Precalculus (6th Edition) Blitzer

The quadratic equation is: $-\frac{3}{4}{{x}^{2}}+6x-11$.
The given quadratic equation is as shown below: $y=a{{x}^{2}}+bx+c$ The given points are $\left( 4,1 \right),\left( 2,-2 \right),\left( 6,-2 \right)$. To find the values of a, b, and c of the quadratic equation, substitute the following pairs in the equation. Substitute $x=4$ and $y=1$ \begin{align} & 1=a{{\left( 4 \right)}^{2}}+b\left( 4 \right)+c \\ & 1=16a+4b+c \end{align} Substitute $x=2$ and $y=-2$ \begin{align} & -2=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\ & -2=4a+2b+c \end{align} Substitute $x=6$ and $y=-2$ \begin{align} & -2=a{{\left( 6 \right)}^{2}}+b\left( 6 \right)+c \\ & -2=36a+6b+c \end{align} Mark the resultant system of equations as shown below: $16a+4b+c=1$ (I) $4a+2b+c=-2$ (II) $36a+6b+c=-2$ (III) By multiplying equation (III) by -1 and eliminating c from equations (I) and (III), we get \begin{align} & \text{ }16a+4b+c=\text{ }1 \\ & \underline{-36a-6b-c\text{ }=\text{ 2}} \\ & \text{ }-\text{20}a-2b\text{ }=\text{ }3 \\ \end{align} (IV) By multiplying the equation (III) by -1 and eliminating c from equations (II) and (III), we get \begin{align} & \text{ }4a+2b+\text{ }c=-2 \\ & \underline{-36a-6b-c=\text{ 2}} \\ & -32a-4b\text{ }=\text{ }0 \\ \end{align} (V) By multiplying the equation (IV) by -2 and then adding equations (IV) and (V), we get the value of a, \begin{align} & -32a-4b=\text{ }0 \\ & \underline{\text{ }40a+4b=-6} \\ & \text{ }8a=-6 \\ \end{align} Now, divide both sides by 8: \begin{align} & \frac{8a}{8}=\frac{-6}{8} \\ & \text{ }a=-\frac{3}{4} \\ \end{align} Substitute the value of a in equation (IV) to get: \begin{align} & -\text{20}\left( -\frac{3}{4} \right)-2b=3\text{ } \\ & 15-2b=3 \end{align} Now, subtract 15 from both sides, \begin{align} & 15-2b-15=3-15 \\ & -2b=-12 \end{align} By dividing both sides by -2, we get \begin{align} & \frac{-2b}{-2}=\frac{-12}{-2} \\ & \text{ }b=6 \\ \end{align} Substitute the values of b and a in equation (II) to get the value of c, \begin{align} & 4\left( -\frac{3}{4} \right)+2\left( 6 \right)+c=-2 \\ & -3+12+c=-2 \\ & 9+c=-2 \end{align} Now, subtract $9$ from both sides and we get \begin{align} & 9+c-9=-2-9 \\ & \text{ }c=-11 \\ \end{align} Thus, the quadratic equation is $-\frac{3}{4}{{x}^{2}}+6x-11$.