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$y^{2}$=$x^{2}(1-x)$

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Toolbox:
  • The strategies for curve tracing
  • (1) Domain-The values of x for which the function is defined
  • (2)Extent -Horizontal (vertical ) extent is determined by the intervals for $x(y)$ for which the curve exists
  • (3) $x=0$ gives the y intercept and y=0 gives the x -intercept
  • (4) If $(0,0)$ satisfies the equation, the curve passes through the orgin
  • Symmetry:
  • (5) Find out whether the curve is symmetrical about any line. The curve is symmetric about
  • (i) x-axis if the equation is unaltered when y is replaced by -y
  • (ii) y-axis if the equation is unaltered when x is replaced by -x
  • (iii)the orgin if the equation is unattened when x is replaced by -x abd y by -y
  • (iv) the line $y=x$ if the equation is uttered when x and y are interchanged
  • (v) the line $ y=-x$ if the eauation is unchanged if x and y are replaced by -y and -x
  • (6) Asymptotes ($\parallel$ to coordinate axes)
  • If $y \to c$ (finite) when $x \to \pm \infty$
  • (or $ x \to k $ (finite ), when $y \to \pm \infty$
  • the line $y=c$(or $x=k$ ) is an asymptole parallel to x-axis (or y-axis)
  • (7)Monotsnicity : Determine the intervals for which the curve is increasing or decreasing, using the first derivative test
  • (8) Special points: Determine the intervals of concavity/convexity and the points of inflection, using the second derivative test.
$y^2=x^2(1-x)$
Step 1:
Existence : The curve does not exist when $x >1,$ ie for $1 < x < \infty $
So exists in $ -\infty < x \leq 1$
Step 2:
(ii) Symmetry: The curve is symmetric about x-axis only
Step 3:
(iii) Asymptotes: The curve does not admit asymptotes since there are no further values of $ x(or\;y)$ for which $y \to \pm \infty(or\; x \to \pm \infty)$
Step 4:
Loops:The curve passes through (0,0) twice and (1,0) once
So a loop is formed between $(0,0)$ and $(1,0)$
answered Aug 19, 2013 by meena.p
 

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