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

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Joined: 27 Jun 2010
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On a two dimensional coordinate plane, the curve y=x^2 - x^3 [#permalink]  20 Sep 2010, 02:32
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Question Stats:

66% (01:42) correct 34% (00:52) wrong based on 128 sessions
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 ?
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 31356
Followers: 5367

Kudos [?]: 62591 [1] , given: 9457

Re: Curve question [#permalink]  20 Sep 2010, 02:39
1
KUDOS
Expert's post
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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Re: On a two dimensional coordinate plane, the curve y=x^2 - x^3 [#permalink]  20 Jul 2014, 02:52
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Re: On a two dimensional coordinate plane, the curve y=x^2 - x^3   [#permalink] 20 Jul 2014, 02:52
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