Solved IGNOU Assignments and Question papers: CURVE TRACING EXAMPLE 3

Given equation of the curve is

${y^2} = \,\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$ ___________________(1)

Symmetry

1.Replace y by -y in the equation of the given curve.

${\left( { - y} \right)^2} = \,\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$

$\Rightarrow {y^2} = \,\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$

which is same as the equation of the given curve.
So, the curve is symmetric about the x-axis.

2.Replace x by -x in the equation of the given curve.

${y^2} = \,\left( { - x - 1} \right)\left( { - x - 2} \right)\left( { - x - 3} \right)$

$\Rightarrow {y^2} = - \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)$

which is not same as the equation of the given curve.
So, the curve is symmetric about the y-axis neither in opposite quadrants.

3.Interchange y by x in the equation of the given curve.

${x^2} = \,\left( {y - 1} \right)\left( {y - 2} \right)\left( {y - 3} \right)$

which is not same as the equation of the given curve.
So, the curve is not symmetric about the line y=x.

4.Interchange y by -x and x by -y in the equation of the given curve.

${\left( { - x} \right)^2} = \,\left( { - y - 1} \right)\left( { - y - 2} \right)\left( { - y - 3} \right)$

$\Rightarrow {x^2} = \, - \left( {y + 1} \right)\left( {y + 2} \right)\left( {y + 3} \right)$

which is not same as the equation of the given curve.
So, the curve is not symmetric about the line y=-x.

Check for origin

Putting (x,y)=(0,0) in the equation of the given curve we get,

$0 = \left( {0 - 1} \right)\left( {0 - 2} \right)\left( {0 - 3} \right)$

0=6.

So the curve does not pass through the origin.

Point of intersection with x-axis and y-axis

Put x=0 in the equation of the given curve.

${y^2} = - 6$

Thus, the curve has no point of intersection with the y-axis.

Put y=0 in the equation of the given curve,

${\left( 0 \right)^2} = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$ .

$x = 1,2,3$ .

The curve cuts the x-axis at (1,0) (2,0) (3,0).

Asymptotes

${y^2} = {x^3} - 6{x^2} + 11x - 6$

Equating the coefficient of highest degree term of y to zero we get,

1=0.

which is not equal.

Hence there are no asymptotes parallel to y-axis.

Equating the coefficient of highest degree term of x to zero we get,

1=0.

which is not equal.

Hence there are no asymptotes parallel to x-axis.

For oblique asymptotes,

Put y=mx+c, in

${y^2} = {x^3} - 6{x^2} + 11x - 6$ _________________________________(2)

${\left( {mx + c} \right)^2} = {x^3} - 6{x^2} + 11x - 6$

${m^2}{x^2} + 2mxc + {c^2} + 6{x^2} + 6 - 3{x^2} - 11x = 0$

There are no real values of m so there are no oblique asymptotes.

Domain

$y = \sqrt {\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}$

y is defined if $\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) \ge 0$

So the domain of y is

$x \in \left[ {1,2} \right]\, \cup \,\left[ {3,\infty } \right)$

Intervals of increase and decrease

Given equation of the curve is

$2y\frac{{dy}}{{dx}} = 3{x^2} - 12x + 11$

The critical points are x=1.423 and x=2.584.

So the curve increases in the interval

$\left( {1,1.423} \right)\, \cup \,\left( {3,\infty } \right)$

and decreases in the interval

$\left( {1.423,2} \right)$ .

Solved IGNOU Assignments and Question papers

Friday, January 24, 2020

CURVE TRACING EXAMPLE 3

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