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

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.

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

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

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 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.

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.

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.

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

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

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:

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

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:


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:


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...


When we connect the points, we see that the graph of y = (x + 1)(x - 1)² + 2 looks something like this.


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

Bunuel Would you please post similar type of questions? I had a hard time figuring out the approach for this one.

The x-axis is shifted downwards to -2 in the new graph. The new graph will intersect x-axis at 1 point.

IMO B

Posted from my mobile device

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

The graph moves up vertically on adding 2, but what will be the number of x-intercepts of the original graph 'y = (x + 1)(x - 1)^2 + 2' ?
Will we consider the point (1,0) be considered as an x intercept as the graph just touches that point changes its direction

If we call the given function as y=f(x) then we are asked about the x intercepts of the graph y=f(x)+2
This just means the existing graph gets shifted upwards by 2 units as y has now become y+2
Since the existing graph has a double root at x=1, which is on the x axis and the new graph is getting shifted upwards it won't touch the x axis at x=1 but will still cross at x=-1. Therefore just once
Hence B

Posted from my mobile device