In the xy plane, at what two points does the graph

[lucky20]

In the xy plane, at what two points does the graph of y = (x+a) (x+b) intersect the x axis?

1. a+b = -1
2. The graph intersects the y axis at (0,-6)

The answer is C.

Can someone explain how to solve this question?

[sanjeev]

Hi,

The graph is y = (x+a) (x+b)

To find out at what point this graph intersect x-axis approach as follows:-

When a graph intersect x-axis, the intersection point would be (x,0).
When a graph intersect y-axis, the intersection point would be (0,y).


Applying the same to equation , we have
=> (x+a) (x+b) = 0
=> x^2 + (a+b)x + ab = 0
The two points can be determined by solving this equation or by finding out the values of a and b.


(1) a+b = -1 , we still dont know the values of a and b. INSUFFICIENT

(2) The graph intersect the y-axis at (0,-6).
So the graph y = (x+a) (x+b) can be rewritten as
-6 = (0 +a) (0+b)
-6 = ab. we still dont know the values of a and b. INSUFFICIENT



Combing (1) and (2) , we have a + b = -1 and ab = -6 ,

The two points would be x^2 + (a+b)x + ab = 0
x^2 -x -6 =0
x^2 -3x +2x -6 =0
x(x-3) +2(x-3) =0
(x-3) (x+2) = 0


So two points where the graph intersect is (3,0) and (-2,0). Hence SUFFICIENT using (1) and (2) so C.

Thanks

[RonPurewal]

Hi,

The graph is y = (x+a) (x+b)

To find out at what point this graph intersect x-axis approach as follows:-

When a graph intersect x-axis, the intersection point would be (x,0).
When a graph intersect y-axis, the intersection point would be (0,y).


Applying the same to equation , we have
=> (x+a) (x+b) = 0
=> x^2 + (a+b)x + ab = 0
The two points can be determined by solving this equation or by finding out the values of a and b.


(1) a+b = -1 , we still dont know the values of a and b. INSUFFICIENT

(2) The graph intersect the y-axis at (0,-6).
So the graph y = (x+a) (x+b) can be rewritten as
-6 = (0 +a) (0+b)
-6 = ab. we still dont know the values of a and b. INSUFFICIENT



Combing (1) and (2) , we have a + b = -1 and ab = -6 ,

The two points would be x^2 + (a+b)x + ab = 0
x^2 -x -6 =0
x^2 -3x +2x -6 =0
x(x-3) +2(x-3) =0
(x-3) (x+2) = 0


So two points where the graph intersect is (3,0) and (-2,0). Hence SUFFICIENT using (1) and (2) so C.

Thanks

well played.

[sudaif]

Hi,

The graph is y = (x+a) (x+b)

To find out at what point this graph intersect x-axis approach as follows:-

When a graph intersect x-axis, the intersection point would be (x,0).
When a graph intersect y-axis, the intersection point would be (0,y).


Applying the same to equation , we have
=> (x+a) (x+b) = 0
=> x^2 + (a+b)x + ab = 0
The two points can be determined by solving this equation or by finding out the values of a and b.


(1) a+b = -1 , we still dont know the values of a and b. INSUFFICIENT

(2) The graph intersect the y-axis at (0,-6).
So the graph y = (x+a) (x+b) can be rewritten as
-6 = (0 +a) (0+b)
-6 = ab. we still dont know the values of a and b. INSUFFICIENT



Combing (1) and (2) , we have a + b = -1 and ab = -6 ,

The two points would be x^2 + (a+b)x + ab = 0
x^2 -x -6 =0
x^2 -3x +2x -6 =0
x(x-3) +2(x-3) =0
(x-3) (x+2) = 0


So two points where the graph intersect is (3,0) and (-2,0). Hence SUFFICIENT using (1) and (2) so C.

Thanks

well played.

When I try to solve scenario statements 1 and 2 combined:
a + b =-1 and ab=-6
I get the quadratic equation a^2 + a - 6 = 0, by substituting b=-6/a into a + b = -1.
through factorization I get to (a-2) (a+3) = 0
which gives me two values for a and thus two values for b
since I DONT have a single, definitive, unique value for each a and b, i marked the answer as E - insufficient.
Ron or someone else, can u please tell me where I went wrong?

[rohit801]

sudaif,
read the problem - "In the xy plane, at what two points does the graph of y = (x+a) (x+b) intersect the x axis?

sure, you get two values of a and b to satisfy the QUAD. what are those: a=2 or -3 and b=2 or -3. Look at the equation when we put these values back in the original one:

y=(x+2)(x-3) or y=(x-3)(x+2)...SAME. question is NOT what specifically a and b are BUT where does the graph intersect the x-axis==> for what points of X is the value of y ZERO. It is clear, then, that Y will be ZERO when x=-2 or x=3.

Hope that helps...

[sudaif]

sudaif,
read the problem - "In the xy plane, at what two points does the graph of y = (x+a) (x+b) intersect the x axis?

sure, you get two values of a and b to satisfy the QUAD. what are those: a=2 or -3 and b=2 or -3. Look at the equation when we put these values back in the original one:

y=(x+2)(x-3) or y=(x-3)(x+2)...SAME. question is NOT what specifically a and b are BUT where does the graph intersect the x-axis==> for what points of X is the value of y ZERO. It is clear, then, that Y will be ZERO when x=-2 or x=3.

Hope that helps...

merci becoup!
for some reason, my brain has been trained to think a certain (wrong) way :)

[RonPurewal]

sudaif,
read the problem - "In the xy plane, at what two points does the graph of y = (x+a) (x+b) intersect the x axis?

sure, you get two values of a and b to satisfy the QUAD. what are those: a=2 or -3 and b=2 or -3. Look at the equation when we put these values back in the original one:

y=(x+2)(x-3) or y=(x-3)(x+2)...SAME. question is NOT what specifically a and b are BUT where does the graph intersect the x-axis==> for what points of X is the value of y ZERO. It is clear, then, that Y will be ZERO when x=-2 or x=3.

Hope that helps...

yep, nice analysis. this problem has been very carefully crafted so that it does not matter which of the two values is designated "a" and which is designated "b".