Graph the function and determine the intervals for which f(x)>0 f(x)=-x^2-2x

Answer 1

-x^2 -2x = x(-x+2) x intercepts are x=0 and x=2

the vertex is found from vertex form -(x^2 +2x+1) +1, (1,1)

interval where f(x)>0 is (0,2)

It's a downward opening parabola with maximum = (1,1) and f(x)>0 between the x intercepts

from (0,0) to (2,0) or from y=0 to y=1