y = (x+a)(x+b) = x^2 + (a+b)x +ab

This is the equation of right side up Parabola. ( Positive general quadratic equation )

y = ax^2 + bx + c ( general positive quadratic equation )

y = a(x – h)^2 + k, then the vertex is the point (h, k). This makes sense, if you think about it. The squared part is always positive (for a right-side-up parabola).

y = ax^2 + bx + c, the vertex (h, k) is found by computing h = –b/2a, and then evaluating y at h to find k.

k = (4ac – b^2) / 4a

Just the background if someone wants to know fundamentals

Okkkkk so the equation we have is

y = x^2 + (a+b)x +ab

vertex (h,k) can be found

h= -(a+b)/2

k=(4ab-(a+b)^2)/4

from (i) we get the vertex as ( -1/2, ab-1/4)

As we do not know ab, we can't determine the point where it corsses the X axis.

From (ii), we know the value of ab=-6

Combining (i) and (ii), we can definitely find the points where the parabola crosses the x-axis. ( I am not finding, but the points would be possibly ( not necessary ) coresponding points on the positive side and negative side of the X axis )

Therefore the answer is C

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Second simple and layman's approach

equation y=x^2+(a+b)x+ab

from (i) y= x^2-x+ab

For y=0, x= ab and also x = 1-ab

as we don't know value of ab, we can't say what are the two values of x

From(ii), we get ab=-6

From (i) and (ii), we can definitely find the values of x where the parabola instersects.

x=-6 and x= 7