## Precalculus (6th Edition) Blitzer

The quadratic equation: ${{x}^{2}}-9x+22$
The quadratic equation is as shown below: $y=a{{x}^{2}}+bx+c$ The points are $\left( 2,8 \right),\ \left( 4,2 \right),\ and\ \left( 6,4 \right).$ The task here is to find the equations in variables a, b and c and solve for the values of a, b, and c. Substitute $x=2$ and $y=8:$ \begin{align} & 8=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\ & 8=4a+2b+c \end{align} Substitute $x=4$ and $y=2:$ \begin{align} & 2=a{{\left( 4 \right)}^{2}}+b\left( 4 \right)+c \\ & 2=16a+4b+c \end{align} Substitute $x=6$ and $y=4:$ \begin{align} & 4=a{{\left( 6 \right)}^{2}}+b\left( 6 \right)+c \\ & 4=36a+6b+c \end{align} The system of equations in variables a, b and c is as shown below: $4a+2b+c=8$ $16a+4b+c=2$ $36a+6b+c=4$ Mark these equations as shown below: $4a+2b+c=8$, Mark it as equation (I) $16a+4b+c=2$, Mark it as equation (II) $36a+6b+c=4$, Mark it as equation (III) By multiplying equation (I) by $-4$ and eliminating a from equations (I) and (III), we get \begin{align} & \text{ }16a+4b+\text{ }c=\text{ }2 \\ & \underline{-16a-8b-4c=-3\text{2}} \\ & \text{ }-\text{4}b-3c=-30 \\ \end{align} Mark this equation as $-\text{4}b-3c=-30$ (IV) By multiplying equation (III) by −9 and eliminating a from equations (II) and (III), we get \begin{align} & \text{ }36a+\text{ }6b+\text{ }c=\text{ }4 \\ & \underline{-36a-18b-9c=-\text{72}} \\ & \text{ }-12b-8c=-68 \\ \end{align} Mark this equation as $-12b-8c=-68$ (V) By multiplying equation (IV) by −3 and then adding equations (IV) and (V), we get the value of c: \begin{align} & -12b-8c=-68 \\ & \underline{-12b+9c=\text{ }90} \\ & \text{ }c=\text{ }22 \\ \end{align} Now, putting the value of c in equation (IV) to get the value of b: \begin{align} & -4b-3\left( 22 \right)=-30 \\ & -4b-66=-30 \end{align} Now, adding 66 to both sides: \begin{align} & -4b-66+66=-30+66 \\ & -4b=36 \end{align} By dividing both sides by −4 we get: \begin{align} & \frac{-4b}{-4}=\frac{36}{-4} \\ & \text{ }b=-9 \end{align} Substitute the values of b and c in equation (I) to get the value of a: \begin{align} & 4a+2\left( -9 \right)+22=8 \\ & 4a-18+22=8 \\ & 4a+4=8 \end{align} Now, subtract $4$ from both sides to get: \begin{align} & 4a+4-4=8-4 \\ & 4a=4 \end{align} By dividing both sides by 4, we get: \begin{align} & \frac{4a}{4}=\frac{4}{4} \\ & \text{ }a=1 \end{align} Hence, the quadratic equation is ${{x}^{2}}-9x+22$.