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Sketching Curves with Asymptotes – Example 2

Now let us look at another example of sketching curves with asymptotes

Example 2: Sketch the curve for y = f(x) = {x^2}/{x^2~-~9}

Step 1: Find the x-intercepts when y = 0

y = 0 when x = 0

(0, 0) is the x-intercept

 

Step 2: y-intercept when x = 0 gives 0

So (0, 0) is the y-intercept.

 

Step 3: Check if the curve is symmetric, i.e. is the function odd or even.

f(x) = {x^2}/{x^2~-~9}

f(-x) =  {(-x)^2}/{(-x)^2~-~9}  = {x^2}/{x^2~-~9}

f(x) = f(-x)

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

 

Step 4: Check for any discontinuities, and find the asymptotes, if any, or the limits

The curve has critical points i.e. it has discontinuities

x^2~-~9~<>~0   doubleright~~~~x~<>~pm 3

At x = 3 and x = -3, the curve is discontinuous. So the vertical asymptotes are at x = 3 and x = -3.

Let us look at the curve around x = 3 and x = -3.

As x right 3 from RHS 3+

lim{x right 3^+}~{x^2}/{x^2~-~9}  = {(3^+)^2}/{(3^+)^2~-~9}  = +/+  =  +

So y~right~infty

As x~right~3 from RHS, y~right~infty

Similarly as x right 3 from LHS (3)

lim{x right 3^−}~{x^2}/{x^2~-~9}  =  {(3^−)^2}/{(3^−)^2~-~9}  =  {+/−}  =   −

As x~right~3^-y~right~- infty

 

As x right -3 from LHS (−3)

y = lim{x right -3^−}~{x^2}/{x^2~-~9}  = {(−3^−)^2}/{(−3^−)^2~-~9}  = +/+  =  +

As x~right~−3 from LHS, y~right~infty

Similarly as x right −3+ from RHS

y = lim{x right -3^+}~{x^2}/{x^2~-~9}  =  {(-3^+)^2}/{(-3^+)^2~-~9}  =  {+/−}  =   −

As x~right~-3 from RHS, y~right~− infty

What happens to the curve as x right~pm infty?

y = f(x) = {x^2}/{x^2~-~9}

y = lim{x right infty}~{{x^2}/{x^2}/{1~-~{9/x^2}}}

As x~right~infty 1/x~right~0

y = lim{x right infty}~{{1}/{1~−~{9/x^2}}}

right~1

f(infty)(infty)^2/{(infty)^2~−~9}  > 1

since numerator > denominator

As x rightinfty  y right~1 from above.

f(− infty)(− infty)^2/{(- infty)^2~-~9}  > 1

since again numerator > denominator

 

Step 5: Find stationary points (put y’ = f'(x) = 0)

dy/dx  =  {(x^2~-~9)(2x)~-~{x^2}.2x}/{(x^2~-~9)^2}

=  {2x^3~-~18x~-~2x^3}/{(x^2~-~9)^2}

=  {-18x}/{(x^2~-~9)^2}

dy/dx  = 0, so {-18x}/{(x^2~-~9)^2}  = 0

or x = 0, y = 0

(0, 0) is a stationary point.

 

Step 6: Find the point of inflection and the nature of the stationary point.

{d^2y}/{dx^2}  =  {d/dx}({-18x}/{(x^2~-~9)^2})

= ~{(-18){(x^2~-~9)^2}~-~(-18x)(2)(x^2~-~9).2x}/({(x^2~-~9)}^4}

= ~{(-18){(x^2~-~9)^2}~+~(72x^2)(x^2~-~9)}/({(x^2~-~9)}^4}

= ~{(-18){(x^2~-~9)}~+~72x^2}/({(x^2~-~9)}^3}

= ~{-18x^2~+~162~+~72x^2}/({(x^2~-~9)}^3}

= ~{54x^2~+~162}/({(x^2~-~9)}^3}

y” at x = 0 is ~{54(0)^2~+~162}/({((0)^2~-~9)}^3}

~{-~162}/({9^3}}

= – 0.222  < 0  (therefore maximum point)

Hence around x = 0, the curve is concave down between x = -3 and x = 3.

When y” = 0 ~{54x^2~+~162}/({(x^2~-~9)}^3}  =  0

54x2 + 162 = 0

x2  = -162/54  =  -3

x2 is negative, so there is no real solution, hence a point of inflection does not exist.

 

Step 7: Now using the points from the above steps.

  • Stationary point (maximum point)  – (0, 0)
  • y-intercept – (0, 0)
  • y = f(x) is symmetric about y-axis

Vertical asymptotes at x = 3 and x = -3

x~right~3^+y~right~infty and x~right~3^-y~right~- infty

x~right~-3^+y~right~infty and x~right~-3^-y~right~infty

Horizontal asymptotes

x~right~pm infty,  y~right~1 from above

curve sketching with asymptotes

See also other examples of sketching curves with asymptotes –

Sketching curves with asymptotes – Example 1

Sketching curves with asymptotes – Example 3

Sketching curves with asymptotes – Example 4