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A polynomial of degree ≤ 2 ? what does this mean.

  1. Apr 5, 2013 #1
    A polynomial of degree ≤ 2 ? what does this mean.


    Would it just be

    a + bt + c t^2 = f(t)

    Or

    at^2 + bt + c = f(t)

    Is there even a difference between the two equations considering the fact that a,b, and c are unknown?
     
  2. jcsd
  3. Apr 5, 2013 #2

    Mentallic

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    Homework Helper

    There's no difference in whether you had

    [tex]f(t)=at^2+bt+c[/tex]

    or

    [tex]f(t)=ct^2+bt+a[/tex]

    or

    [tex]f(t)=xt^2+yt+z[/tex]

    But the first is customary, and the last is using letters that usually denote variables as opposed to constants, so unless you have a good reason to otherwise deviate from the first, just stick with that.

    Also, with any polynomial of degree n, the leading coefficient (coefficient of tn) must be non-zero, else the polynomial will no longer be degree n. Since you have a polynomial of degree [itex]\leq[/itex] 2, that means the leading coefficient of t2 does not have to be non-zero. You could even have all coefficients equal to 0 and thus simply have f(t)=0.
     
  4. Apr 5, 2013 #3
    So what is the difference between a polynomial with degree = 2 and a polynomial with degree ≤ 2 or in general what is the difference between a polynomial with degree = 2 vs a polynomial with degree ≤ n ?
     
  5. Apr 5, 2013 #4

    Mark44

    Staff: Mentor

    f(t) = at2 + bt + c, a 2nd-degree polynomial, also called a quadratic polynomial.
    Degree ≤ 2 would also include 1st degree polynomials, such as g(t) = at + b, or zero-degree polynomials, such as h(t) = a.
    I assume you mean degree = n vs. degree ≤ n. An nth degree polynomial has to have a term in which the variable has an exponent of n. A polynomial of degree ≤ n includes lower-degree polynomials.
     
  6. Apr 5, 2013 #5
    Yes sorry for the typo. I meant to say degree = n. The reason I started this thread was with regards to a problem in my linear algebra class where the problem states:

    Find all polynomials f(t) of degree ≤ 2 whose graphs run through the points (1,3) and (2,6) , such that f`(1) = 1 .

    When I started to solve the problem I used the form f(t) = a + bt + ct^2 for my polynomial and after solving the matrices I got c = 2. However when I checked the solutions in the back of the book they had a = 2 which makes sense because they used f(t) = at^2 + bt + c . So what confused me was, how is one suppose to know which form to use to get the right a or c even though both essentially yield the same variables considering 2 is the value of the variable in front of the t^2 term. In general you mentioned a polynomial of degree ≤ n includes lower-degree polynomials. According to that statement, how would I know to use f(t) = at^2 + bt + c or f(t) = at + c when setting up my system of equations?
     
  7. Apr 5, 2013 #6

    Mentallic

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    Since the polynomial is of degree [itex]\leq[/itex] 2, that means the degree can at most be 2, so you should have [itex]f(t)=at^2+bt+c[/itex] which also ensures that even if the polynomial is just degree 1, then a=0. If you used [itex]f(t)=at+b[/itex] or [itex]f(t)=a[/itex] then you're assuming the equation must be of that form, and you'll soon find that there is no possible solution to the question if you begin with the assumption that the polynomials are of degree [itex]\leq[/itex] 1, or equivalently, [itex]f(t)=at+b[/itex].
     
  8. Apr 6, 2013 #7

    HallsofIvy

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    If f(t) is a "polynomial of degree 2 or less" it can be written in the form [itex]at^2+ bt+ c[/itex] where a, b, and c can be any numbers.

    If f(t) is a "polynomial of degree 2" it can be written in the form [itex]at^2+ bt+ c[/itex] where a, b, and c can be any numbers- except that a cannot be 0.
     
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