Quadratic, standard and standard form of a curve

  • Thread starter lucifer_x
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  • #1
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the curve points are:
a) (25, 43)
b) (28, 46)
c) (32, 47)
d) (35, 45)

would i use y=mx+b i think thats a straight line
 

Answers and Replies

  • #2
gabbagabbahey
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You are given four points on the curve, so in general, the smallest degree polynomial that would accommodate them is a cubic polynomial:

[tex]y(x)=Ax^3+Bx^2+Cx+D[/tex]

Where you can determine the constants [itex]A[/itex], [itex]B[/itex], [itex]C[/itex] and [itex]D[/itex] by plugging in the given points and solving the resulting system of 4 equations.
 
  • #3
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so how would i put the points of A into

[tex]y(x)=Ax^3+Bx^2+Cx+D[/tex]


cause there are 2 points and only 1 A
 
  • #4
151
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The points are not a straight-line. Try graphing them to see this for yourself. They look like a curve to me- probably quadratic and opening downwards (ie [tex]-x^2[/tex]).

If you are studying Quadratic, standard and standard form of a curve, my guess is that you only need [tex]y(x) = Ax^2 + Bx + C[/tex].
 
  • #5
gabbagabbahey
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my guess is that you only need [tex]y(x) = Ax^2 + Bx + C[/tex].
Unfortunately the points don't all lie on the same quadratic, so a cubic is necessary.
 
  • #6
gabbagabbahey
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so how would i put the points of A into

[tex]y(x)=Ax^3+Bx^2+Cx+D[/tex]


cause there are 2 points and only 1 A
I thought a), b) c) and d) were each points on the curve in the form [itex](x_1,y_1)=(25,43)[/itex] etc.....is this assumption incorrect?
 
  • #7
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ya thats what it is, there are 4 points on the curve
so like

[itex](x_1,y_1)=(25,43)[/itex]

is what i was thinking

but thats only one point
 
  • #8
gabbagabbahey
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Okay, well each point gives you an equation. For example, the first point tells you:

[tex]43=A(25)^3+B(25)^2+C(25)+D[/tex]

And so using all 4 points you get 4 equations and 4 unknowns ([itex]A[/itex], [itex]B[/itex], [itex]C[/itex], and [itex]D[/itex]) allowing you to solve for them.
 
  • #9
15
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ohhh ok that makes sense thank you for making it more clear

so the almost final will look like this

[tex]43=A(15625)+B(625)+C(25)+D[/tex]

then you divide them ??
 
  • #10
151
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You need to setup a system of four equations, then you can solve by substitution, elimination, or matrix methods to find A,B,C, and D. So it should like this like:

Equation 1: [tex]y_1(x)=Ax^3_1+Bx^2_1+Cx_1+D[/tex]

Equation 2: [tex]y_2(x)=Ax^3_2+Bx^2_2+Cx_2+D[/tex]

Equation 3: [tex]y_3(x)=Ax^3_3+Bx^2_3+Cx_3+D[/tex]

Equation 4: [tex]y_4(x)=Ax^3_4+Bx^2_4+Cx_4+D[/tex]

Plug in your coordinates for your (x,y) values above and then solve the system.
 

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