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

Question

https://gmatclub.com/forum/download/file.php?id=21232 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 Untitled.png

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=21232 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): https://gmatclub.com/forum/download/file.php?id=21242 Thus the graph of f(x) + 2 intersect the x-axis at one point. Answer: B. graph (1).png

Bunuel · Answer

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.

imhimanshu · Answer

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

ankushbassi · Answer

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?

Bunuel · Answer

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

SD007 · Answer

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 !

prashantmanit · Answer

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.

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=21242

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.

PrakharGMAT · Answer

Hi Bunuel / chetan2u , I am not able to understand, how can you say that the graph move up 2 units up. I can see that in second equation there is "+2" but still how could you say. If the equation would have been y = (x + 1)(x - 1)^2 - 2. Then will the graph will move 2 stps down..?? If the equation would have been x = (y + 1)(y - 1)^2 and some graph was given and we need to find the intercept of equation x = (y + 1)(y - 1)^2 + 2 or x = (y + 1)(y - 1)^2 -2. Then how will the graph move. I have gone through the link math-coordinate-geometry-87652.html but it does not provide anything related to the movement of graph. Can you please provide basic understanding on the movement of graph so that if I come across any weired equation in exam I can make a rough idea of it. I can't take this question lightly because its an official GMAT Prep question. Please assist. Thanks and Regards. Prakhar

chetan2u · Answer

Hi,

let  y = (x + 1)(x - 1)^2 ...
and  y_1 = (x + 1)(x - 1)^2 +2 = y+2 ......

so when  x= 0... y = (0+1)(0-1)^2 = 1  and  y_1 = (0+1)(0-1)^2 + 2 = 1+2 = 3 
when  x=1... y = (1+1)(1-1)^2 = 0 and  y_1=0+2 = 2 ..

so for earlier existing values of y,  y_1  is 2 more so the values of y has shifted two points up in new graph although the graph remains the same..

YES

x = (y + 1)(y - 1)^2 and  x_1 = (y + 1)(y - 1)^2 + 2 .... the graph will move two steps right

DRAW the graph for all four and you will strengthen the concept on this

Hope it is clear

keats · Answer

There is a smart way to do these problems. I'd like to explain it for a few scenarios:

Let y = f(x) and lets assume f(x) as given in this question (f(x) = (x+1)(x-1)²)

Case 1: f(x) → f(x) + a 
Move the graph of f(x) by a points  in upward direction  on the Y-axis, keeping its shape same

Case 2: f(x) → f(x) - a 
Move the graph of f(x) by a points  in downward direction  on the Y-axis, keeping its shape same

Case 3: f(x) → f(x+a) 
Move the graph of f(x) by a points  in the left direction  on the X-axis, keeping its shape same

Case 4: f(x) → f(x-a) 
Move the graph of f(x) by a points  in the right direction  on the X-axis, keeping its shape same
(Yes, in case 3 and 4 the graph moves in opposite direction. If you have f(x+a) it moves to the left side(negative X-axis), and if it is f(x-a), it moves to right side(positive X-axis))

Case 5: f(x) → -f(x) 
Take the image of f(x) in Y-axis as mirror

Case 6: f(x) → f(-x) 
Turn the graph of f(x) by 180 degree about X-axis

EMPOWERgmatRichC · Answer

Hi All,

Even though this graphing question looks complex, it's actually built on a real simple concept...

Graphing an equation is a fairly straightforward process when the equation is in "slope-intercept format" (e.g. y = mx + b); you plug in a value for X, do the calculation and get the value of Y, then graph the point (X,Y). Repeat as much as necessary.

Notice that the two given equations are almost identical? The only difference is that the second equation "adds 2" to the calculation. In the initial equation, you plug in a value for X and get a value for Y. In the second equation, you plug in the SAME value for X, but you have to "add 2" to the end result, so your Y is "2 greater" than before.

The effect of "adding 2" to Y means that the graph will look the same, but it will be "shifted up" 2 spots (since all of the values of Y will be 2 greater than they were before).

With the initial drawing, there are 2 X-intercepts. Shifting the entire graph "up" 2 spots will give us a picture with just 1 X-intercept.

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

BrentGMATPrepNow · Answer

Let's find some points that lie on each of the curves. 
So, for each equation, we'll find a pair of values (an x-value and a y-value) that satisfy each equation. 
We'll do so by plugging in some x-values and calculating the corresponding y-values.

Let's start with x =  0 
Plug x =  0  into the FIRST equation to get: y = ( 0  + 1)( 0  - 1)² =  1 
So, the point ( 0 ,  1 ) lies ON the curve defined by y = (x + 1)(x - 1)²

Now, plug x =  0  into the SECOND equation to get: y = ( 0  + 1)( 0  - 1)² + 2 =  3 
So, the point ( 0 ,  3 ) lies ON the curve defined by y = (x + 1)(x - 1)² + 2

Add the point ( 0 ,  3 ) to our graph to get: 
 https://i.imgur.com/or8bDzu.png

Notice that the point ( 0 ,  3 ) is 2 UNITS directly ABOVE the point ( 0 ,  1 ) 
---------------------------------------------

Let's try another x-value....
Try x =  1 
Plug x =  1  into the FIRST equation to get: y = ( 1  + 1)( 1  - 1)² =  0 
So, the point ( 1 ,  0 ) lies ON the curve defined by y = (x + 1)(x - 1)²

Now, plug x =  1  into the SECOND equation to get: y = ( 1  + 1)( 1  - 1)² + 2 =  2 
So, the point ( 1 ,  2 ) lies ON the curve defined by y = (x + 1)(x - 1)² + 2

Add the point ( 1 ,  2 ) to our graph to get: 
 https://i.imgur.com/MApa5fh.png

Notice that the point ( 1 ,  2 ) is 2 UNITS directly ABOVE the point ( 1 ,  0 ) 
---------------------------------------------

At this point, we should recognize that the graph of y = (x + 1)(x - 1)² + 2 is very similar to the graph of y = (x + 1)(x - 1)²
The only difference is that the graph of y = (x + 1)(x - 1)² + 2 is SHIFTED UP 2 units.

So, to graph the curve y = (x + 1)(x - 1)² + 2, we can just take every point on the curve y = (x + 1)(x - 1)² and move it UP 2 units...
 https://i.imgur.com/FYHNDpE.png

When we connect the points, we see that the graph of y = (x + 1)(x - 1)² + 2 looks something like this. 
 https://i.imgur.com/FoxY6Yc.png

From our sketch, we can see that the graph of y = (x + 1)(x - 1)² + 2 intercepts the x-axis ONCE

Answer: B

Cheers,
Brent

Kritisood · Answer

Ian Stewart had explained the basics behind this on another discussed question on the platform: Link to the explanation/question: https://gmatclub.com/forum/the-line-rep ... l#p1545268

GDT · Answer

VeritasKarishma

Can you pls explain me why the graph moves in opposite direction in case 3 & 4?

Also in case 5, if f(1)=2 then -f(1)=-2. So the x axis acts as mirror isn't it?

Thanks in advance!

KarishmaB · Answer

Try putting in some constants and then draw the graph to get a good understanding of how the graph moves.

Whatever the graph of f(x) is, the graph of f(x + 2) will move to the left because now you need to put x = -2 to get f(0), instead of putting x = 0. So the whole graph moves to left.

As for when you are given f(x), -f(x) will be the mirror image in x axis
and f(-x) will be the reflection about the y axis.

KarishmaB · Answer

This should help: Screenshot 2020-05-27 at 16.31.17.png

aveek86 · Answer

It's an equation of Parabola.

Instead of putting Y=0 (as the question intends you to do/think) put X=0 and you will find Y=1 for the first equation and Y=3 for the second. Similarly, put X=1 and -1; you will find Y=0 for the first equation and Y=2 for the second. And since these two curves are parallel you can now easily plot those points and identify the nature of the new curve.

Thus, it cuts X-axis only at one point.

Voila!!

siddhantvarma · Answer

Can I expect such a question in the Focus edition now that geometry is not there?

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

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