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)
Alternate approach:
The question stem is asking for the x-intercepts.
An x-interceps occurs when y=0.
Statement 1: a+b = -1
Case 1: a=1, b=-2
Substituting a=1 and b=-2 into y= (x+a)(x+b), we get:
y = (x+1)(x-2).
Here, y=0 when x=-1 or x=2.
Thus, the x-intercepts are -1 and 2.
Case 2: a=2, b=-3
Substituting a=2 and b=-3 into y= (x+a)(x+b), we get:
y = (x+2)(x-3).
Here, y=0 when x=-2 or x=3.
Thus, the x-intercepts are -2 and 3.
Since the x-intercepts are different in each case, INSUFFICIENT.
Statement 2: The graph intersects the y-axis at (0, -6)
Substituting (0, -6) into y = (x+a)(a+b), we get:
-6 = (0+a)(0+b)
-6 = ab.
Case 2 also satisfies statement 2, since ab=-6 if a=2 and b=-3.
In Case 2, the x-intercepts are -2 and 3.
Case 3: a=1, b=-6
Substituting a=1 and b=-6 into y= (x+a)(x+b), we get:
y = (x+1)(x-6).
Here, y=0 when x=-1 or x=6.
Thus, the x-intercepts are -1 and 6.
Since the x-intercepts are different in each case, INSUFFICIENT.
Statements combined:
Case 2 satisfies both statements.
In Case 2, the x-intercepts are -2 and 3.
Check whether any other combination for x and y will satisfy both ab=-6 and a+b = -1.
Factor pairs for ab=-6 other than a=2 and b=-3:
a=-6, b=1
a=-3, b=2
a=-2, b=3
a=-1, b=6
a=1, b=-6
a=3, b=-2.
Only the option in red satisfies a+b=-1.
Case 4: a=-3, b=2
Substituting a=-3 and b=2 into y= (x+a)(x+b), we get:
y = (x-3)(x+2).
Here, y=0 when x=3 or x=-2.
Thus, the x-intercepts are -2 and 3.
Since Cases 2 and 4 yield the same x-intercepts, SUFFICIENT.
The correct answer is
C.
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