Quadratic, standard and standard form of a curve

[lucifer_x]

the curve points are:
a) (25, 43)
b) (28, 46)
c) (32, 47)
d) (35, 45)

would i use y=mx+b i think that's a straight line

 

[gabbagabbahey]

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

y(x)=Ax^3+Bx^2+Cx+D

Where you can determine the constants A, B, C and D by plugging in the given points and solving the resulting system of 4 equations.

 

[lucifer_x]

so how would i put the points of A into

y(x)=Ax^3+Bx^2+Cx+D


cause there are 2 points and only 1 A

 

[Bacat]

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 -x^2).

If you are studying Quadratic, standard and standard form of a curve, my guess is that you only need y(x) = Ax^2 + Bx + C.

 

[gabbagabbahey]

my guess is that you only need y(x) = Ax^2 + Bx + C.

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

 

[gabbagabbahey]

so how would i put the points of A into

y(x)=Ax^3+Bx^2+Cx+Dcause there are 2 points and only 1 A

I thought a), b) c) and d) were each points on the curve in the form (x_1,y_1)=(25,43) etc...is this assumption incorrect?

 

[lucifer_x]

ya that's what it is, there are 4 points on the curve
so like

(x_1,y_1)=(25,43)

is what i was thinking

but that's only one point

 

[gabbagabbahey]

Okay, well each point gives you an equation. For example, the first point tells you:

43=A(25)^3+B(25)^2+C(25)+D

And so using all 4 points you get 4 equations and 4 unknowns (A, B, C, and D) allowing you to solve for them.

 

[lucifer_x]

ohhh ok that makes sense thank you for making it more clear

so the almost final will look like this

43=A(15625)+B(625)+C(25)+D

then you divide them ??

 

[Bacat]

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: y_1(x)=Ax^3_1+Bx^2_1+Cx_1+D

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

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

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

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