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# On a two dimensional coordinate plane, the curve y=x^2 - x^3

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Intern
Joined: 27 Jun 2010
Posts: 33
On a two dimensional coordinate plane, the curve y=x^2 - x^3  [#permalink]

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20 Sep 2010, 02:32
11
00:00

Difficulty:

25% (medium)

Question Stats:

68% (01:05) correct 32% (01:16) wrong based on 315 sessions

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On a two dimensional coordinate plane, the curve y=x^2 - x^3 intersects the x-axis at how many points?

A. 0
B. 1
C. 2
D. 3
E. 4

I got the answer as B,

The way i solved the question was to find at how many points a curve touches x axis is by putting y = 0 , when we put y=0 then we get the equation x^2 - x^3 =0

So x^2 = x^3, so cancelling x on both the sides we get x = 1

so the curve touches x axis at x=1 so the answer is B,

But the official answer is C, wat is wrong my method of solving ?
Math Expert
Joined: 02 Sep 2009
Posts: 53063

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20 Sep 2010, 02:39
2
1
sachinrelan wrote:
On a two dimensional coordinate plane, the curve y=x^2 - x^3 intersects the x-axis at how many points?

A. 0
B. 1
C. 2
D. 3
E. 4

I got the answer as B,

The way i solved the question was to find at how many points a curve touches x axis is by putting y = 0 , when we put y=0 then we get the equation x^2 - x^3 =0

So x^2 = x^3, so cancelling x on both the sides we get x = 1

so the curve touches x axis at x=1 so the answer is B,

But the official answer is C, wat is wrong my method of solving ?

X-intercepts of a graph of a function, in our case $$y=x^2-x^3$$ is the values of $$x$$ for $$y=0$$.

$$y=0$$ --> $$x^2-x^3=0$$ --> $$x^2(1-x)=0$$ --> two solutions: $$x=0$$ and $$x=1$$.

The way you are doing is wrong as when reducing (dividing) the equation $$x^2=x^3$$ by $$x^2$$ you are assuming that $$x\neq{0}$$, so excluding this solution.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Hope it's clear.
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Joined: 23 Jan 2016
Posts: 182
Location: India
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Re: On a two dimensional coordinate plane, the curve y=x^2 - x^3  [#permalink]

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29 Oct 2016, 01:12
Bunuel, in mgmat tells us that the number of times a parabola touches the x axis can be figured out calucating the value of b^2-4ac for any equation in the form of ax^2+bx+c. can that concept be applied for the equaiton of this question - x^2-x^3? Let me know your thoughts, thank you.
Math Expert
Joined: 02 Sep 2009
Posts: 53063
Re: On a two dimensional coordinate plane, the curve y=x^2 - x^3  [#permalink]

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30 Oct 2016, 00:28
1
TheLordCommander wrote:
Bunuel, in mgmat tells us that the number of times a parabola touches the x axis can be figured out calucating the value of b^2-4ac for any equation in the form of ax^2+bx+c. can that concept be applied for the equaiton of this question - x^2-x^3? Let me know your thoughts, thank you.

No, this is a discriminant formula which can only be applied to the quadratic equations.
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Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 373
Re: On a two dimensional coordinate plane, the curve y=x^2 - x^3  [#permalink]

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31 Oct 2016, 13:27
The answer is 2 because of the following:

Manipulate the equation --> y = (x^2) - (x^3) --> y = (x^2)(1-x)

We then set y = 0 to find where the x-intercepts are --> 0 = (x^2)(1-x) --> This means that x intercepts are 0 and 1.

Thus C is the correct answer.
Non-Human User
Joined: 09 Sep 2013
Posts: 9880
Re: On a two dimensional coordinate plane, the curve y=x^2 - x^3  [#permalink]

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20 Oct 2018, 01:56
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: On a two dimensional coordinate plane, the curve y=x^2 - x^3   [#permalink] 20 Oct 2018, 01:56
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