Question

How do you sketch the curve `y=x^2/(x^2+9)` by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Answer 1

The minimum is `=(0,0)`
The points of inflexions are `=(-sqrt3,1/4)` and `=(sqrt3,1/4)`
The intercept is `=(0,0)`
The horizontal asymptote is `y=1`

Explanation:

Let `f(x)=x^2/(x^2+9)`

`f(-x)=x^2/(x^2+9)`

`f(x)=f(-x)`

The curve is symmetric about the y-axis

The derivative of a quotient is

`(u/v)'=(u'v-uv')/(v^2)`

We start by calculating the first derivative

`y=x^2/(x^2+9)`

`u=x^2`, `=>`, `u'=2x`

`v=x^2+9`, `=>`, `v'=2x`

`dy/dx=(2x(x^2+9)-2x(x^2))/(x^2+9)^2`

`=(2x^3+18x-2x^3)/(x^2+9)^2`

`=(18x)/(x^2+9)`

The critical values are when `dy/dx=0`

`(18x)/(x^2+9)^2=0`

When `x=0`

We can build a chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaa)` `0` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `dy/dx` `color(white)(aaaaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaa)` `↘` `color(white)(aaa)` `0` `color(white)(aaaa)` `↗`

Now, we calculate the second derivative

`u=18x`, `=>`, `u'=18`

`v=(x^2+9)^2`, `=>`, `v'=2(x^2+9)*2x`

`(d^2y)/dx^2=(18(x^2+9)^2-18x*(4x(x^2+9)))/(x^2+9)^4`

`=18(((x^2+9)^2-4x^2(x^2+9)))/(x^2+9)^4`

`=(18(x^2+9)((x^2+9)-4x^2))/(x^2+9)^4`

`=(18(9-3x^2))/(x^2+9)^3`

`=(54(3-x^2))/(x^2+9)^3`

`(d^2y)/dx^2=0` when `x =-sqrt3` and `x=sqrt3`

The points of inflexions are `(-sqrt3, 1/4)` and `(sqrt3,1/4)`

We can build the chart

`color(white)(aa)` `Interval` `color(white)(aaaa)` `]-oo,-sqrt3[` `color(white)(aaaa)` `]-sqrt3,sqrt3[` `color(white)(aaaa)` `]sqrt3,+oo[`

`color(white)(aa)` `(d^2y)/dx^2` `color(white)(aaaaaaaaaaaaa)` `-` `color(white)(aaaaaaaaaaaa)` `+` `color(white)(aaaaaaaaa)` `-`

`color(white)(aa)` `y` `color(white)(aaaaaaaaaaaaaaaa)` `nn` `color(white)(aaaaaaaaaaaa)` `uu` `color(white)(aaaaaaaaa)` `nn`

`lim_(x->+-oo)y=lim_(x->+-oo)x^2/x^2=1`

The horizontal asymptote is `y=1`

graph{(y-(x^2)/(x^2+9))(y-1)=0 [-7.02, 7.024, -3.51, 3.51]}