# math: find an equation from a graph

Discussion in 'Homework Help' started by bug13, Apr 29, 2012.

1. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
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as seen in the pic, how do I find the equation from the graph, I am lost with these two graph.

thanks

Feb 5, 2010
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951
3. ### blah2222 Well-Known Member

May 3, 2010
554
33
(ii)

You have four pieces of information to use:

1) y = 0 at x = 0
2) y = 0 at x = 3
3) dy/dx = 0 at x = 0
4) dy/dx = 0 at around x = 2.5

Since y = 0 at x = 0, 3, they must be factors of y and in the form:

$y = Ax^P(x - 3)^Q$

dy/dx must be in the form of:

$\frac{dy}{dx} = Bx^{P - 1}(x - 2.5)^R$

(iv)

Right off the bat this looks like an exponential graph but flipped in the x-axis, so with that in mind it has the form:

$y = Ae^{-x} + C$

At x = 0, y = 2 so this concludes that A could be 2 and C = 0:

$y = 2e^{-x}$

Testing y(3) we get 0.0996 which is pretty much 0.1 and there you have it.

bug13 likes this.
4. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
1,208
38
Thanks blah2222, your solution gives me a new way to look at this problem, I now learn I can use information like these:

That's something I have never thought off before

Feb 13, 2012
1,208
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6. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
My first impression is that it is a cubic expression of the form:

$
y = ax^3+bx^2+cx+d
$

And you know the following:

y goes to +oo as x goes to -oo
y goes to -oo as x goes to +oo
y' = dy/dx = 0 at x = 0
y = 0 at x=0
y = 0 at x=3

Since y(0) = d, we know that d = 0.

Since y'(0) = 0, we know that c = 0.

Hence we are left with

$
y = (ax+b)x^2
$

From y(3) = 0, we know that b = -3a. so now we have

$
y = a(x-3)x^2
$

We also know that a<0 in order to get the tails to go in the right direction.

Is this consistent with the answers offered in the book?

Assuming this is correct, the slope goes to zero at x=0 and x=2 (not x=2.5).

Since there is no scaling information on the y-axis, we have to leave 'a' as an arbitrary scale factor that must be negative.

If I were more up to speed on cubics, I could probably have noted that there is a root at +3 (that much was obvious) but that there was a double root at zero (it was obvious there was one, but I couldn't remember if it required two there). I suspected that there had to be a double root at either 0 or 3 because if there weren't, then it would have meant that the other root had to be complex. But, IIRC, on a real function the complex roots have to come in complex conjugate pairs and since there are only three roots and we know two of them, then this isn't possible. Given that the third root either has to be a double at 0 or 3, I suspected it was at zero.

But hopefully you can see how you can still get the form of the equation by looking at the zero crossings and the asymptotic behavior of both the function and the derivatives to figure things out without having to recall a bunch of specifics about roots and such.

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7. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
1,208
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That's the answer I get too,
, and then I am stucked, I don't have the answer yet, it's an per exam exercise, I will need to wait until Monday to ask my tutor for the answer.

I assumed I will need to get the value of a too, but I am out of ideas.

Thanks for your answer, I have never come across the idea of double root, that's something learned

8. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
You can't get the value of 'a' beyond noting that it must be negative. The sign of 'a' determines whether the graph you get is the one on the page or one that is mirrored about the x-axis. But the magnitude of 'a' determines the vertical scaling and there is absolutely no vertical scaling information given in the graph. You don't know if the peak at x=2 occurs at a y value of 1 or 1000 or 0.0001. Given that value, you can determine the value of 'a', otherwise it has to remain a negative constant of arbitrary magnitude.

9. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
1,208
38
The answer I got from my tutor is:

Y=-x^2(x-3)

But I think it's incomplete

10. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
Yes, arguably it is incomplete because this equation peaks at a specific value, namely 4, at x=2 while the graph peaks at an arbitrary and unknown value. In many respects, it is like saying that the (indefinite) integral of 2x is x^2 and not x^2+C. Forgeting the arbitrary scaling factor goes unnoticed for many people, but doing so is wrong and can get you into trouble.

11. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
1,208
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My plan is stick to my own answer but not to argue with my tutor

12. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
Good plan! Sure wish I had been smart enough to follow that route when I was at that point in my life.

13. ### steveb Senior Member

Jul 3, 2008
2,433
469
The answer is not incomplete. It is the question that is incomplete. The question needs to be clear about assumptions for the function. However, the assumptions that lead to this answer are very reasonalbe ones to make in the absence of a rigorously clear question.

Last edited: Apr 30, 2012
14. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
I'm gonna disagree with you on this one. The question may be incomplete, but the answer still needs to match the question. If someone asks me how much gasoline they need to drive 1000 miles in their car, I can tell them that they need 1000mi/M where M is their car's gas mileage. It would not be correct for me to arbitrarily set "M equal to 1" (1mpg, that is) and say that the "complete answer" is 1000 gallons. Or, using the example I used earlier, if the question is to take the integral of 2x (perhaps stated something like, "What function of x has a slope that is everywhere equal to 2x?", an answer of x^2 can't be claimed to be a "complete answer" to an incomplete question. If the question is defined only to within an arbitrary constant, then the answer needs to reflect the existance and influence of that constant.

15. ### steveb Senior Member

Jul 3, 2008
2,433
469
I can find many other answers to the question that are not the same as the one you suggested. So, your answer does not match the question any better than the tutor's does. You have made specific assumptions about the question and hence feel your answer is complete. The tutor did the same. The fact that your answer is more general than the tutor's does not mean that someone else does not have an answer that is more general than yours.

The second question with the exponential decay is also wide open to interpretation without some constraints, or more information given.

For the second one, I would suggest the following, but there are alternatives.

$y={{2}\over{20^{x/3}}}$

16. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
I'll grant you that. In any question like that there is virtually always the unstated goal of using the "simplest" solution possible, but as I remarked in another thread, what is the metric by which "simple" is judged? Having said that, the heart of most of these types of questions is whether or not the person can make the "correct" initial assumption about the basic form of the underlying equation.

I'm not sure I'll grant you that, but I definitely understand your point.

That's what I got, though I may have come at it from a different direction. I started with a generic decaying exponential and fit the two points. The result reduces to your solution. That same solution can reasonably be arrived at by inspection.

One thing that I think is reasonable to expect from any proposed answer to most questions like this (and I'm referring to an academic setting such as this) is that the anwers match the information that IS given exactly. Not almost, but exactly. Also, I think it is pretty important not to infer information that is not there as part of the solution. That last statement is almost certainly too broad to capture what I mean, so what I mean by that is doing something like saying that the zero crossing on a graph appears to be about 2/3 of the way from 4 and 7 and so we will impose the constraint that y(6)=0.

17. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
1,208
38
the answer given by my tutor is:

he starts off with the assumption of it's exponential equation with a base of e, however, I personally like the answer of:

this answer makes more sense to me, can you explain how did you get this answer please? thanks

Wu

PS: my tutor's solution

18. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
If you believe that solutions to these types of questions should provide exact matches to the data values given, then this is wrong. If you are willing to settle for numerical approximations that are close, then this is almost certainly close enough.

An exponential equation is an exponential equation, the choice of base is arbitrary and merely introduces a scaling factor.

Your tutor was just a step away. They have:

$
y=2e^{-kx}

k = \frac{ln \left( \frac{0.1}{2} \right)}{-3}
$

But at this point they grabbed for their calculator instead of working the problem just a bit further.

$
k \ = \ \frac{-ln (0.1/2) }{3} \ = \ \frac{ln (2/0.1) }{3} \ = \ \frac{ln (20) }{3}

y=2e^{\frac{-ln(20)x }{3}}
y=2 {\left( e^{ln20} \right)}^{\frac{-x}{3}}
y=2 {\left(20)}^{\frac{-x}{3}}
y=\frac{2}{20^{x/3}}

$

19. ### steveb Senior Member

Jul 3, 2008
2,433
469
The tutor's method is correct, but the answer is not mathematically exact, hence making it wrong in most math contexts. It might be good enough for engineering or science purposes, but the constant k is not exactly equal to one.

I got my answer by the same method as your tutor, but I did not make any approximations. Let's pick up where he went wrong.

$y=2 {\rm e}^{{{x}\over{3}}{\rm ln}(0.05)}$

Now use a property of exponents.

$y=2 ({\rm e}^{{\rm ln}(0.05)})^{{{x}\over{3}}}$

Now, note that ln and exp are inverse functions.

$y=2 (0.05)^{{{x}\over{3}}}={{2}\over{20^{x/3}}}$

20. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
1,208
38
To WBahn:

Point taken, thanks for pointing this out, sometime I do a bit too captious

To WBahn and steveb:

Thanks for the solutions, that's something I wasn't aware of before, appreciate it!!