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#

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]}

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Answer 2

To sketch the curve ( y = \frac{x^2}{x^2+9} ), follow these steps:

  1. Intercepts:

    • y-intercept: Set ( x = 0 ) and solve for ( y ).
    • x-intercept: Set ( y = 0 ) and solve for ( x ).
  2. Asymptotes:

    • Vertical asymptotes occur where the denominator equals zero, so set ( x^2 + 9 = 0 ) and solve for ( x ).
    • Horizontal asymptotes occur as ( x ) approaches positive or negative infinity. Use limits to find them.
  3. Local Maximum and Minimum:

    • Find critical points by taking the derivative of ( y ), setting it equal to zero, and solving for ( x ).
    • Use the second derivative test or first derivative test to determine if these points are local maxima or minima.
  4. Inflection Points:

    • Find the second derivative of ( y ) and set it equal to zero to find inflection points.

Once you have found all these points, sketch the curve by plotting the intercepts, asymptotes, local maximum and minimum points, and inflection points, and then connecting them smoothly.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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