## Precalculus (6th Edition) Blitzer

The required equation is, $1682=16a+4b+c$.
It is provided that: x in $y=a{{x}^{2}}+bx+c$ is 4; The value of y is 1682. Substitute the value of x in the given equation: \begin{align} & 1682=a{{\left( 4 \right)}^{2}}+4b+c \\ & 1682=16a+4b+c \\ \end{align} Thus, the solution is $1682=16a+4b+c$.