Given that a parabola of the form [tex]y=ax^2+bx+c[/tex] passing through the points (2, -11), (-2, -23) and (4, -53)
Thus, substituting the points we have:
[tex]-11=(2)^2a+2b+c \\ \\ \Rightarrow-11=4a+2b+c\ .\ .\ .\ (1) \\ \\
-23=(-2)^2a+(-2)b+c \\ \\ \Rightarrow -23=4a-2b+c\ .\ .\ .\ (2) \\ \\
-53=(4)^2a+4b+c \\ \\ \Rightarrow-53=16a+4b+c\ .\ .\ .\ (3)[/tex]
We solve equations (1), (2) and (3) simulataneously. (There are many mays it can be solved but I will use row reduction method here).
We form the augumented matrix for equations (1), (2) and (3) and perform elementary row operations as follows:
[tex] \left[\begin{array}{ccc}4&2&1\\4&-2&1\\16&4&1\end{array}\right| \left.\begin{array}{c}-11\\-23\\-53\end{array}\right] \ \ \ \ \ \frac{1}{4} R_1\rightarrow R_1 \\ \\ \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\4&-2&1\\16&4&1\end{array}\right| \left.\begin{array}{c}- \frac{11}{4} \\-23\\-53\end{array}\right] \ \ \ \ \ {{-4R_1+R_2\rightarrow R_2} \atop {-16R_1+R_3\rightarrow R_3}}[/tex]
[tex] \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\0&-4&0\\0&-4&-3\end{array}\right| \left.\begin{array}{c}- \frac{11}{4} \\-12\\-9\end{array}\right] \ \ \ \ \ - \frac{1}{4} R_2 \\ \\ \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\0&1&0\\0&-4&-3\end{array}\right| \left.\begin{array}{c}- \frac{11}{4} \\3\\-9\end{array}\right] \ \ \ \ \ {{ -\frac{1}{2} R_2+R_1\rightarrow R_1} \atop {4R_2+R_3\rightarrow R_3}}[/tex]
[tex] \left[\begin{array}{ccc}1&0& \frac{1}{4} \\0&1&0\\0&0&-3\end{array}\right| \left.\begin{array}{c}- \frac{17}{4} \\3\\3\end{array}\right] \ \ \ \ \ - \frac{1}{3} R_3 \\ \\ \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right| \left.\begin{array}{c}-4\\3\\-1\end{array}\right] \ \ \ \ \ - \frac{1}{4} R_3+R_1\rightarrow R_1[/tex]
Thus, a = -4, b = 3, c = -1
Therefore, the required polynomial is [tex]y=-4x^2+3x-1[/tex]
