## Precalculus (6th Edition) Blitzer

The quadratic function is $y=2{{x}^{2}}-x+3$.
Let us consider the given equation: $y=a{{x}^{2}}+bx+c$ …… (I) If the graph of $y=a{{x}^{2}}+bx+c$ passes through the point $\left( -1,6 \right),\left( 1,4 \right)\text{ and }\left( 2,9 \right)$ then it must satisfy the given equation. Therefore, putting the value of $x=-1\text{ and }y=6$ in equation (I): $6=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c$ $6=a-b+c$ …… (II) Putting the value of $x=1\text{ and }y=4$: $4=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c$ $4=a+b+c$ …… (III) Now, putting the value of $x=2,y=9$ $9=4a+2b+c$ …… (IV) Now, subtract equation (III) from equation (II): \begin{align} & a-b+c=6 \\ & a+b+c=4 \\ & \underline{-\text{ }-\text{ }-\text{ }-\text{ }} \\ & -2b=2 \\ \end{align} Further simplify, $b=-1$ Putting the value of $b=-1$ in equation (IV): \begin{align} & 4a+2\left( -1 \right)+c=9 \\ & 4a-2+c=9 \\ \end{align} $4a+c=11$ …… (V) Now, adding equation (II) and equation (III): \begin{align} & a-b+c=6 \\ & \underline{a+b+c=4} \\ & 2a+2c=10 \\ \end{align} $a+c=5$ …… (VI) Now, subtract the equation (VI) from equation (V): \begin{align} & 4a+c=11 \\ & a+c=5 \\ & \underleftarrow{-\text{ }-\text{ }-\text{ }} \\ & 3a=6 \\ \end{align} Further simplify, $a=2$ Putting in the value of $a=2$ in equation (VI): \begin{align} & 2+c=5 \\ & c=3 \end{align} Now, putting in the values of $a,b\text{ and }c$ in the equation $y=a{{x}^{2}}+bx+c$ then the equation is, $y=2{{x}^{2}}-x+3$ Thus, the quadratic function is $y=2{{x}^{2}}-x+3$.