# Quadratic, standard and standard form of a curve

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

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:

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

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.

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

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
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Gold Member
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
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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
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?

ya thats what it is, there are 4 points on the curve
so like

$(x_1,y_1)=(25,43)$

is what i was thinking

but thats only one point

gabbagabbahey
Homework Helper
Gold Member
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.

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 ??

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.