Sketch the graph of the following; `y = x^3/(x+1)^2` Sketch the graph of the following; y= (x^3)/(x+1)^(2)) a) Find all x-intercepts b) Find all y-intercepts c) All max/min (x,y coordinates.) d)...

Sketch the graph of the following;

`y = x^3/(x+1)^2`

Sketch the graph of the following;

y= (x^3)/(x+1)^(2))

a) Find all x-intercepts

b) Find all y-intercepts

c) All max/min (x,y coordinates.)

d) Any Vertical Asymptote

e) Any Horizontal Asymptote

f) Any Slant Asmptote

g) All points of inflection (x,y)

h) Arrows to show graph continuous & overall shape

 

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mathsworkmusic | (Level 2) Educator

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We have the function `y = x^3/(x+1)^2`

a) `x` intercepts are the values of `x` when `y =0`.

    The only value of `x` for which this is true is `x = 0`

b) `y` intercepts are the values of `y` when `x =0`. When `x=0``y=0`.

    Since this is a function (one-one), that is the only `y` intercept

c) To find the extrema, differentiate the function

   `(dy)/(dx) = (3x^2)/(x+1)^2 + x^3((-2))/(x+1)^3 = (3x^2(x+1) - 2x^3)/(x+1)^3`

`= (3x^3 + 3x^2 - 2x^3)/(x+1)^3 = (x^2(x + 3))/(x+1)^3`

The extrema are at `x` such that `(dy)/(dx) = 0`

So the extrema are at `x=0` and `x = -3`

At `x=-3`, `y = (-3)^3/(-3 +1)^2 = -27/4`

The coordinates of the extrema are `(0,0)` and `(-3,-27/4)`

To find whether they are maxima, minima or points of inflection check the sign of `(d^2y)/(dx^2)`

Now, `(d^2y)/(dx^2) = (d/(dx))(dy)/(dx) = (3x^2+6x)/(x+1)^3 +x^2(x+3)((-3))/(x+1)^4`

When `x=0` this equals 0 `implies` point of inflection at `(0,0)`

When `x=-3` this equals `(27-18)/(-8) = -9/8` `implies` maximum at `(-3,-27/4)`

d) A vertical asymptote is a value of `x` such that `y -> +-oo`.

   As `x -> -1``y -> -1/0`  so `y -> -oo`

   So there is a vertical asymptote at `x =-1`

e) A horizontal asymptote is a value of `y` such that `x -> oo`

` `` ` ` `` ` ` ` ` ` ` `` `` `` `     There are no horizontal asymptotes

f) A slant asymptote is when `(x,y)` tends to a line other than` ` a horizontal or vertical one.` `

  Rewrite `y =x^3/(x+1)^2`  as `y =x/(1/x^2(x+1)^2) = x/(1+1/x)`

  Now, as `x -> oo``1/x -> 0` and `y -> x/1 = x` (similarly for `x -> -oo`)

  Therefore the line `y=x` is a slant asymptote

g) Points of inflection are when `(d^2y)/(dx^2) = 0`

   Here this is when `(3x^2 + 6x)/(x+1)^3 +x^2(x+3)(-3)/(x+1)^4 = 0`

   ie when `((3x^2 + 6x)(x+1) - 3x^2(x+3))/(x+1)^4 = 0`

   `implies 3x^3 + 9x^2 + 6x - 3x^3 - 9x^2 = 0` `implies 6x =0`

  There is only one point of inflection, at `(0,0)`

h) The graph is continuous to the left from `x = -1` and continuous to  the right from `x = -1`. There is a singularity at `x = -1`.

    The graph is very similar to `y = x` except near the origin where there is a discontinuity, a maxmimum and a point of inflection.

 

sketch of graph y = x^3/(x+1)^2 looks like this

 

 

 

 

  

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