`y = x^2/(x + 8)` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

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Chapter 4, Review - Problem 25 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`y=x^2/(x+8)`

a) Asymptotes

Vertical asymptotes are the zeros of the denominator

`x+8=0rArrx=-8`

Vertical asymptote is x=-8

Degree of numerator=2

Degree of denominator=1

Degree of numerator=1+Degree of denominator, so the asymptote is a slant asymptote of the form y=mx+b

For a rational function the slant asymptote is the quotient of the polynomial division.

`x^2/(x+8)=x-8+64/(x+8)`

 the slant asymptote is y=x-8

b) Maxima/Minima

`y'=((x+8)2x-x^2)/(x+8)^2`

`y'=(2x^2+16x-x^2)/(x+8)^2`

`y'=(x(x+16))/(x+8)^2`

Let's find critical numbers by solving x for y'=0,

`x(x+16)=0rArrx=0 , x=-16`

`y(-16)=(-16)^2/(-16+8)=-32`

`y(0)=0^2/(0+8)=0`

Local maximum=-32 at x=-16

Local minimum=0 at x=0

c) Inflection points

`y''=((x+8)^2(2x+16)-(x^2+16x)(2)(x+8))/(x+8)^4`

`y''=(2x^2+16x+16x+128-2x^2-32x)/(x+8)^3`

`y''=128/(x+8)^3`

`128/(x+8)^3=0`

there is no solution for x , so there are no inflection points.

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