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Home > Calculus > Curve Sketching > Sketching Curves with Asymptotes – Example 3

Sketching Curves with Asymptotes – Example 3

Now let us look at another method of curve sketching with asymptotes. This method involves dividing the number plane into regions by drawing dotted lines through the critical points. A critical point is a x-intercept point or a point where there is a vertical asymptote.

The following steps are involved in this method of sketching curves with asymptotes:

Step 1: Draw dotted lines through critical points. The critical point/s will divide the number planes to regions. Name each region as region1, region2, and so on.

Step 2: Draw lines through the critical points as y  =  x + critical points

Step 3: Now look at the sign of the curve in each region. A graph is positive if it is above the x-axis.

Step 4: Find any horizontal asymptotes i.e. when x right~pm infty, find the limiting value of y.

Use the above information to sketch the graph. If you need more accurate sketching, you can also follow these additional steps:

Step 5: Find the stationary points and its nature

Step 6: Find the point of inflection – if it exists – and check for its concavity

Step 7: Sketch the curve

 

Example of sketching curves with asymptotes:

Sketch the curve for y = {x}/{x^2~-~1}

Step 1: The critical points are:

x-intercept : x = 0, y = 0 (0, 0)

Vertical asymptotes at x = pm 1  because x² – 1 ≠ 0

The number plane is divided into 4 regions as shown in the diagram below:

regions with critical points

 

Step 2: Now draw lines through the critical points as shown below.

The lines are y = x, y = x + 1 and y = x – 1

lines through critical points

 

Step 3: Find the sign of y.

Region I:

y = x
y = x + 1
y = x – 1
y = {x}/{x^2~-~1}  = {x}/{(x~-~1)(x~+~1)}= {-}/{-~*~-}  =  {-}/{+}   =  –

In region I, the curve is below the x-axis

Region II:

y = x
y = x + 1 +
y = x – 1
y = {-}/{+~*~-}  =  {-}/{-}   =  +

In region II, the curve is above the x-axis

Region III:

y = x +
y = x + 1 +
y = x – 1
y = {+}/{+~*~-}  =  {+}/{-}   =  –

In region III, the curve is below the x-axis

Region IV:

y = x +
y = x + 1 +
y = x – 1 +
y = {+}/{+~*~+}  =  {+}/{+}   =  +

In region IV, the curve is above the x-axis

 

Step 4: Let’s find the horizontal asymptotes

We can find them by finding y when x right~pm infty

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

When x right~infty, 1/x~right 0. So

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

=  0+

When x right~infty, y~right 0+ from above.

When x right~- infty, 1/x~right 0. So

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

=  0- from below

 

Step 5: We next need to find the stationary points

dy/dx  =  {(x^2~-~1)(1)~-~{x}.2x}/{(x^2~-~1)^2}

=  {x^2~-~1~-~2x^2}/{(x^2~-~1)^2}

=  {-x^2~-~1}/{(x^2~-~1)^2}

y’ = 0 for stationary point.

When y’ = 0, -x2 – 1 = 0

Or x2 = -1

So x has no real value. Hence there is no stationary points.

 

Step 6: We now check for any points of inflection.

y” = {d^2y}/{dx^2}  =  {d/dx}({-x^2~-~1}/{(x^2~-~1)^2})

= ~{(-2x){(x^2~-~1)^2}~+~2.(x^2~+~1)(x^2~-~1).2x}/({(x^2~-~1)}^4}

= ~{(x^2~-~1){(-2x.(x^2~-~1)}~+~(4x)(x^2~+~1))}/({(x^2~-~1)}^4}

= ~{(-2x^3~+~2x~+~4x^3~+4x)}/({(x^2~-~1)}^3}

= ~{(2x^3~+~6x)}/({(x^2~-~1)}^3}

= ~{(2x(x^2~+~6))}/({(x^2~-~1)}^3}

To obtain the point of inflection, y” = 0, so

~{2x(x^2~+~6)}/({(x^2~-~1)}^3}  =  0

right  x = 0  or x2 + 6 = 0

x2 + 6 = 0  does not give a real solution, hence it cannot be a point of inflection.

Let us check if x = 0 is a point of inflection by checking for y” at points adjacent to 0, viz -1/2 and 1/2.

f”(-1/2)  =  ~{2~*~{-1/2}({-1/2}^2~+~6)}/({({-1/2}^2~-~1)}^3}  =  ~{-1({1/4}~+~6)}/({({1/4}~-~1)}^3}

=  ~{({-25/4})}/({({-3/4})}^3}  =  a positive number

f”(1/2)  =  ~{2~*~{1/2}({1/2}^2~+~6)}/({({1/2}^2~-~1)}^3}  =  ~{({1/4}~+~6)}/({({1/4}~-~1)}^3}

=  ~{({25/4})}/({({-3/4})}^3}  =  a negative number

This can be shown in the table below:

x {-1/2} 0 {1/2}
f”(x) + 0

The concavity of the curve changes, hence x = 0 is a point of inflection

 

Step 7: Using the points and region, we sketch the graph

(0, 0) is a point of inflection

(0, 0) is also the x-intercept

x = pm1 is vertical asymptotes

y = 0 is a horizontal asymptote

curve sketching with asymptotes using critical points

See also other examples of sketching curves with asymptotes –

Sketching curves with asymptotes – Example 1

Sketching curves with asymptotes – Example 2

Sketching curves with asymptotes – Example 4