PS problem on graph:X intercept

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avanti@gmat: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Tue May 20, 2014 3:01 am; Thanked: 1 times; Followed by:1 members

PS problem on graph:X intercept

by avanti@gmat » Sat Jul 19, 2014 7:53 am

Figure shows the graph of y=(x+1)(x-1)^2.At how many points do the graph of y=(x+1)(x-1)^2+2 intercept x axis?
1)None
2)One
3)Two
4)Three
5)four
[url=https://postimg.cc/image/tykqqdvm5/][img]https://s30.postimg.cc/tykqqdvm5/graph.jpg[/img][/url]

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Source: Beat The GMAT — Problem Solving | Sat Jul 19, 2014 7:53 am

GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pm; Location: New Delhi, India; Thanked: 205 times; Followed by:24 members

by GMATinsight » Sat Jul 19, 2014 8:01 am

avanti@gmat wrote:Figure shows the graph of y=(x+1)(x-1)^2.At how many points do the graph of y=(x+1)(x-1)^2+2 intercept x axis?
1)None
2)One
3)Two
4)Three
5)four

For all points of intersection

(x+1)(x-1)^2+2 = 0

As Graph is just going 2 Units upward Therefore ONE point of intersection in Negative X-axis

Answer: Option B

Last edited by GMATinsight on Sat Jul 19, 2014 9:03 am, edited 2 times in total.

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pm; Location: New Delhi, India; Thanked: 205 times; Followed by:24 members

by GMATinsight » Sat Jul 19, 2014 8:03 am

Graphically you can understand this solution by SHIFTING OF GRAPH

The Graph of second Equation is obtained by Shifting the graph of first equation by 2 Units towards positive Y-axis (upward)

Therefore only one Point of intersection will remain the the graph is infinitely extended in third quadrant and intersecting X-axis on Negative X-axis.

Therefore 1 point of Intersection

Answer: Option B

Last edited by GMATinsight on Sat Jul 19, 2014 9:01 am, edited 2 times in total.

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avanti@gmat: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Tue May 20, 2014 3:01 am; Thanked: 1 times; Followed by:1 members

by avanti@gmat » Sat Jul 19, 2014 8:08 am

But OA given is one point.
What I solved is that I put y=0 in 2nd equation and got x=-3 and -1 and hence marked two points of intersection as answer choice but the correct is one point of intersection.
Considering what you said about shifting of Graph in Y+, then there is one point of intersection I suppose.

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pm; Location: New Delhi, India; Thanked: 205 times; Followed by:24 members

by GMATinsight » Sat Jul 19, 2014 8:26 am

You Are right about it. When you see the graphs then they are approaching each other in Negative X Axis and that's where it will intersect X-axis at ONE point.

Answer: Option B

Last edited by GMATinsight on Sat Jul 19, 2014 9:00 am, edited 2 times in total.

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avanti@gmat: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Tue May 20, 2014 3:01 am; Thanked: 1 times; Followed by:1 members

by avanti@gmat » Sat Jul 19, 2014 8:32 am

But i dint actually get how we are getting 2 points ie x=-3,-1 when we put y=0

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pm; Location: New Delhi, India; Thanked: 205 times; Followed by:24 members

by GMATinsight » Sat Jul 19, 2014 8:36 am

avanti@gmat wrote:But i dint actually get how we are getting 2 points ie x=-3,-1 when we put y=0

These are the roots of the equation i.e. points at which graph intersects X-Axis and not the point where the two graphs intersect

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avanti@gmat: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Tue May 20, 2014 3:01 am; Thanked: 1 times; Followed by:1 members

by avanti@gmat » Sat Jul 19, 2014 8:40 am

[quote]
These are the roots of the equation i.e. points at which graph intersects X-Axis and not the point where the two graphs intersect[/quote]

but the question only asks about the point of intersection of graph and not of the two graphs!!!

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pm; Location: New Delhi, India; Thanked: 205 times; Followed by:24 members

by GMATinsight » Sat Jul 19, 2014 8:59 am

I read the question wrong... My Bad..... Please ignore my previous comment on this question...

y=(x+1)(x-1)^2+2

@x = -3, y =(-3+1)(-3-1)^2 +2 = (-2)(-4)^2 +2 which is not equal to Zero
@x = -1, y =(-1+1)(-3-1)^2 +2 = 2 which is not equal to Zero

Therefore these are not the points where the new graph will intersect X-Axis

For solution, y=(x+1)(x-1)^2+2 = 0

(x+1)(x-1)^2 = -2
(x+1) must be Negative as (x-1)^2 has to be positive
the value may not be integer therefore we are in trouble

But there will be just one point as the graph has shifted two units upward

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avanti@gmat: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Tue May 20, 2014 3:01 am; Thanked: 1 times; Followed by:1 members

by avanti@gmat » Sat Jul 19, 2014 9:04 am

[quote="GMATinsight"]I read the question wrong... My Bad..... Please ignore my previous comment on this question...

y=(x+1)(x-1)^2+2

@x = -3, y =(-3+1)(-3-1)^2 +2 = (-2)(-4)^2 +2 which is not equal to Zero
@x = -1, y =(-1+1)(-3-1)^2 +2 = 2 which is not equal to Zero

[color=red]Therefore these are not the points where the new graph will intersect X-Axis[/color]

For solution, y=(x+1)(x-1)^2+2 = 0

(x+1)(x-1)^2 = -2
(x+1) must be Negative as (x-1)^2 has to be positive
the value may not be integer therefore we are in trouble

But there will be just one point as the graph has shifted two units upward[/quote]

Yes I found out that the roots x=-3,-1 doesnot satisfy the equation but dint get the further course prior. I thought that the point of intersection is an integer and missed the que.
Well now i got it. Thank You.