## Precalculus (6th Edition) Blitzer

The required quadratic equation is $y=3{{x}^{2}}-4x+5$.
We know that the graph is passing through the point $\left( 1,4 \right);$ the equation becomes: $a+b+c=4$ Now, the graph is passing through the point $\left( 3,20 \right);$ the equation becomes: $9a+3b+c=20$ Then the graph is passing through the point $\left( -2,25 \right);$ the equation becomes: $4a-2b+c=25$ Subtract equation $a+b+c=4$ from equation $9a+3b+c=20$, given that \begin{align} & 8a+2b=16 \\ & b=8-4a \end{align} Subtract equation $a+b+c=4$ from equation $4a-2b+c=25$ to get: \begin{align} & 3a-3b=21 \\ & a-b=7 \end{align} Substitute the value of b from equation $b=8-4a$ in equation $a-b=7$ as given below: \begin{align} & a-\left( 8-4a \right)=7 \\ & a-8+4a=7 \\ & 5a=15 \\ & a=3 \end{align} From equation $b=8-4a$, \begin{align} & b=8-4\times 3 \\ & =-4 \end{align} From equation $9a+3b+c=20$, \begin{align} & c=4-a-b \\ & =4-3+4 \\ & =5 \end{align} Therefore, $a=3,b=-4,\text{ and }c=5$; put these values in the quadratic equation $y=a{{x}^{2}}+bx+c$ as given below: $y=3{{x}^{2}}-4x+5$ Hence, the quadratic equation is $y=3{{x}^{2}}-4x+5$.