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



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

Answer: B.

Attachment:
graph (1).png
[/quote]

Hey @buenel, is this vertical shift only in cases of quadratic/cubic functions? when i do this for y=x to y=x+2, I get a different shift
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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy [#permalink]
BrentGMATPrepNow wrote:
imhimanshu wrote:

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


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


Great explanation BrentGMATPrepNow. Is it only positive x in Quadrant I matter here? How about the -x intercept in Quandrant II here?
Could you help clarify? 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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Re: The figure shows the graph of y = (x + 1)(x - 1)^2 in the xy [#permalink]
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