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Home > Calculus > Curve Sketching > Curve Sketching with Asymptotes – Example 4

Curve Sketching with Asymptotes – Example 4

Let’s look at another example of curve sketching with asymptotes.

y = {x^2}/{x^2~+~4}

Step 1: We get the x-intercept when y = 0. So x2 = 0, or x = 0

So x intercept is (0, 0)

We get y-intercept when x = 0. Hence y = 0, and the y-intercept is also (0, 0).

Let’s do a quick check of the type of function: f(x) = f(-x).

So this is an even function, and will be symmetric about the y-axis.

 

Step 2: Since x2 + 4 > 0, there is no vertical asymptote here in the real domain.

 

Step 3: Let’s now find the horizontal asymptotes by looking at the function as x approaches infty and - infty.

lim{x right infty}~{{x^2}/{x^2~+~4}}  =  lim{1/x right 0}~~{{x^2/x^2}/{{x^2/x^2}~+~{4/x^2}}}

lim{1/x right 0}~{1/{1~+~{4/x^2}}}

=  1

lim{x right - infty}~{{x^2}/{x^2~+~4}}  =  lim{-1/x right 0}~~{{x^2/x^2}/{{x^2/x^2}~+~{4/x^2}}}

lim{-1/x right 0}~{1/{1~+~{4/x^2}}}

=  1

 

Step 4: Now we need to find the stationary points

dy/dx  =  {(x^2~+~4).(2x)~-~{x^2}.2x}/{(x^2~+~4)^2}

=  {2x^3~+~8x~-~2x^3}/{(x^2~+~4)^2}

=  {8x}/{(x^2~+~4)^2}

y’ = 0 for stationary point. Hence 8x = 0, or x = 0.

(0, 0) is a stationary point.

 

Step 5: Next we need to find the point/s of inflection, if any

y” = ~{(x^2~+~4)^2~(8)~-~8x{(2(x^2~+~4)}.2x)}/({(x^2~+~4)}^4}

= ~{(x^2~+~4){(8(x^2~+~4)}~-~32x^2)}/({(x^2~+~4)}^4}

= ~{(8(x^2~+~4)}~-~32x^2}/({(x^2~+~4)}^3}

= ~{8x^2~+~32~-~32x^2}/({(x^2~+~4)}^3}

= ~{-24x^2~+~32}/({(x^2~+~4)}^3}

When y” = 0, 24x2 = 32

So x2  =  4/3

Hence x  =  pm 2/{sqrt 3}

Now we need to check for the change of concavity at x = 2/{sqrt 3} and x = -2/{sqrt 3}

 

x 1 2/{sqrt 3} 2
f”(x) {-24~+~32}/25 0 {-24~*~4~+~32}/64
f”(x) = > 0 0 -1

The concavity of the curve changes, so x = 2/{sqrt 3} is a point of inflection.

What about x = -2/{sqrt 3}

x -2 -2/{sqrt 3} -1
f”(x) 8/25 0 -1
f”(x) = > 0 0 < 0

The concavity of the curve changes, so x = -2/{sqrt 3} is also a point of inflection.

At x = 2/{sqrt 3},  y  = {(2/{sqrt 3})^2}/{{(2/{sqrt 3})^2}~+~4}

~{4/3}/{{4/3}~+~4}

1/4

At x = -2/{sqrt 3},  y  = {(-2/{sqrt 3})^2}/{{(-2/{sqrt 3})^2}~+~4}

~{4/3}/{{4/3}~+~4}

1/4

Therefore (2/{sqrt 3}~,~{1/4}) and (-2/{sqrt 3}~,~{1/4}) are points of inflections.

 

Step 7: Finally we need to check the nature of the stationary point at (0, 0) and draw the graph.

y” at x = 0 is  y”  =  32/{4^2}  =  2  > 0

Hence (0, 0) is the minimum turning point.

Now using the points (0, 0) as the minimum turning points, and the points of inflection being (2/{sqrt 3}~,~{1/4}) and (-2/{sqrt 3}~,~{1/4}), we find that as x right~pm infty, y right  1+

curve sketching with asymptotes

See also other examples of curve sketching with asymptotes –

Curve sketching with asymptotes – Example 1

Curve sketching with asymptotes – Example 2

Curve sketching with asymptotes – Example 3