Question #215180

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


1
Expert's answer
2021-07-09T08:47:03-0400

1. Equation of the curve : 

x=(y1)(y2)(y5)x=(y-1)(y-2)(y-5)




x=(y1)(y27y+10)x=(y-1)(y^2-7y+10)

x=y38y2+17y10x=y^3-8y^2+17y-10

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

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


2. Symmetry

y:-y:


x=(y1)(y2)(y5)x=(-y-1)(-y-2)(-y-5)

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

The curve is not symmetrical about xx -axis.


x:-x:


x=(y1)(y2)(y5)-x=(y-1)(y-2)(y-5)

x=(y1)(y2)(y5)x=-(y-1)(y-2)(y-5)

The curve is not symmetrical about yy -axis.


x,y-x, -y :



x=(y1)(y2)(y5)-x=(-y-1)(-y-2)(-y-5)

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

The curve is not symmetrical about the origin.


Interchange xx and y:y:


y=(x1)(x2)(x5)y=(x-1)(x-2)(x-5)

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


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


3. Intersection(s)

Oy:x=0Oy: x=0


0=(y1)(y2)(y5)0=(y-1)(y-2)(y-5)

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



Ox:y=0Ox: y=0

x=(01)(02)(05)x=(0-1)(0-2)(0-5)

Point(10,0).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.

yy\to-\infin as xx\to-\infin


yy\to\infin as xx\to\infin


There is no slant (oblique) asymptote.


6. Local maximum or minimum.


x=y38y2+17y10x=y^3-8y^2+17y-10

Differentiate both sides with respect to yy

xy=3y216y+17x_y'=3y^2 -16y+17

Critical number(s)

x=0=>3y216y+17=0x'=0=>3y^2 -16y+17=0

y=8±133y=\dfrac{8\pm\sqrt{13}}{3}

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


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


7. Sketch the graph







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