Quadratic, standard and standard form of a curve

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.

 

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].

 

Unfortunately the points don't all lie on the same quadratic, so a cubic is necessary.

 

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?

 

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.

 

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.