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# Curve Sketching with Asymptotes – Example 4

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

y = 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 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, 24x2 = 32

So x2  = 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+ 