## Precalculus (6th Edition) Blitzer

The quadratic function is $f\left( x \right)={{x}^{2}}-x+3$.
To obtain the quadratic function, at first we will find the coefficients a, b, c. Since, $f\left( x \right)=a{{x}^{2}}+bx+c$. So, $f\left( -1 \right)=5$ gives, \begin{align} & f\left( -1 \right)=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c \\ & 5=a-b+c \\ & a-b+c=5 \end{align} (I) $f\left( 1 \right)=3$ gives, \begin{align} & f\left( 1 \right)=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c \\ & 3=a+b+c \\ & a+b+c=3 \end{align} (II) $f\left( 2 \right)=5$ gives, \begin{align} & f\left( 2 \right)=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\ & 5=4a+2b+c \\ & 4a+2b+c=5 \end{align} (III) Equations (I), (II) and (III) give, \begin{align} & a-b+c=5 \\ & a+b+c=3 \\ & 4a+2b+c=5 \end{align} Use the Gauss elimination method to obtain the solution of the above system of equations. The matrix corresponding to the system of equations is as follows: $\left[ \begin{matrix} 1 & -1 & 1 & 5 \\ 1 & 1 & 1 & 3 \\ 4 & 2 & 1 & 5 \\ \end{matrix} \right]$ The matrix will be converted into the echelon form by using the elementary row transformation. ${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -4 \right){{R}_{1}}$ give, $\left[ \begin{matrix} 1 & -1 & 1 & 5 \\ 0 & 2 & 0 & -2 \\ 0 & 6 & -3 & -15 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{2}{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -1 & 1 & 5 \\ 0 & 1 & 0 & -1 \\ 0 & 6 & -3 & -15 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\left( -6 \right){{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -1 & 1 & 5 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & -3 & -9 \\ \end{matrix} \right]$ ${{R}_{3}}\to -\frac{1}{3}{{R}_{3}}$ $\left[ \begin{matrix} 1 & -1 & 1 & 5 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ \end{matrix} \right]$ Since, the matrix is in the echelon form, so, express the system of linear equations corresponding to the echelon form of the matrix as follows: $a-b+c=5$ (IV) $b=-1$ (V) $c=3$ (VI) Apply back substitution method: Equation (VI) gives, $c=3$ Equation (V) gives, $b=-1$ Substitute the values of b and c in the equation (IV) as follows \begin{align} & a-\left( -1 \right)+\left( 3 \right)=5 \\ & a=1 \end{align} Now, substitute the values of a, b, and c in the function $f\left( x \right)=a{{x}^{2}}+bx+c$ ,this gives, \begin{align} & f\left( x \right)=\left( 1 \right){{x}^{2}}+\left( -1 \right)x+\left( 3 \right) \\ & ={{x}^{2}}-x+3 \end{align} Hence, the required quadratic function is: $f\left( x \right)={{x}^{2}}-x+3$