f(x)=-(x+5)(x+2)
f(x)=-(x^2+5x+2x+10)
f(x)=-(x^2+7x+10)
f(x)=-x^2-7x-10
df/dx=-2x-7 and d2f/dx2=-2
Since d2f/dx2, acceleration, is a constant negative, when df/dx, velocity, equals zero, f(x) is at an absolute maximum.
df/dx=0 when -2x-7=0, -2x=7, x=-3.5
f(-3.5)=22.25
The end behavior, as x approaches ±oo is simply the behavior of y as the leading term when x approaches ±oo.
-x^2 as x approaches ±oo approaches -oo
So the range of this function is y=(-oo, 22.5]
