Calculus Vertical Asymptotes
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Graphing Special Cases
Vertical Asymptotes

 

 A special case:

F(X) = X3
 
 The first derivative of F(X) = X 3 is:
 

F'(X) = 3X2
 
 To find where the derivative is zero, set F'(X) = 0
 

F'(X) = 0 = 3X2

X = 0
 
 The second derivative of F(X) = X 3 is:
 

 F''(X) = 6X
 
 Now lets see what the second derivative is doing
 where the slope of the original equation is zero.
 That is, let's see if it is concave up (a smile) or concave down (a frown):
 The slope was zero when X = 0 so:
 

 F''(X) = 6X

 F''(0) = 6(0)

 F''(0) = 0
 
 OK, so this is why this one is a special case.
 At the point where the slope is zero, the graph is not curving up or down.
 It has a point of inflection.
 
 Let's do a little more testing.
 Pick a point on either side of the point of inflection
 and see what the slope is doing at those points...
 
F'(X) =  3X2 F'(X) =  3X2
F'(X) =  3(-1)2 F'(X) =  3(1)2
F'(X) =  3 F'(X) =  3

the slope is positive (up)

the slope is positive (up)
 
 So the slope is positive for all X values except when X = 0
 That would make the graph look something like this:

 
 Let's check and see what the second derivative is doing at X = -1 and X = 1 ...
 
F''(X) =  6X F''(X) =  6X
F''(X) =  6(-1) F''(X) =  6(1)
F''(X) =  - 6 F''(X) =  6

concave down

concave up
 
 Which means the graph looks like this ...

 
 The frown part gets just to where the slope is zero,
 and then it turns into the smile part.
 
 A point here is that just because there is a point where the slope is zero, 
 doesn't mean that there HAS to be a point that is a relative minimum or maximum.
 
 Another special case happens when the denominator of an equation
 is equal to zero.
 This happens a lot if there is a variable in the denominator.
 
 If we can factor out the terms that make the denominator zero,
 the graph will have a hole.
 In the equation:
F(x) =  X3
X
 
 The denominator will equal zero when X = 0.
 
 We can factor an X out of the numerator and the denominator,
 so that this equation becomes:
 

 F(X) = X2
 
 This is just a parabola.
 But the original equation has a problem when X = 0,
 so the graph of the equation is a parabola with a hole at X = 0:
 

 
 
 The other side of the story is when the zero in the denominator
 can't be factored out.
 When that happens, 
 there will be a vertical asymptote at the variable value 
that makes the denominator equal zero.
 In the equation:

F(X) =

 1 also known as F(X) = X-1
X
 
 The X in the denominator can't be factored out.
 That means there will be a vertical asymptote when X = 0

 
 Let's do some calculus on this equation:
 The first and second derivatives of this equation are:
 
F'(x) = -1

also known as F'(X) = -X-2
X2
     
F''(x) = 2

also known as F''(X) = 2X-3
X3
 
 There is no place where any of these functions (F(X), F'(X). F''(X)) will equal zero.
 Since there are only numbers, not variables, on top.
 But all of them are undefined at X = 0.
 
 We can just pick any point to the left of zero, and any point to the right of zero.
 Then check for slope and concavity at these points.
 All points on that side of the X = 0 point will have the same slope and concavity.
 
 Here we go. Always use the easiest points you can.
 Let's use -1 and +1
 First -1 ...
F'(X) = -1 F''(X) = 2
X2 X3
 
F'(-1) = -1 F''(-1) = 2
(-1)2 (-1)3
 
F'(-1) = -1 F''(-1) = 2
1 -1
 
F'(-1) = -1 F''(-1) = -2
 
the slope is negative (down) concave down (frown)
 

 Now +1 ... 
F'(X) = -1 F''(X) = 2
X2 X3
F'(1) = -1 F''(1) = 2
(1)2 (1)3
F'(1) = -1 F''(1) = 2
1 1
F'(1) = -1 F''(1) = 2
the slope is negative (down) concave up (smile)
 
 So the graph looks like this:

 
 

   copyright 2005 Bruce Kirkpatrick
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