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

y =

**Step 1**: We get the x-intercept when y = 0. So x^{2} = 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 x^{2} + 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 and .

=

=

= 1

=

=

= 1

**Step 4**: Now we need to find the stationary points

=

=

=

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” =

=

=

=

=

When y” = 0, 24x^{2} = 32

So x^{2} =

Hence x =

Now we need to check for the change of concavity at x = and x =

x | 1 | 2 | |

f”(x) | 0 | ||

f”(x) = | > 0 | 0 | -1 |

The concavity of the curve changes, so x = is a point of inflection.

What about x =

x | -2 | -1 | |

f”(x) | 0 | -1 | |

f”(x) = | > 0 | 0 | < 0 |

The concavity of the curve changes, so x = is also a point of inflection.

At x = , y =

=

=

At x = , y =

=

=

Therefore and 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” = = 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 and , we find that as x , y 1+

See also other examples of curve sketching with asymptotes –

Curve sketching with asymptotes – Example 1