Question 1013975
It depends on what you want to know.
.
y=-x^2-10x
y=(-1x)(x+10)
.
To find the roots, let y=0
(-1x)(x+10)=0
-1x=0 --OR-- x+10=0
x=0 --OR-- x=-10 
Intercepts x axis at 0 and -10.
.
To find the x value at maximum(or minimum):
Set first derivative=0
.
f(x)=-x^2-10x . (original equation)
.
f'(x)=-2x-10 . (first derivative)
0=-2x-10
10=-2x
-5=x  
.
The sign of the second derivative tells if 
it has a maximum or minimum (if negative, 
it has maximum, if positive it has minimum)
f''(x)=-2 . (second derivative)
is negative, so there is a maximum where x=-5.
. 
Find y where x=-5:
.
f(x)=-x^2-10x
f(-5)=-(-5)^2-10(-5)=-25=50=25
The maximum is at the point (-5,25), 
crosses x-axis at -10 and 0
.
GRAPH:
.
{{{ graph( 800, 800, -15, 15, -3, 28, -x^2-10x) }}}