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

1 Answer
Mar 16, 2017

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]}