#### Answer

maximum, $21$

#### 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 $f(x)=-x^2+10x-4$ with $f(x) = ax^2+bx+c$, we get: $a=-1, b=10,c=-4; a \lt 0$. 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: $x=\displaystyle \frac{-b}{2a}=\frac{-(10)}{2(-1)}=5$
Therefore, the maximum value of the graph is: $f(5)=(-5)^2+(10)(5)-4=21$