## Precalculus (6th Edition) Blitzer

The quadratic equation is $2{{x}^{2}}+x-5$.
The standard quadratic equation is shown below: $y=a{{x}^{2}}+bx+c$ The points are $\left( -1,-4 \right),\left( 1,-2 \right),\left( 2,5 \right)$. In order to get the quadratic function, find the values of a, b, and c, and substitute the values of the ordered pairs in the quadratic function. Put $x=-1$ and $y=-4$ \begin{align} & -4=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c \\ & -4=a-b+c \end{align} (I) Put $x=1$ and $y=-2$ \begin{align} & -2=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c \\ & -2=a+b+c \end{align} (II) Put $x=2$ and $y=5$ \begin{align} & 5=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\ & 5=4a+2b+c \end{align} (III) Get the system of equations as follows: \begin{align} & a-b+c=-4 \\ & a+b+c=-2 \\ & 4a+2b+c=5 \end{align} Eliminate a from equations (I) and (II) to get; \begin{align} & \text{ }a-b+c=-4 \\ & -a-b-c=2 \\ \end{align} $-2b=-2$ Divide both sides by $-2$ to get: \begin{align} & \frac{-2b}{-2}=\frac{-2}{-2} \\ & b=1 \end{align} Putting in the value of b in equation (I): we get \begin{align} & a-\left( 1 \right)+c=-4 \\ & a+c-1=-4 \end{align} Now, adding 1 to both sides to get: \begin{align} & a+c-1+1=-4+1 \\ & a+c=-3 \end{align} (IV) Putting the value of b in equation (III) to get: \begin{align} & 4a+2\left( 1 \right)+c=5 \\ & 4a+2+c=5 \end{align} Subtract 2 from both sides to get, \begin{align} & 4a+c+2-2=5-2 \\ & 4a+c=3 \end{align} (V) By eliminating c from equations (IV) and (V) \begin{align} & 4a+c=\text{3} \\ & -a-c=\text{3} \end{align} $3a\text{ }=\text{ }6$ (VI) By dividing both sides by 3: we get, \begin{align} & \frac{3a}{3}=\frac{6}{3} \\ & a=2 \end{align} Putting the value of a in equation (IV) to find the value of c: we get $2+c=-3$ Now, subtract 2 from both sides to get, \begin{align} & 2+c-2=-3-2 \\ & c=-5 \end{align} Now putting the values of a, b and c to get the equation $2{{x}^{2}}+x-5$. Hence, the quadratic equation is $2{{x}^{2}}+x-5$.