Answer :
We will investigate how to construct an equation of a parabola given three points.
A parabola is described as a polynomial relationship of highest order 2. The general equation of a parabola is given as follows:
[tex]y=ax^2\text{ + bx + c}[/tex]Where,
[tex]a\text{ , b , and c are constants}[/tex]To determine the values of these constants we will utilize a set of pair off coordinates given as follows:
[tex](\text{ -1 , 14 ) , ( 0 , 7 ) , ( 1 , 6 )}[/tex]The process off evaluating the three constants ( a,b, and c ) is to plug in the respective coordinates in the general form of a parabolic function. Then develop a system of equations which can be solved simultaneously.
Using the pair ( 0 , 7 ). We will plug in the respective coordinates in the general form and evaluate:
[tex]\begin{gathered} 7\text{ = a}\cdot(0)^2\text{ + b}\cdot(0)\text{ + c} \\ c\text{ = 7} \end{gathered}[/tex]Taking the pairs ( -1 , 14 ) and ( 1 , 6 ). We get a pair of equations as follows:
[tex]\begin{gathered} 14\text{ = a}\cdot(-1)^2+b\cdot(-1)\text{ + 7} \\ 7\text{ = a - b }\ldots\text{ Eq1} \\ \\ 6\text{ = a}\cdot(1)^2+b\cdot(1)\text{ + 7} \\ -1\text{ = a + b }\ldots\text{ Eq2} \end{gathered}[/tex]Now once we have developed a system of linear equations. We will solve them simultaneously by elimination:
[tex]\begin{gathered} Eq\text{ 1 + Eq2} \\ a\text{ - b = 7} \\ a+b\text{ = -1} \\ ======= \\ 2a\text{ = 6} \\ a\text{ = 3} \\ ======= \end{gathered}[/tex]Now use back substitution to plug in the value of constant ( a = 3 ) into either of the equations ( Eq1 or Eq2) as follows:
[tex]\begin{gathered} Eq2\colon-1\text{ = 3 + b} \\ b\text{ = -4} \end{gathered}[/tex]We have the values of the three constants:
[tex]a\text{ = 3 , b = -4 , c = 7}[/tex]The equation of the parabola turns out to be:
[tex]y=3x^2-4x\text{ + 7}[/tex]