are right! Me being overhasty, contradicting my entish motto..

Anyhow, it seems that the OP has gotten the gist of the idea..

hurliehoo

Sorry I'm having another look at this and I'm not so sure I understand it now. Some questions about the method ...

What does the arrow signify ?

How can 0.5 = -0.5/7.5^2 ?

Mentallic

The arrow is just like another line, you're re-arranging the equation.

So rather than

[tex]x+2=5[/tex]
[tex]x=3[/tex]

We instead have

[tex]x+2=5 \rightarrow x=3[/tex]


It's not, take a look at the arrow

hurliehoo

Ok in that case I am definitely doing something wrong here. Using my understanding of the above method :

1--> max occurs at (3.75, 0.5). Therefore if x = -b / 2a then b = -7.5a

2--> 0 = (7.5^2)a - 7.5a + c therefore c = 7.5a - (7.5^2)a

3--> y = a((x^2) - (7.5^2) + 7.5)

4--> 0.5 = a((3.75^2) - (7.5^2) + 7.5)

so a = -.0144

and I can then get b and c from there, however this seems to be completely wrong.

Mentallic

What's this line? Particularly, what is the a in front doing there?

The equation is [tex]y=ax^2+bx+c[/tex]

and you've already found that
[tex]b=-7.5a[/tex]
and
[tex]c=7.5a - (7.5^2)a[/tex]
[tex]c=7.5(1-7.5)a[/tex]
[tex]c=7.5(-6.5)a[/tex]
[tex]c=-48.75a[/tex]

So now, knowing that the point (3.75,0.5) satisfies the equation, plug this into the general form of the equation and use these values of b and c to find a.

[tex]y=ax^2+bx+c[/tex]

[tex]0.5=a(3.75^2)-7.5a(3.75)-48.75a[/tex]

You would then solve to find a, then use another point like (0,0) to find c (which would be 0) and then find b with (7.5,0).

This is a pretty tedious method though. You already know that the parabola cuts the x-axis at 0 and 7.5 so you should instantly turn it into a factored parabola of the form [tex]y=kx(x-7.5)[/tex] for some constant value k, which you can easily find by plugging in the other known point (3.75,0.5).
Note that plugging in (0,0) or (7.5,0) won't work because then you end up with 0=0 and that doesn't help find the value of k.

hurliehoo

Regarding my first working, which I posted, of the first (agreed tedious) method I forgot to include the x for b (duh), however the next time I tried I got a similar working to yours ... this didn't work due though, I think because c=-48.75a makes it non-zero. So even when c=0, which it has to for the other two coordinates, I got y = -0.00796x^2 + 0.0597x, which when plugging in x=3.75 gives y=0.11194 instead of 0.5

Anyway ... your second method was infinitely simpler and does indeed work very well. Thanks!

Mentallic

Oh yes that's right, I totally missed that part too!

We found that b=-7.5a and used the point (7.5,0)

[tex]y=ax^2+bx+c[/tex]

[tex]0=a(7.5)^2+(-7.5a)(7.5)+c[/tex]

[tex]c=0[/tex]

We screwed up the bx part, we substituted b=-7.5a but forgot about the x=7.5 haha ^^

Yeah, better if you stick to the simpler method... saves you easily making an inevitable mistake like we both did