## Precalculus (6th Edition) Blitzer

The quadratic function is $f\left( x \right)=-{{x}^{2}}+x+2$.
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, $f\left( -2 \right)=-4$ gives, \begin{align} & f\left( -2 \right)=a{{\left( -2 \right)}^{2}}+b\left( -2 \right)+c \\ & -4=4a-2b+c \\ & 4a-2b+c=-4 \end{align} (I) $f\left( 1 \right)=2$ gives, \begin{align} & f\left( 1 \right)=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c \\ & 2=a+b+c \\ & a+b+c=2 \end{align} (II) $f\left( 2 \right)=0$ gives, \begin{align} & f\left( 2 \right)=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\ & 0=4a+2b+c \\ & 4a+2b+c=0 \end{align} (III) Equations (I), (II) and (III) give, \begin{align} & 4a-2b+c=-4 \\ & a+b+c=2 \\ & 4a+2b+c=0 \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: $\left[ \begin{matrix} 4 & -2 & 1 & -4 \\ 1 & 1 & 1 & 2 \\ 4 & 2 & 1 & 0 \\ \end{matrix} \right]$ ${{R}_{1}}\to \frac{1}{4}{{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & -\frac{1}{2} & \frac{1}{4} & -1 \\ 1 & 1 & 1 & 2 \\ 4 & 2 & 1 & 0 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -4 \right){{R}_{1}}$ give, $\left[ \begin{matrix} 1 & -\frac{1}{2} & \frac{1}{4} & -1 \\ 0 & \frac{3}{2} & \frac{3}{4} & 3 \\ 0 & 4 & 0 & 4 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{2}{3}{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -\frac{1}{2} & \frac{1}{4} & -1 \\ 0 & 1 & \frac{1}{2} & 2 \\ 0 & 4 & 0 & 4 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}+\left( -4 \right){{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & -\frac{1}{2} & \frac{1}{4} & -1 \\ 0 & 1 & \frac{1}{2} & 2 \\ 0 & 0 & -2 & -4 \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{-1}{2}{{R}_{3}}$ $\left[ \begin{matrix} 1 & -\frac{1}{2} & \frac{1}{4} & -1 \\ 0 & 1 & \frac{1}{2} & 2 \\ 0 & 0 & 1 & 2 \\ \end{matrix} \right]$ Since the matrix is in the echelon form, so the system of the linear equations corresponding to the echelon form of the matrix is given as below: $a-\frac{1}{2}b+\frac{1}{4}c=-1$ (IV) $b+\frac{1}{2}c=2$ (V) $c=2$ (VI) Apply back the substitution method: Equation (VI) gives, $c=2$ Put the value of c in the equation (V) as follows: It can be further simplified as, $b=1$ Put the values of b and c in the equation (IV) to get, $a-\frac{1}{2}\left( 1 \right)+\frac{1}{4}\left( 2 \right)=-1$ It can be further simplified as, $a=-1$ 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( 2 \right) \\ & =-{{x}^{2}}+x+2 \end{align} Hence, the required quadratic function is $f\left( x \right)=-{{x}^{2}}+x+2$.