## 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 $\left( -2,7 \right),\left( 1,-2 \right)\text{ and }\left( 2,3 \right)$ then it must satisfy the given equation. Therefore, putting in the value of $x=-2,y=7$ in equation (I): $7=a{{\left( -2 \right)}^{2}}+b\left( -2 \right)+c$ $7=4a-2b+c$ …… (II) Now putting in the value of $x=1\text{ and }y=-2$: $-2=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c$ $-2=a+b+c$ …… (III) Putting the value of $x=2\text{ and }y=3$: $3=4a+2b+c$ …… (IV) Subtract equation (II) from equation (I): \begin{align} & 4a-2b+c=7 \\ & a+b+c=-2 \\ & \underline{-\text{ }-\text{ }-\text{ + }} \\ & 3a-3b=9 \\ \end{align} Further simplify: $a-b=3$ …… (V) Now, subtract equation (III) from equation (I): \begin{align} & 4a-2b+c=7 \\ & 4a+2b+c=3 \\ & \underline{-\text{ }-\text{ }-\text{ }-\text{ }} \\ & -4b=4 \\ & \text{ }b=-1 \\ \end{align} Putting in the value of $b=-1$ in equation (V): \begin{align} & a-\left( -1 \right)=3 \\ & a=2 \end{align} Putting in the value of $b=-1,a=2$ in equation (II): \begin{align} & 2+\left( -1 \right)+c=-2 \\ & 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 $y=2{{x}^{2}}-x-3$.