These 3 points are on a parabola defining the edge of a ski:

(-4,1)
(-2,0.94)
(0,1)
The general form for the equation of a parabola is .


1.
Use the x- and y-values of 1 to build a linear equation with 3 variables: A, B, and C.
Repeat this process with the other 2 points to build a 2nd linear equation.
Record all three equations in the box below.

2.
Build a matrix equation that represents this system of equations. Record your matrix equation here.

3.
Use a graphing calculator or other graphing utility to find the inverse of the coefficient matrix. Record your result here.

4.
Use the inverse matrix to solve the system of equations. Record the equation of the parabola here.

Respuesta :

The equation of the parabola is y = 0.015x^2 + 0.06x + 1

The linear equations

A parabola is represented as:

y = ax^2 + bx + c

The points are given as:

(-4,1), (-2,0.94) and (0,1)

When these points are substituted in y = ax^2 + bx + c, we have the following linear equations

16a - 4b + c = 1

4a - 2b + c = 0.94

c = 1

The matrix that represent the system of equations

In (a), we have:

16a - 4b + c = 1

4a - 2b + c = 0.94

c = 1

Remove the variables (leaving the coefficients)

a      b       c

16    -4      1        1

4      -2     1        0.94

0      0     1          1

So, the matrix representation of the system of equations is:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right] \left[\begin{array}{c}1&0.94&1\end{array}\right][/tex]

The inverse of the coefficient matrix

Using a graphing calculator, we have the inverse of the coefficient matrix to be

[tex]\left[\begin{array}{ccc}0.125&-0.25&0.125\\0.25&-1&0.75\\0&0&1\end{array}\right][/tex]

The solution to the system of equations

Using a graphing calculator, we have the solution to the system of equations to be

a = 0.015

b = 0.06

c = 1

Recall that:

y = ax^2 + bx + c

So, we have:

y = 0.015x^2 + 0.06x + 1

Hence, the equation of the parabola is y = 0.015x^2 + 0.06x + 1

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