## Precalculus (6th Edition) Blitzer

The quadratic equation is $y={{x}^{2}}-3$
Let the quadratic function be $y=a{{x}^{2}}+bx+c$. Since it passes through the points $\left( -1,-2 \right),\left( 2,1 \right),\left( -2,1 \right)$, we put these points one by one in the quadratic function. Since it passes through the point $\left( -1,-2 \right)$, the equation becomes: \begin{align} & -2=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c \\ & a-b+c=-2 \end{align} (I) Next it passes through the point $\left( 2,1 \right)$, so the equation becomes: \begin{align} & 1=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\ & 4a+2b+c=1 \end{align} (II) It also passes through the point $\left( -2,1 \right)$, so the equation becomes: \begin{align} & 1=a{{\left( -2 \right)}^{2}}+b\left( -2 \right)+c \\ & 4a-2b+c=1 \end{align} (III) Then consider Equations (I), (II), and (III), and represent them as a system of linear equations: \begin{align} & a-b+c=-2 \\ & 4a+2b+c=1 \\ & 4a-2b+c=1 \end{align} And the augmented matrix is $\left[ \begin{matrix} 1 & -1 & 1 & -2 \\ 4 & 2 & 1 & 1 \\ 4 & -2 & 1 & 1 \\ \end{matrix} \right]$ Use the Gaussian elimination method to find the values of a, b, and c. To find the row echelon form of the matrix, perform the elementary row operations. ${{R}_{2}}\to {{R}_{2}}+\left( -4 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -4 \right){{R}_{1}}$ Which gives $\left[ \begin{matrix} 1 & -1 & 1 & -2 \\ 0 & 6 & -3 & 9 \\ 0 & 2 & -3 & 9 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{6}{{R}_{2}}$ Which gives $\left[ \begin{matrix} 1 & -1 & 1 & -2 \\ 0 & 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 2 & -3 & 9 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\left( -2 \right){{R}_{2}}$ That gives $\left[ \begin{matrix} 1 & -1 & 1 & -2 \\ 0 & 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 0 & -2 & 6 \\ \end{matrix} \right]$ ${{R}_{3}}\to -\frac{1}{2}{{R}_{3}}$ Which gives $\left[ \begin{matrix} 1 & -1 & 1 & -2 \\ 0 & 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 0 & 1 & -3 \\ \end{matrix} \right]$ The above matrix is in row echelon form; represent the system of equations from the last matrix: $a-b+c=-2$ (IV) $b-\frac{1}{2}c=\frac{3}{2}$ (V) $c=-3$ (VI) Then use the back substitution method and obtain the values of a, b, c: $c=-3,b=0,\text{ and }a=1$. Therefore, the quadratic function is $y=1{{x}^{2}}+0x-3$ Thus, the quadratic function is $y={{x}^{2}}-3$.