The following 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:
Ax^2 + Bx + C= y

Required:
a. Use the x- and y-values of 1 of the points to build a linear equation with 3 variables: A, B, and C.
b. Record your equation here. Repeat this process with 1 of the other 2 points to build a 2nd linear equation.
c. Record your equation here. Repeat this process with the other point to build a 3rd equation.
d. Record your equation here. Build a matrix equation that represents this system of equations.
e. Record your matrix equation here. Use a graphing calculator or other graphing utility to find the inverse of the coefficient matrix.
f. Record your result here. Use the inverse matrix to solve the system of equations. Record the equation of the parabola here.

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

e. The inverse of the coefficient matrix looks like this:

[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]

f. The equation of the parabola is: [tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]

a. In order to build a linear equation from the given points, we need to substitute them into the general form of the equation.

Let's take the first point (-4,1). When substituting it into the general form of the quadratic equation we end up with:

[tex](-4)^{2}A+(-4)B+C=1[/tex]

which yields:

[tex]16A-4B+C=1[/tex]

b. Let's take the second point (-2,0.94). When substituting it into the general form of the quadratic equation we end up with:

[tex](-2)^{2}A+(-2)B+C=0.94[/tex]

which yields:

[tex]4A-2B+C=0.94[/tex]

c. Let's take the third point (0,1). When substituting it into the general form of the quadratic equation we end up with:

[tex](0)^{2}A+(0)B+C=1[/tex]

which yields:

[tex]0A+0B+C=1[/tex]

d. A matrix equation consists on three matrices. The first matrix contains the coefficients (this is the numbers on the left side of the linear equations). Make sure to write them in the right order, this is, the numbers next to the A's should go on the first column, the numbers next to the B's should go on the second column and the numbers next to the C's should go on the third column.

The equations are the following:

16A-4B+C=1

4A-2B+C=0.94

0A+0B+C=1

So the coefficient matrix looks like this:

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

Next we have the matrix that has the variables, in this case our variables are the letters A, B and C. So the matrix looks like this:

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right][/tex]

and finally the matrix with the answers to the equations, in this case 1, 0.94 and 1:

[tex]\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

so if we put it all together we end up with the following matrix equation:

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

e. When inputing the coefficient matrix in our graphing calculator we end up with the following inverse matrix:

[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]

Inputing matrices and calculating their inverses depends on the model of a calculator you are using. You can refer to the user's manual on how to do that.

f. Our matrix equation has the following general form:

AX=B

where:

A=Coefficient matrix

X=Variables matrix

B= Answers matrix

In order to solve this type of equations, we can make use of the inverse of the coefficient matrix to end up with an equation that looks like this:

[tex]X=A^{-1}B[/tex]

Be careful with the order in which you are doing the multiplication, if A and B change places, then the multiplication will not work and you will not get the answer you need. So when solving this equation we get:

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]*\left[\begin{array}{c}1\\\frac{47}{50}\\1\end{array}\right][/tex]

(Notice that I changed 0.94 for the fraction 47/50 you can get this number by dividing 94/100 and simplifying the fraction)

So, in order to do the multiplication, we need to multiply each row of the coefficient matrix by the answer matrix and add the results. Like this:

[tex]\frac{1}{8}*1+(-\frac{1}{4})(\frac{47}{50})+\frac{1}{8}*1[/tex]

[tex]\frac{1}{8}-\frac{47}{200}+\frac{1}{8}=\frac{3}{200}[/tex]

So the first number for the answer matrix is [tex]\frac{3}{200}[/tex]

[tex]\frac{1}{4}*1+(-1)(\frac{47}{50})+\frac{3}{4}*1[/tex]

[tex]\frac{1}{4}-\frac{47}{50}+\frac{3}{4}=\frac{3}{50}[/tex]

So the second number for the answer matrix is [tex]\frac{3}{50}[/tex]

[tex]0*1+0(\frac{47}{50})+1*1[/tex]

[tex]0+0+1=1[/tex]

So the third number for the answer matrix is 1

In the end, the matrix equation has the following answer.

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}\frac{3}{200}\\\frac{3}{50}\\1\end{array}\right][/tex]

which means that:

[tex]A=\frac{3}{200}[/tex]

[tex]B=\frac{3}{50}[/tex]

and C=1

so, when substituting these answers in the general form of the equation of the parabola we get:

[tex]Ax^{2}+Bx+C=y[/tex]

[tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]

For further information, you can go to the following link:

https://brainly.com/question/12628757?referrer=searchResults