Answer
maximum, $13$
Work Step by Step
1) The graph of the quadratic function $f(x) = ax^2+bx+c$ is a parabola that opens:
(a) Upward when $a \gt 0$ and its vertex has a minimum value.
(b) Downward when $a \lt 0$ and its vertex has a maximum value.
2) The coordinates of the vertex of a quadratic function $f(x) = ax^2+bx+c$ are given by: $\displaystyle(\frac{-b}{2a}, f(-\frac{2}{a}))$
On comparing $f(x)=-3x^2+12x+1$ with $f(x) = ax^2+bx+c$, we get: $a=-3, b=12,c=1; a \lt 0$. We can see that the given function shows a graph of a parabola that opens downward. So, its vertex has a maximum value.
Thus, the maximum value at $x$ can be expressed as: $x=\displaystyle \frac{-b}{2a}=\frac{-(12)}{2(-3)}=2$
Therefore, the maximum value of the graph is: $f(2)=(-3)(2)^2+(12)(2)+1=13$
