The equation of the parabola is y = 0.015x^2 + 0.06x + 1
<h3>The linear equations</h3>
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
<h3>The matrix that represent the system of equations</h3>
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:
![\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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D16%26-4%261%5C%5C4%26-2%261%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%260.94%261%5Cend%7Barray%7D%5Cright%5D)
<h3>The inverse of the coefficient matrix</h3>
Using a graphing calculator, we have the inverse of the coefficient matrix to be
![\left[\begin{array}{ccc}0.125&-0.25&0.125\\0.25&-1&0.75\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.125%26-0.25%260.125%5C%5C0.25%26-1%260.75%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
<h3>The solution to the
system of equations</h3>
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
Read more about matrix at:
brainly.com/question/9967572
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