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)
OA: C
DS: Coordinates
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- rishab1988
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IMO C
Here is why:
Question is where y intersects x axis. When y intersects x axis t will be 0 ( because the coordinate of the point will be (0,x))
0=x^2+(a+b)x+ab
This is a quadratic equation. We need a+b and ab to determine the value of x.
1) a+b=-1
x^2 -x +ab =0
now and b could be anything for eg a =2 b=-1 gives a+b=-1 and ab=-2
Whereas a=-1/2 and b=-1/2 gives ab =1/4
Hence we can't solve the quadratic.
Eliminate A and D
2) when x =0 y=-6
substitute this in prompt to get : -6=ab
Then x^2+(a+b)x+ab=0 becomes x^2+(a+b)x-6=0
since a=-2 and b=3 gives ab=-6 and a+b=1 and a=-6 and b=1 gives a+b=-5
We can't solve the quadratic.Hence insufficient
Eliminate B
Combining 1 and 2
ab=-6 a+b=-1
The quadratic becomes x^2-x-6=0
We can solve this equation.Hence sufficient. The question is about two points (if it were for 1 point.again insufficient).
Answer C
What is the OA?
Here is why:
Question is where y intersects x axis. When y intersects x axis t will be 0 ( because the coordinate of the point will be (0,x))
0=x^2+(a+b)x+ab
This is a quadratic equation. We need a+b and ab to determine the value of x.
1) a+b=-1
x^2 -x +ab =0
now and b could be anything for eg a =2 b=-1 gives a+b=-1 and ab=-2
Whereas a=-1/2 and b=-1/2 gives ab =1/4
Hence we can't solve the quadratic.
Eliminate A and D
2) when x =0 y=-6
substitute this in prompt to get : -6=ab
Then x^2+(a+b)x+ab=0 becomes x^2+(a+b)x-6=0
since a=-2 and b=3 gives ab=-6 and a+b=1 and a=-6 and b=1 gives a+b=-5
We can't solve the quadratic.Hence insufficient
Eliminate B
Combining 1 and 2
ab=-6 a+b=-1
The quadratic becomes x^2-x-6=0
We can solve this equation.Hence sufficient. The question is about two points (if it were for 1 point.again insufficient).
Answer C
What is the OA?
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Target question: At which two points of the graph does y=(x+a)(x+b) intersect the x-axis?haidgmat wrote: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)
OA: C
IMPORTANT: Let's examine the point where a line (or curve) crosses the x-axis. At the point of intersection, the point is on the x-axis, which means that the y-coordinate of that point is 0. So, for example, to find where the line y=2x+3 crosses the x-axis, we let y=0 and solve for x. We get: 0 = 2x+3
When we solve this for x, we get x= -3/2.
So, the line y=2x+3 crosses the x-axis at (-3/2, 0)
Likewise, to determine the point where y = (x + a)(x + b) crosses the x axis, let y=0 and solve for x.
We get: 0 = (x + a)(x + b), which means x=-a or x=-b
This means that y = (x + a)(x + b) crosses the x axis at (-a, 0) and (-b, 0)
So, to solve this question, we need the values of a and b
Aside: y = (x + a)(x + b) is actually a parabola. This explains why it crosses the x axis at two points.
Now let's rephrase the target question...
REPHRASED target question: What are the values of a and b?
Statement 1: a + b = -1
There's no way we can use this to determine the values of a and b.
Since we can answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: The line intercepts the y axis at (0,-6)
This tells us that when x = 0, y = -6
When we plug x = 0 and y = -6 into the equation y = (x + a)(x + b), we get -6 = (0 + a)(0 + b), which tells us that ab=-6
In other words, statement 2 is a fancy way to tell us that ab = -6
Since there's no way we can use this information to determine the values of a and b, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined:
Statement 1 tells us that a+b = -1
Statement 2 tells us that ab = -6
Rewrite equation 1 as a = -1 - b
Then take equation 2 and replace a with (-1 - b) to get: (-1 - b)(b) = -6
Expand: -b - b^2 = -6
Set equal to zero: b^2 + b - 6 = 0
Factor: (b+3)(b-2) = 0
So, b= -3 or b= 2
When b = -3, a = 2 and when b = 2, a = -3
In both cases, the two points of intersection are (3, 0) and (-2, 0)
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
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CONCEPT: Intersections of Quadratic graph on x-Axis = roots of equation
Roots of teh given equation are , x = -1 and x = -b
To answer the question we need th evalues of a and b
STatement 1: a + b= -1
a = 0 and b = -1 or a = -2 and b = 1 (Inconsistent solutions hence
NOT SUFFICIENT
STatement 2: The graph intersects the y-axis at (0, -6)
i.e y-intercept of teh graph = -6
but, y=(x+a)(x+b) = x^2 + (a+b)x + ab
i.e. ab = -6
NOT SUFFICIENT
COmbining the statements
a + b= -1 and ab = -6
We can deduce that a = -3 and b = 2 hence
SUFFICIENT
Answer: Option C
Roots of teh given equation are , x = -1 and x = -b
To answer the question we need th evalues of a and b
STatement 1: a + b= -1
a = 0 and b = -1 or a = -2 and b = 1 (Inconsistent solutions hence
NOT SUFFICIENT
STatement 2: The graph intersects the y-axis at (0, -6)
i.e y-intercept of teh graph = -6
but, y=(x+a)(x+b) = x^2 + (a+b)x + ab
i.e. ab = -6
NOT SUFFICIENT
COmbining the statements
a + b= -1 and ab = -6
We can deduce that a = -3 and b = 2 hence
SUFFICIENT
Answer: Option C
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