Quadratic, standard and standard form of a curve

• lucifer_x
In summary, the conversation is about using four given points to determine a curve and the necessary steps to solve for the coefficients of a cubic polynomial. This is done by setting up a system of equations and solving for the unknown coefficients.
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

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 $$y(x) = Ax^2 + Bx + C$$.

Bacat said:
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.

lucifer_x said:
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 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

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.

Related to Quadratic, standard and standard form of a curve

A quadratic curve is a type of curve that can be represented by a second-degree polynomial equation of the form y = ax^2 + bx + c. It is a U-shaped curve that can have either a maximum or minimum point, depending on the values of the coefficients a, b, and c.

What is the Standard Form of a Quadratic Curve?

The standard form of a quadratic curve is y = ax^2 + bx + c, where a, b, and c are constants. This form is also known as the general form and is used to easily identify the values of the coefficients and the overall shape of the curve.

What is the Standard Form of a Circle?

The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius. This form can also be rewritten as x^2 + y^2 + 2gx + 2fy + c = 0, where g = -h and f = -k. This form is useful for identifying the center and radius of a circle.

How do you Convert from Standard Form to Vertex Form of a Quadratic Curve?

To convert from standard form to vertex form of a quadratic curve, we need to complete the square. This involves adding and subtracting a constant term to the standard form equation, such that the resulting equation can be factored into the form y = a(x - h)^2 + k, where (h,k) is the vertex of the curve.

What is the Difference between Standard Form and Vertex Form of a Quadratic Curve?

The main difference between standard form and vertex form of a quadratic curve is the way they are written. Standard form is in the form of y = ax^2 + bx + c, while vertex form is in the form of y = a(x - h)^2 + k. Standard form is useful for identifying the x-intercepts of a quadratic curve, while vertex form is useful for identifying the vertex of the curve.

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