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]

y=(x+a)*(x+b);

now this is a parabola which will intersect the x-axis at the point(-a,0) and (-b,0);

1) a+b=-1; here different results are possible depending upon different values of a and b; for example a=-3 and b=2; a=-4, b=3; a=-5, b=4; etc..
hence 1 alone is not sufficient to answer the question..!!!

2) graph intersect the y-axis at(0,-6);
now when graph intersect y-axis its x co-ordinate will be zero;
putting x=0 in the equation y=(x+a)*(x+b); we have;
y=a*b;
and also a*b=-6;
again here different values are possible for example; a=-6 b=1; a=-3, b=2 hence 2 alone is not sufficient to answer the question.!!

combining 1 and 2 we have;

a*b=-6;
and a+b=-1;
the only possible value for a and b can be; a=-3, b=2 or a=2 and b=-3 {which both means same as question is asking about the point at which curve intersects the x-axis and not about the value of and b;}

hence answer should be C

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]

The graph crosses the x axis when y=0.
In the equation y = (x + a)(x + b), y=0 when either (x+a) = 0 or (x+b) = 0.
In order to determine which values of x will make the factors above equal 0, we need to know what combination of values is represented by a and b.

Statement 1: a+b = -1.
The equation above has an infinite number of solutions.
Insufficient.

Statement 2: The graph intersects the y axis at (0,-6).
Plugging (0,-6) into y = (x + a)(x + b), we get:
-6 = (0+a)(0+b)
-6 = ab.
The equation above has an infinite number of solutions.
Insufficient.

Statements 1 and 2 combined:
a+b = -1 can be rewritten as b= -a-1.
Plugging b = -a-1 into ab = -6, we get:
a(-a-1) = -6.
-a² - a + 6 = 0.
a² + a - 6 = 0.
(a+3)(a-2) = 0.
a=-3 or a=2.

Plugging a=-3 into ab=-6, we get b=2.
Plugging a=2 into ab=-6, we get b=-3.
In each case, we get the same combination of values: -3 and 2.
Thus, we know that the two factors of the graph are (x-3) and (x+2) and that the graph crosses the x axis when x=3 and x=-2.
Sufficient.

The correct answer is C.

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]