The function can also be written f(x)=-(x^2+2x-1)=-(x^2+2x+1-2)=-(x+1)^2+2. The vertex can be seen to be at x=-1 (the line x=-1 is the line of symmetry), y=2 (the point (-1,2)). The zeroes would be the solution of -(x+1)^2+2=0 which is the same as (x+1)^2=2, so x+1=+sqrt(2). The zeroes will show as x intercepts because the curve (parabola) cuts the x axis at the two points -1+sqrt(2) and -1-sqrt(2) (0.4142 and -2.4142). The quadratic formula if applied to -x^2-2x+1 gives a=-1, b=-2 and c=1, so x=(2+sqrt(4-4(-1)(1))/(-2)=(2+sqrt(4+4))/(-2)=2+2sqrt(2))/(-2)=-1+sqrt(2). The y intercept is 1 (when x=0). The curve is an inverted U shape.