Chapter 4 - Section 4.3 - Quadratic Functions and Their Properties - 4.3 Assess Your Understanding - Page 300: 59

See below.

Let's compare $f(x)=-x^2+10x-4$ to $f(x)=ax^2+bx+c$. We can see that a=-1, b=10, c=-4. $a\lt0$, hence the graph opens down, hence its vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{10}{2\cdot(-1)}=5.$ Hence the maximum value is $f(5)=-(5)^2+10(5)-4=21.$