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The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy

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New post Updated on: 08 Jul 2015, 10:58
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The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy-plane. At how many points does the graph of y = (x + 1)(x - 1)^2 + 2 intercept the x-axis?

A. None
B. One
C. Two
D. Three
E. Four

Attachment:
Untitled.png
Untitled.png [ 13.18 KiB | Viewed 52531 times ]

Originally posted by imhimanshu on 30 Sep 2013, 07:16.
Last edited by Bunuel on 08 Jul 2015, 10:58, edited 2 times in total.
Renamed the topic and edited the question.
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 30 Sep 2013, 08:16
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Image
The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy-plane. At how many points does the graph of y = (x + 1)(x - 1)^2 + 2 intercept the x-axis?

A. None
B. One
C. Two
D. Three
E. Four

The graph of f(x) + 2 is shifted 2 units up compared to f(x):

Image

Thus the graph of f(x) + 2 intersect the x-axis at one point.

Answer: B.

Attachment:
graph (1).png
graph (1).png [ 8.64 KiB | Viewed 51794 times ]

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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 30 Sep 2013, 10:26
Thanks Bunuel for the reply.
Frankly speaking, I don't have much idea about how quadratic behaves graphically. Can you suggest something to build my knowledge on this topic.

Thanks
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 01 Oct 2013, 01:58
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imhimanshu wrote:
Thanks Bunuel for the reply.
Frankly speaking, I don't have much idea about how quadratic behaves graphically. Can you suggest something to build my knowledge on this topic.

Thanks
Himanshu


Consider this: each y of \(y=(x + 1)(x - 1)^2 + 2\) is two greater than corresponding y of \(y=(x + 1)(x - 1)^2\). The value of y defines vertical position of a point. Thus \(y=(x + 1)(x - 1)^2 + 2\) is simply shifted 2 units up compared to \(y=(x + 1)(x - 1)^2\).

Next, notice that \(y=(x + 1)(x - 1)^2=x^3 - x^2 - x + 1\), thus it' not a quadratic function, its cubic.

As for quadratic functions, check the last chapter here: math-coordinate-geometry-87652.html

Hope this helps.
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 30 May 2014, 11:30
Thanks for the solution Bunuel.
However,when we equate (x + 1)(x - 1)^2 +2=0 for x intercept, it does not give any value for x.Any inputs on this please?
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 30 May 2014, 11:52
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ankushbassi wrote:
Thanks for the solution Bunuel.
However,when we equate (x + 1)(x - 1)^2 +2=0 for x intercept, it does not give any value for x.Any inputs on this please?


Actually there is: \(x=\frac{1}{3}(1-\frac{4}{\sqrt[3]{35-3\sqrt{129}}}-\sqrt[3]{35-3\sqrt{129}})\approx{-1.36}\).
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 11 Jan 2015, 20:37
@Bunnel, I solved it the following way, do you think there is any problem with the approach? Thanks

The graph intersect x-axis when y = 0, so --> (x+1)(x-1)^2 +2 = 0 ---> (x+1) = -2 OR (x-1)^2 = -2

Since you can't take the square root of -ve number, therefore it only has one solution i.e. x = -1
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 12 Jan 2015, 00:47
bkhan wrote:
@Bunnel, I solved it the following way, do you think there is any problem with the approach? Thanks

The graph intersect x-axis when y = 0, so --> (x+1)(x-1)^2 +2 = 0 ---> (x+1) = -2 OR (x-1)^2 = -2

Since you can't take the square root of -ve number, therefore it only has one solution i.e. x = -1


Why should x+1 or (x-1)^2 be -2?
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 12 Jan 2015, 13:14
Yes, I guess I can't just equate with -2. You can only do that with 0. I've have to expand and adjust 2 if I have to solve using algebra.
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 02 Jun 2015, 09:06
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imhimanshu wrote:
Attachment:
Untitled.png
The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy-plane. At how many points does the graph of y = (x + 1)(x - 1)^2 + 2 intercept the x-axis?

A. None
B. One
C. Two
D. Three
E. Four


I solved it this way :

First observation : Both the equations look similar it's just the y intercept that is changing. Therefore the graph would remain same but will shift up or down given the sign of the intercept (here it is +2).

Put the value of x=0 in the first equation you get y=1
Now put the value of x=0 in the second equation you get y=2

Hence the graph is moving one unit above its actual place in the Y axis. Visibly you can notice it will no longer touch the X axis twice but only once.

Hope this helps.

Cheers !
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 30 Jun 2015, 04:02
SD007 wrote:
imhimanshu wrote:
Attachment:
Untitled.png
The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy-plane. At how many points does the graph of y = (x + 1)(x - 1)^2 + 2 intercept the x-axis?

A. None
B. One
C. Two
D. Three
E. Four


I solved it this way :

First observation : Both the equations look similar it's just the y intercept that is changing. Therefore the graph would remain same but will shift up or down given the sign of the intercept (here it is +2).

Put the value of x=0 in the first equation you get y=1
Now put the value of x=0 in the second equation you get y=2

Hence the graph is moving one unit above its actual place in the Y axis. Visibly you can notice it will no longer touch the X axis twice but only once.

Hope this helps.

Cheers !



Hi,

Though I agree with solution, but for the concept sake, is it not possible for graph to move up at y=3 and still touch the x xis at sm other point apart from x= -1 or y it even touches x= -1 when x=-1 doesnt satisfy the equation.

Please guide.
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 08 Jul 2015, 10:36
Hi Bunuel,

Though I agree with solution, but for the concept sake, is it not possible for graph to move up at y=3 and still touch the x axis at sm other point apart from x= -1 ?

2. why it even touches x= -1 when x=-1 doesnt satisfy the equation ?

Please guide.
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 08 Jul 2015, 10:56
prashantmanit wrote:
Hi Bunuel,

Though I agree with solution, but for the concept sake, is it not possible for graph to move up at y=3 and still touch the x axis at sm other point apart from x= -1 ?

2. why it even touches x= -1 when x=-1 doesnt satisfy the equation ?

Please guide.


Image

1. No. When you shift blue graph 2 units up it does not change the shape, it simply moves up.

2. Blue (original) graph intersect x-axis at two points: at x= -1 and x = 1. x-intercept(s) of a graph is the value(s) of x when y = 0, so x-intercepts of y = (x + 1)(x - 1)^2 are values of x such that (x + 1)(x - 1)^2 = 0, so x = -1 or x = 1.
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 07 Sep 2015, 21:21
Bunuel wrote:
ankushbassi wrote:
Thanks for the solution Bunuel.
However,when we equate (x + 1)(x - 1)^2 +2=0 for x intercept, it does not give any value for x.Any inputs on this please?


Actually there is: \(x=\frac{1}{3}(1-\frac{4}{\sqrt[3]{35-3\sqrt{129}}}-\sqrt[3]{35-3\sqrt{129}})\approx{-1.36}\).


Hi Bunuel,

I am unable to find the value of X in the equation below:
(x+1)(X-1)^2+2= 0 (When on the X-axis)
Now solving above we get to: (X+1)(X^2+1-2X)+2=0
Now X^3+X-2X^2+X^2+1-2X+2=0
X^3-X^2-X+3=0
x(X^2-X-1)+3=0
How to go further.

Thanks
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New post 07 Sep 2015, 21:44
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


The figure shows the graph of y=(x+1)(x-1)^2 in the xy-plane. At how many points does the graph of y=(x+1)(x-1)^2 +2 intercept the x-axis?
a) none
b) one
c) two
d) three
e) four

The Problem actually asks the number of intercepts not the exact intercepting point. So let's consider (x+1)(x-1)^2 +2=0 and change the equation into (x+1)(x-1)^2 =-2. Then the problem is changed to find the number of intercepts between y=(x+1)(x-1)^2 and y=-2. Looking graph you would easily find that y=(x+1)(x-1)^2 intercepts y=-2(which is the line parallel to the x-axis passing (0,-2)) only once. So the answer is B.
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The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 11 Mar 2016, 04:35
I used the quadratic formula to solve this question and I got the answer right. Was I lucky, or can I just use the quadratic formula to figure out the number of real roots of this problem. Considering that this apparently is a cubic function.
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The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 11 Mar 2016, 05:52
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saiesta wrote:
I used the quadratic formula to solve this question and I got the answer right. Was I lucky, or can I just use the quadratic formula to figure out the number of real roots of this problem. Considering that this apparently is a cubic function.


I am curious to see how did you employ the QUADRATIC formula to a CUBIC equation. If you used the quadratic formula to come up with something like (x-a)(x-b) then yes, you can apply the same to break down a cubic equation to its corresponding roots but for any other reason, this method will not be correct.

The most straightforward way is to use the given graph and know that polynomials of the form f(x) = x^n and f(x)= x^n+constant will be the same except the fact that the graph of f(x)=x^n will be moved by the amount = constant (up or down will depend upon the sign of the constant!).

Thus y=(x+1)(x-1)^2+2 will have the graph of y=(x+1)(x-1)^2 shifted UP by 2 units and hence the x-axis intersection will be at 1 point.

Hope this helps.
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 11 Mar 2016, 07:00
Engr2012 wrote:
saiesta wrote:
I used the quadratic formula to solve this question and I got the answer right. Was I lucky, or can I just use the quadratic formula to figure out the number of real roots of this problem. Considering that this apparently is a cubic function.


I am curious to see how did you employ the QUADRATIC formula to a CUBIC equation. If you used the quadratic formula to come up with something like (x-a)(x-b) then yes, you can apply the same to break down a cubic equation to its corresponding roots but for any other reason, this method will not be correct.

The most straightforward way is to use the given graph and know that polynomials of the form f(x) = x^n and f(x)= x^n+constant will be the same except the fact that the graph of f(x)=x^n will be moved by the amount = constant (up or down will depend upon the sign of the constant!).

Thus y=(x+1)(x-1)^2+2 will have the graph of y=(x+1)(x-1)^2 shifted UP by 2 units and hence the x-axis intersection will be at 1 point.

Hope this helps.


Thank you for the explanation, I understand now. Quant is not my strongest side but I have been improving. I used the discriminant of the quadratic formula \(b^2 - 4ac\) but now that I know that I can't apply this to a cubic function could I use this discriminant instead \(b^2 - 3ac\)?
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 11 Mar 2016, 07:26
saiesta wrote:
Engr2012 wrote:
saiesta wrote:
I used the quadratic formula to solve this question and I got the answer right. Was I lucky, or can I just use the quadratic formula to figure out the number of real roots of this problem. Considering that this apparently is a cubic function.


I am curious to see how did you employ the QUADRATIC formula to a CUBIC equation. If you used the quadratic formula to come up with something like (x-a)(x-b) then yes, you can apply the same to break down a cubic equation to its corresponding roots but for any other reason, this method will not be correct.

The most straightforward way is to use the given graph and know that polynomials of the form f(x) = x^n and f(x)= x^n+constant will be the same except the fact that the graph of f(x)=x^n will be moved by the amount = constant (up or down will depend upon the sign of the constant!).

Thus y=(x+1)(x-1)^2+2 will have the graph of y=(x+1)(x-1)^2 shifted UP by 2 units and hence the x-axis intersection will be at 1 point.

Hope this helps.


Thank you for the explanation, I understand now. Quant is not my strongest side but I have been improving. I used the discriminant of the quadratic formula \(b^2 - 4ac\) but now that I know that I can't apply this to a cubic function could I use this discriminant instead \(b^2 - 3ac\)?


Where did you get that the discriminant of a cubic polynomial is \(b^2 - 3ac\) ? It is not correct.

For a cubic polynomial, ax^3+bx^2+cx+d , the discriminant is = \(b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd\) , weird and not useful for GMAT to remember.

Using any other method but the graphical method is going to be lengthy and time consuming.

Hope this helps.
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The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy  [#permalink]

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New post 18 May 2016, 03:00
Bunuel
I am curious to know if question is twisted like

y = (x + 1)(x - 1)^2 in the xy-plane. At how many points does the graph of y = (x + 1)(x - 1)^2 -2 intercept the x-axis?

PS : I am looking for scenarios, where it can cut x axis more than once. :shock:
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