## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$a=6, b=0,c=2$
When the vertex of a graph is at $(h, k)$, then the general form for the quadratic function can be expressed as: $f(x) = a(x-h)^2+k~~~(1)$ We are given that the vertex is at $(h,k)=(0,2)$ Therefore, $f(x)= a(x-0)^2+k \implies f(x) =ax^2 +2 ~~~(2)$ Plug the point $(1,8)$ into equation (2) to obtain: $8 = a(1)^2+2 \implies a=6$ Thus, the equation of the function is: $f(x)=6x^2+2$ On comparing $f(x)=6x^2+2$ with $f(x) = ax^2+bx+c$, we get: $a=6, b=0,c=2$