In the xy-plane, at how many points do line y = 6 and curve y = |x^2-

Question

In the xy-plane, at how many points do line y = 6 and curve  y=|x^2-x-6|  intersect?

A. 0
B. 1
C. 2
D. 3
E. 4

gmatCuck · Answer

We need to substitute y=6 in the other equation to find their points of intersection

Modulus indicates we can take x2 - x - 6 = 6 or x2 - x - 6 = - 6 (i. e. y=6 and y= - 6)

There are four points of intersection just by solving the equations

However, the question asks only about line y = 6 and not y = - 6

So we choose option C

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prachisaraf · Answer

I believe that if the equation did not have a modulus, then we would consider only positive value of y and thus, the answer would have been 2. Can someone post their working of how they arrived at 4 ?

Regor60 · Answer

Substitute 6 for Y:

6 = |X^2-X-6|

So the value within the absolute value can be either 6 or -6.

X^2-X-6 = 6, so

X^2-X-12=0 and

X=-3 or 4

Similarly,

X^2-X-6=-6 so

X^2-X=0 and

X(X-1)=0, so

X=1 or 0

Total points: X=4,-3,1,0

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Regor60 · Answer

There is no Y=-6 because Y is defined as positive.

However, this does not eliminate the value within the absolute value from being -6 because the action of the absolute value renders it positive

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=xBFh9WL_zgc

In the xy-plane, at how many points do line y = 6 and curve  y=|x^2 - x - 6|  intersect

Line and the curve intersect, so we can equate both of them to find the point of intersection

=>  |x^2 - x - 6| = 6

We will have two cases

-   Case 1:   x^2 - x - 6 = 6  
=>  x^2 - x - 12 = 0 
=>  x^2 – 4x + 3x - 12 = 0 
=> x*(x - 4) + 3*(x - 4) = 0
=> (x - 4) * (x + 3) = 0
=>  x = -3, 4

-   Case 2:   x^2 - x - 6 = -6

=>  x^2 - x = 0 
=> x*(x - 1) = 0

=>  x = 0, 1

=> We will have 4 solutions.

So,  Answer will be E 
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

https://www.youtube.com/watch?v=huFiBk8PaGY

mohammedturk · Answer

because the f(x) is in absolute value so the negative sign would be positive which would change the curve graph and make it reflect to the positive quadrants

In the xy-plane, at how many points do line y = 6 and curve y = |x^2-

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We will have two cases
-Case 1: \(x^2 - x - 6 = 6\) => \(x^2 - x - 12 = 0\) => \(x^2 – 4x + 3x - 12 = 0\) => x(x - 4) + 3(x - 4) = 0 => (x - 4) * (x + 3) = 0 => x = -3, 4	-Case 2: \(x^2 - x - 6 = -6\) => \(x^2 - x = 0\) => x*(x - 1) = 0 => x = 0, 1

GMAT Problem Solving (PS) Questions

In the xy-plane, at how many points do line y = 6 and curve y = |x^2-

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards