Answer to Question #215180 in Calculus for Manoranjan Kumar

Question

Trace the curve x = (y – 1) (y – 2) (y – 5)

Answer 1

1. Equation of the curve : 

$$x=(y-1)(y-2)(y-5)$$

$$x=(y-1)(y^2-7y+10)$$

$$x=y^3-8y^2+17y-10$$

Domain: $(-\infin, \infin).$

Range: $(-\infin, \infin).$

2. Symmetry

$-y:$

$$x=(-y-1)(-y-2)(-y-5)$$

$$x=-(y+1)(y+2)(y+5)$$

The curve is not symmetrical about $x$ -axis.

$-x:$

$$-x=(y-1)(y-2)(y-5)$$

$$x=-(y-1)(y-2)(y-5)$$

The curve is not symmetrical about $y$ -axis.

$-x, -y$ :

$$-x=(-y-1)(-y-2)(-y-5)$$

$$x=(y+1)(y+2)(y+5)$$

The curve is not symmetrical about the origin.

Interchange $x$ and $y:$

$$y=(x-1)(x-2)(x-5)$$

The curve is not symmetrical about the line $y=x.$

The curve is not symmetric with respect to any line or axis.

3. Intersection(s)

$Oy: x=0$

$$0=(y-1)(y-2)(y-5)$$

$Point(0,1), Point(0, 2), Point(0, 5)$

$Ox: y=0$

$$x=(0-1)(0-2)(0-5)$$

$Point(-10,0).$

4. Tangent at Origin :

The curve does not go through the origin. So there is no tangent at the origin.

5. There does not exist any horizontal or vertical asymptote.

$y\to-\infin$ as $x\to-\infin$

$y\to\infin$ as $x\to\infin$

There is no slant (oblique) asymptote.

6. Local maximum or minimum.

$$x=y^3-8y^2+17y-10$$

Differentiate both sides with respect to $y$

$$x_y'=3y^2 -16y+17$$

Critical number(s)

$$x'=0=>3y^2 -16y+17=0$$

$$y=\dfrac{8\pm\sqrt{13}}{3}$$

$Point(-6.065, 3.869)$ is a local minimum.

$Point(0.879, 1.465)$ is a local maximum.

7. Sketch the graph