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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

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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

by BTGmoderatorDC » Mon Jun 08, 2020 7:02 pm

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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


OA B

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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy-plane. At how many points does the graph of

by Jay@ManhattanReview » Mon Jun 08, 2020 9:21 pm
BTGmoderatorDC wrote: ↑
Mon Jun 08, 2020 7:02 pm
 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

OA B

Source: GMAT Prep
Note that if a function is y = f(x), then a function y = f(x) + 2 will move 2 units up on the y-axis.

We see the graph of y = (x + 1)(x - 1)^2, intercepting x-axis at two points [(–1, 0) & (1, 0)], then y = (x + 1)(x - 1)^2 + 2 would move up 2 units. This way the interfering point (1, 0) would move up, thereby not intercepting positive x-axis. However, the graph will still intercept negative x-axis, slightly moving to left of point (1, 0). See the attached image.

So, the y = (x + 1)(x - 1)^2 would intercept at only one point.

The correct answer: B

Hope this helps!

-Jay

Graph: Courtesy Google
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy-plane. At how many points does the graph of

by Jay@ManhattanReview » Mon Jun 08, 2020 9:35 pm
BTGmoderatorDC wrote: ↑
Mon Jun 08, 2020 7:02 pm
 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

OA B

Source: GMAT Prep
I already submitted the solution, following graphical analysis.

Let's do this problem algebraically.

Since the graph of y = (x + 1)(x - 1)^2 + 2 intercepts the x-axis, the y-coordinate would be 0.

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

(x + 1)(x – 1)^2 = –2

We see that the product of (x + 1) & (x - 1)^2 is a negative number. Since (x - 1)^2 is a non-negative number, (x + 1) must be negative. Or, x + 1 < 0 => x < –1. So the y = (x + 1)(x - 1)^2 + 2 intercepts the x-axis at only ONE point.

Needless to state that the funtion y = (x + 1)(x - 1)^2 + 2 cannot intercept positive x-axis, else y would not be 0.

The correct answer: B

Hope this helps!

-Jay
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy-plane. At how many points does the graph of

by Brent@GMATPrepNow » Tue Jun 09, 2020 5:05 am
BTGmoderatorDC wrote: ↑
Mon Jun 08, 2020 7:02 pm
 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


OA B

Source: GMAT Prep
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:
Image

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:
Image

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...
Image

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

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
Brent Hanneson - Creator of GMATPrepNow.com
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