## Precalculus (6th Edition) Blitzer

The resultant quadratic equation is ${{x}^{2}}-6x+8$.
The given quadratic equation is as shown below: $y=a{{x}^{2}}+bx+c$ The given points are $\left( 1,3 \right),\left( 3,-1 \right),\left( 4,0 \right)$. In order to find the quadratic equation, find the values of a, b, and c. Putting in the value of $x=1$ and $y=3$: we get, \begin{align} & 3=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c \\ & 3=a+b+c \end{align} Putting in the value of $x=3$ and $y=-1$: we get, \begin{align} & -1=a{{\left( 3 \right)}^{2}}+b\left( 3 \right)+c \\ & -1=9a+3b+c \end{align} Putting the value of $x=4$ and $y=0$: we get, \begin{align} & 0=a{{\left( 4 \right)}^{2}}+b\left( 4 \right)+c \\ & 0=16a+4b+c \end{align} The resultant system of equations having variable a, b and c is as shown below: \begin{align} & a+b+c=3 \\ & 9a+3b+c=-1 \\ & 16a+4b+c=0 \end{align} By multiplying equation (I) with $-9$ and eliminate a equation (I) and (II): we get, \begin{align} & 9a+3b+\text{ }c=-1 \\ & -9a-9b-9c=-27 \end{align} $-6b-8c=-28$ (IV) Multiply equation (I) with -16 and eliminate a equation (I) and (III): we get, \begin{align} & 16a+\text{ }4b+\text{ }c=0 \\ & -16a-16b-16c=-48 \end{align} $-12b-15c=-48$ (V) Now, multiply the equation (IV) by 2 and then add to get the value of c, \begin{align} & -12b-15c=-48 \\ & 12b+16c=56 \end{align} $c=8$ Putting in the value of c in equation (IV), we get, \begin{align} & -6b-8\left( 8 \right)=-28 \\ & -6b-64=-28 \end{align} By adding 64 to both sides: we get, \begin{align} & -6b-64+64=-28+64 \\ & -6b=36 \end{align} Now, divide both sides by $-6$ to get the value of b, \begin{align} & \frac{-6b}{-6}=\frac{36}{-6} \\ & \text{ }b=-6 \\ \end{align} Putting the values of b and c in equation (I) to get the value of a, \begin{align} & a+\left( -6 \right)+8=3 \\ & a+2=3 \end{align} Subtract 2 from both sides to get, \begin{align} & a+2-2=3-2 \\ & \text{ }a=1 \\ \end{align} Now substitute the values of a, b and c to get the equation ${{x}^{2}}-6x+8$. Hence, the quadratic equation is ${{x}^{2}}-6x+8$.