you need to find x-intercept when y=0, y=x^2+x(a+b)+ab
st(1) tells you a+b=-1. By using our question 0=x^2+x(-1)+ab, Not sufficient since we don't know ab;
st(2) suggests that y-intercept is -6 from coordinates (0,-6) <> y-intercept is when x=0. So, -6=(0+a)(0+b) or ab=-6 BUT we need to find x-intercept. Not Sufficient, as we don't know what is x-intercept;
Combined st(1&2): by substituting ab=-6 from st(2) into our original question simplified by st(1) we get 0=x^2-x-6 and solve for x -> (x-3)(x+2)=0. So x can be 3 or -2. We can stop here for DS and answer 'c', BUT if we wish to find the coordinates of x-intercept we continue
let's check the y-intercept validity, x must be 0
1) x=3, points (3;0) and (0;-6) -> slope=2 and the line equation is -6=2x-6, x=0 OK
2) x=-2, points (-2;0) and (0;-6) -> slope=-3 and the line equation is -6=-3x-6, x=0 OK
p.s. the graph of this function is not line BUT parabola. I have been testing the intercepts by using the line equation of linear functions and it's not linear function but quadratic function.
i believe the source of this question is a bit step forward from GMAT to other math test, as i've never run across the quadratic function parameters on GMAT.
Akansha wrote:In the xy plane, at what point does y = (x + a)(x + b) cross the x axis?
a. a + b = -1
b. graph intersects y axis at (0, -6)
The correct ans is C but I had opted for E. can someone explain why it would be C?[/spoiler]