Answer:
Procedure:
1) Form a system of 3 linear equations based on the two zeroes and a point.
2) Solve the resulting system by analytical methods.
3) Substitute all coefficients.
Step-by-step explanation:
A quadratic function is a polynomial of the form:
[tex]y = a\cdot x^{2}+b\cdot x + c[/tex] (1)
Where:
[tex]x[/tex] - Independent variable.
[tex]y[/tex] - Dependent variable.
[tex]a[/tex], [tex]b[/tex], [tex]c[/tex] - Coefficients.
A value of [tex]x[/tex] is a zero of the quadratic function if and only if [tex]y = 0[/tex]. By Fundamental Theorem of Algebra, quadratic functions with real coefficients may have two real solutions. We know the following three points: [tex]A(x,y) = (r_{1}, 0)[/tex], [tex]B(x,y) = (r_{2},0)[/tex] and [tex]C(x,y) = (x,y)[/tex]
Based on such information, we form the following system of linear equations:
[tex]a\cdot r_{1}^{2}+b\cdot r_{1} + c = 0[/tex] (2)
[tex]a\cdot r_{2}^{2}+b\cdot r_{2} + c = 0[/tex] (3)
[tex]a\cdot x^{2} + b\cdot x + c = y[/tex] (4)
There are several forms of solving the system of equations. We decide to solve for all coefficients by determinants:
[tex]a = \frac{\left|\begin{array}{ccc}0&r_{1}&1\\0&r_{2}&1\\y&x&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }[/tex]
[tex]a = \frac{y\cdot r_{1}-y\cdot r_{2}}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x+x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}[/tex]
[tex]a = \frac{y\cdot (r_{1}-r_{2})}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}[/tex]
[tex]b = \frac{\left|\begin{array}{ccc}r_{1}^{2}&0&1\\r_{2}^{2}&0&1\\x^{2}&y&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }[/tex]
[tex]b = \frac{(r_{2}^{2}-r_{1}^{2})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}[/tex]
[tex]c = \frac{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&0\\r_{2}^{2}&r_{2}&0\\x^{2}&x&y\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }[/tex]
[tex]c = \frac{(r_{1}^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x + x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}[/tex]
And finally we obtain the equation of the quadratic function given two zeroes and a point.
