Home > Calculus > Curve Sketching > Sketching Curves with Asymptotes – Example 2

# 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) =

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) =

f(-x) =    =

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

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 3 from RHS 3+

=   =  =  +

So

As  from RHS,

Similarly as x 3 from LHS (3)

=    =    =   −

As

As x -3 from LHS (−3)

y =  =   =  =  +

As  from LHS,

Similarly as x −3+ from RHS

y =  =    =    =   −

As  from RHS,

What happens to the curve as x ?

y = f(x) =

y =

As

y =

> 1

since numerator > denominator

As x  y  from above.

> 1

since again numerator > denominator

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

=

=

=

= 0, so   = 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.

=

=

=

=

=

=

y” at x = 0 is

= – 0.222  < 0  (therefore maximum point)

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

When y” = 0   =  0

54x2 + 162 = 0

x2  =  =  -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

and

and

Horizontal asymptotes

,   from above