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Finding a polynomial when only given five points 
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#1
Sep1408, 05:37 PM

P: 12

i just enrolled at the linear algebra class at my university. and after out first test the professor gave us this "calculator project" (meaning just using the calculator in some way to get the answer) that has 4 linear algebra application problems. i figured out the last three but i am stuck on the first one. it is in two parts.
a. determine the polynomial whose graph passes through the points (1/2, 75/16), (0,6), (2/3, 220/81), (3,48), and (4,210) b. the second question just says to sketch an accurate graph of the polynomial. i went and asked the professor the other day how we might go about finding the answer. and she hinted around having a equation in the fourth degree (ax^4+bx^3+cx^2+dx+e if i'm not mistaken) all the other questions involved me making a matrix out of the system of equations and putting them in reduced row echelon form to find the answer so i'm suspecting that it must be done that way for this one too b/c that is all we have learned in the class so far. any helps or hints would be appreciated guys. 


#2
Sep1408, 06:08 PM

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What would you get if you plugged in the 5 data points into ax^4+bx^3+cx^2+dx+e ?



#3
Sep1408, 07:21 PM

P: 12

ok i solved the matrix i got in the end and found the equation to be
y=x^4+x^37x^2x+6 the question says to name the type of polynomial. is there anywhere i can find graphs related to this equation to show me the different kinds of polynomials and their names. i graphed it on my calculator and it showed me what the graph looks like but i don't know what the name of this one is. any ideas where i can find somehting like this? 


#4
Sep1408, 08:49 PM

PF Gold
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Finding a polynomial when only given five points



#5
Sep1408, 10:40 PM

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I'm curious as to what your prof told you about the order of the polynomial. Could it have been done easily if you weren't told as a hint that the polynomial was 4th order?
As for graphing, just apply the first and second derivative test to find all the turning points. 


#6
Sep1408, 10:51 PM

P: 12

she said she wants it to be named. such as a slope, parabola, etc. i just can't find what equations of the fourth degree are named



#7
Sep1408, 10:53 PM

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Graph it and see if you can recognise the shape of the graph.



#8
Sep1508, 12:40 AM

P: 12

i have graphed it and its kinda weird looking and have not seen one like it before. here is the function if anyone wants to graph it and see if they know
y=x^4+x^37x^2x+6 


#9
Sep1508, 08:00 AM

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P: 2,616

Well I reread your OP and saw that all that is required is a graph of the function. I graphed the function you got (I didn't check if you got that function correctly) online and saw that it has four real roots, but that I can't think of any name for the graph except that it is a graph of a fourth order polynomial with 4 real roots.



#10
Sep1508, 08:11 AM

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I find it hard to believe that your teacher said "she said she wants it to be named. such as a slope, parabola, etc. i just can't find what equations of the fourth degree are named". "slope" is not a kind of graph. "Linear" or "straight line", perhaps.
In any case, danalo has already told you: "* Degree 4  quartic". 


#11
Sep1608, 11:23 AM

P: 5

i am on that same problem and still cannot figure it out. I am confused on how to put it in the calculator. help needed.



#12
Sep1608, 06:47 PM

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???? Did you just create a new user account?
And what do you mean "how to put it in the calculator" ? 


#13
Sep1608, 06:58 PM

P: 5

what i mean is i believe i solved for the equation but when i put it in the calculator it shows me a function curve not a polynomial



#14
Sep1608, 07:04 PM

P: 5

yes i just started an acount today



#15
Sep1608, 07:07 PM

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P: 2,616

Well actually I meant to ask "Did you just create a duplicated account in addition to your old "cougarsoccer" account ?"
And what do you think a polynomial curve means? A polynomial is a function is it not? What were you expecting instead? 


#16
Sep1608, 07:12 PM

P: 5

no icreated my own account. it just so happens that i have the same question as cougar. yes a poly is a function so that makes sense. The equation that you use is y=ex^0+dx^1+cx^2+bx^3+ax^4 ?? plug all five points in?



#17
Sep1608, 07:16 PM

HW Helper
P: 2,616

Well yes. That's how you find the unknown coefficients. Then you have to graph it. This was explained above.



#18
Sep1608, 07:43 PM

P: 5

so this curve is a quartic? where do you find that info?



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