Curve Sketching
To sketch a curve, consider this four-part analysis that’ll give you all the information you need.
Test the Function
Find where f  x   0 . This tells you the x-intercepts (or roots). By setting x  0 , we can determine the yintercept. Finally, find any vertical asymptotes or perhaps a hole in the graph. Look for discontinuities.
Test the First Derivative
Find where f   x   0 . This gives you the critical points. We can determine whether the curve is rising or
falling, as well as where the maxima and minima are. It’s also possible to determine if the curve has any points
where it’s nondifferentiable.
Test the Second Derivative
Find where f   x   0 . This shows you the where any (possible) points of inflection are. These points are
where the graph of the function changes concavity. Then we can determine where the graph curves upward and
where it curves downward.
Test the End Behavior
Look at what the general shape of the graph will be, based on the values of y for very large values of  x .
Look for horizontal asymptotes.
Below is a chart for the end behavior of polynomial functions.
Here are things to remember:
1. When f   x   0 , the curve is rising. When f   x   0 , the curve is falling. When f   x   0 or f   x 
does not exist, the curve is at a critical point.
2. When f   x   0 , the curve is “concave up.” When f   x   0 , the curve is “concave down.” When
f   x   0 , the curve is at a point of inflection.
3. The y-coordinates of each critical point are found by plugging the x-value into the original function.
1
Example 1
Sketch the graph of f  x   x3 12x .
Step 1 Test the Function.
Find intercepts.
x-intercepts
y-intercept
x3  12 x  0

x( x 2  12)  0
x x  12
 x 
y  03  12  0  0

12  0
x  0, x   12
There are no asymptotes (polynomial functions do not have asymptotes).
x-intercepts:
 0, 0 ,
y-intercept:
 0, 0

 
12, 0 ,
12, 0

Step 2 Test the First Derivative.
Find critical numbers by taking the derivative of the function.
f   x   3x2 12
Set the derivative equal to zero and solve for x:
3x 2  12  0
3 x2  4  0
3  x  2  x  2   0
x  2
Next, plug x  2 and x  2 into the original equation.
y   2   12  2   8  24  16
3
y   2   12  2   8  24  16
3
Coordinates of critical points are at  2, 16  and
 2, 16 .
2
Find where the curve is increasing and decreasing.
f
Inc
Dec
Inc
___________-2________________2____________
f
+++++++
-------------
+++++++
Step 3 Test the Second Derivative.
Now take the second derivative to find any points of inflection.
f   x   6 x
Set the derivative equal to zero and solve for x:
6x  0
x0
The curve has a point of inflection at  0, 0 .
Determine concavity.
f
Concave Down
Concave Up
___________________0____________________
f 
----------------
++++++++++++
Now plug the critical values (found in step 2) into the second derivative to determine whether that
critical number is a maximum or a minimum. (We are using the Second Derivative Test here.)
f   2  6  2  12 . This is positive, so the curve has a minimum at  2, 16  and the curve is concave
up at that point. f   2  6  2  12 . This is negative, so the curve has a maximum at  2, 16 and
the curve is concave down at that point.
Step 4 Test the End Behavior.
From College Algebra, we know that a third-degree polynomial with a positive leading coefficient has
end behavior that starts negative and ends positive.
3
Put all this information into a chart:
f x 
f x 
f x 
+
-
Increasing, Concave Down
0
-
Relative Maximum at  2, 16 
-
-
Decreasing, Concave Down
-
0
Point of Inflection at 0, 0
-
+
Decreasing, Concave Up
0
+
Relative Minimum at 2,  16
+
+
Increasing, Concave Up
   x  2
x  2
16
2 x 0
x0
0
0 x2
x2
-16
2 x
x-intercepts:
 0, 0 ,

 
12, 0 ,
12, 0

Characteristics of the Graph
y-intercept:
 0, 0
Armed with this information, we can now plot the graph.
4
Example 2
Sketch the graph of f  x   x4  2 x3  2 x2  1.
Step 1 Test the Function.
Find intercepts.
If the equation doesn’t factor easily, it’s best not to bother to find the function’s roots.
y-intercept
y  f  0   0  2 0  2 0  1  1
4
3
2
There are no asymptotes (polynomial functions do not have asymptotes).
x-intercepts:
Not easy to find. Maybe use a calculator or don’t bother.
y-intercept:
 0, 1
Step 2 Test the First Derivative.
Find critical points by taking the derivative of the function.
f   x   4 x3  6 x 2  4 x
Set the derivative equal to zero and solve for x:
4 x3  6 x 2  4 x  0


2 x 2 x 2  3x  2  0
2 x  2 x  1 x  2   0
x  0,
Next, plug x  0 , x 
1
, 2
2
1
, and x  2 into the original equation.
2
y  f  0   0  2 0  2 0  1  1
4
3
4
2
3
2
13
1 1
1
1
y  f       2   2  1 
16
2 2
2
2
y  f  2    2   2  2   2  2   1  7
4
3
 1 13 
Critical points are at  0, 1 ,  ,  , and
 2 16 
2
 2,  7  .
5
Find where the curve is increasing and decreasing.
f
Dec
Inc
Dec
Inc
___________-2________________0_________________1/2____________
f
----------
+++++++++
--------------
++++++++
Step 3 Test the Second Derivative.
Now take the second derivative to find any points of inflection.
f   x   12x2  12x  4
Set the derivative equal to zero and solve for x:
12 x 2  12 x  4  0


4 3x 2  3x  1  0
3x 2  3x  1  0
3  21
6
x  0.26,  1.26
x
Now find the y-coordinates for the points above:
y  f  0.26    0.26   2  0.26   2  0.26   1  0.90
4
3
2
y  f  1.26    1.26   2  1.26   2  1.26   1  3.66
4
3
2
The curve has two points of inflection at  0.26, 0.90 and  1.26,  3.66
Now plug the critical values (found in step 2) into the second derivative to determine whether we have a
maximum or a minimum. (We are using the Second Derivative Test here.)
f   0   12  0   12  0   4  4 .
2
This is negative, so the curve has a maximum at  0, 1 and the curve is concave down at that point.
2
1
1
1
 1 13 
f     12    12    4  5 . This is positive, so the curve has a minimum at  ,
 and the
2
2
2
 2 16 
curve is concave up there.
f   2   12  2   12  2   4  20 . This is positive, so the curve has a minimum at  2,  7  and the
2
curve is concave up there.
6
Determine concavity.
f
Concave Up
Concave Down
Concave Up
___________________-1.26____________________0.26________________
f 
++++++++++++
-----------------
++++++++++
Step 4 Test the End Behavior.
From Algebra II (or College Algebra), we know that a fourth-degree polynomial with a positive leading
coefficient has end behavior that starts positive and ends positive.
Put all this information into a chart:
f x 
   x  2
x  2
-7
 2  x  1.26
x  1.26
-3.66
 1.26  x  0
x0
1
0  x  0.26
x  0.26
0.90
0.26  x  0.5
x  0.5
0.5  x  
x-intercepts:
13/16
f x 
f x 
-
+
Decreasing, Concave Up
0
+
Relative Minimum at  2,  7
+
+
Increasing, Concave Up
+
0
Point of Inflection at  1.26,  3.66
+
-
Increasing, Concave Down
0
-
Relative Maximum at 0,1
-
+
Decreasing, Concave Down
-
0
Point of Inflection at  0.26, 0.90
-
+
Decreasing, Concave Up
Characteristics of the Graph
0
 1 13 
Relative Minimum at  , 
 2 16 
+
Increasing, Concave Up
Used a calculator to find:  2.692, 0 ,  0.594, 0 .
y-intercept:
 0, 1
7
We can now plot the curve.
8
Example 3
Sketch the graph of f  x   2  x2/ 3 .
Step 1 Test the Function.
Find intercepts:
x-intercepts
y-intercept
2  x2/ 3  0
x2/ 3  2

x2/ 3

3/ 2
f 0  2  0
 23/ 2
2/3
2
x  23/ 2  2 2
There are no asymptotes (there are no places where the function is undefined).
x-intercepts:
2
  2
y-intercept:
 0, 2
2, 0 ,
2, 0

Step 2 Test the First Derivative
Find critical points by taking the derivative of the function.
2
2
f   x    x 1/ 3   3
3
3 x
Set the derivative equal to zero and solve for x:

2
3
3 x
0
We have a problem here. There are no values of x for which the derivative is zero. However, at x  0 ,
the derivative is undefined. Therefore, x  0 is a critical number of the function, and the curve has a
cusp at  0, 2 . There are no other critical points.
We can see that when x is negative, the derivative is positive. Therefore, the curve is rising to the left of
zero.
We can also see that when x is positive, the derivative is negative. Therefore, the curve is falling to the
right of zero.
9
f
Inc
Dec
____________0____________
f
+++++++
----------
Step 3 Test the Second Derivative.
2
Now take the second derivative to find any points of inflection. Use f   x    x 1/ 3 .
3
f   x  
2 4 / 3
2
2
x
 4/3 
9
9x
9 x1/ 3
 
4
2

9
 x
3
4
Set the derivative equal to zero and solve for x:
2
9
 x
3
4
0
Again, there are no values of x where this is zero. There are no inflection points. In fact, the second
derivative is positive at all values of x except 0. Therefore, the graph is everywhere concave up.
f
Concave Up
Concave Up
___________________ 0 ____________________
f 
++++++++++++
++++++++++++
Step 4 Test the End Behavior.
 


Notice that when x is a very large positive number lim , the expression 2  x 2 / 3 is a large negative
x 
 


number lim . Also, when x is a very large negative number, the expression 2  x 2 / 3 is a very large
x  
negative number. We can write these facts using correct mathematical notation as follows:


lim 2  x 2 / 3  
x 
and


lim 2  x 2 / 3  
x 
10
Put all this information into a chart:
f x 
  x  0
x0
2
0 x
x-intercepts:
2
  2
2, 0 ,
f x 
f x 
+
+
Does Not
Exist
Does Not
Exist
-
+
2, 0

Characteristics of the Graph
Increasing, Concave Up
Relative Maximum at 0, 2 , which is a
cusp in the graph
Decreasing, Concave Up
y-intercept:
 0, 2
We can now plot the curve.
11
Example 4
Sketch the graph of f  x  
3x
.
x2
Step 1 Test the Function.
Find intercepts. A fraction can only equal zero when the numerator is zero. So, to find the x-intercepts,
set the numerator equal to 0 and solve for x.
x-intercepts
3x
0
x2
3x  0
y-intercept
f  0 
x0
x-intercepts:
 0, 0
y-intercept:
 0, 0
30
0  2

0
0
2
Next, look for asymptotes. The denominator is undefined at x  2 , so there is a vertical asymptote at
x  2 . Take the left-hand and right-hand limits of the function at x  2 :
lim
x 2
3x
 
x2
and
lim
x 2
3x
 
x2
This means that the graph rises infinitely to the left of x  2 and falls infinitely to the right of x  2 .
Step 2 Test the First Derivative
Find critical points by taking the derivative of the function.
f x  
x  23  3x 1  3x  6  3x  6
 x  2 2
 x  2 2
 x  2 2
Set the derivative equal to zero and solve for x:
6
 x  2
2
0
There are no values of x for which the derivative is zero. Because the numerator is 6 and the
denominator is squared, the derivative will always be positive (the curve is always rising). You should
note that the derivative is undefined at x  2 , but you already know that there’s an asymptote at x  2
so you don’t need to examine this point further.
12
Here’s a visual:
f
Inc
Inc
____________ -2 ____________
f
+++++++
+++++++
Step 3 Test the Second Derivative.
Now take the second derivative to find any points of inflection. Use f   x  
 x  2  0   6  2  x  2
2
f   x  
2 2
 x  2  



12  x  2 
 x  2
4

6
 x  2
2
.
12
 x  2
3
Set the derivative equal to zero and solve for x:
12
 x  2
3
0
Again, there are no values of x where this expression is zero (because the numerator is never zero).
When x  2 , the expression is positive and so the graph is concave up.
When x  2 , the expression is negative and so the graph is concave down.
f
Concave Up
Concave Down
___________________ -2 ____________________
f 
++++++++++++
-----------------
Step 4 Test the End Behavior.
3x
3x
3
3
As x moves infinitely to the right, we have lim
 lim x  lim

3.
x  x  2
x  x
x

2
2 1 0

1
x x
x
3x
3x
x  lim 3  3  3 .
As x moves infinitely to the left, we have lim
 lim
x  x  2
x  x
2 x
2 1 0

1
x x
x
Therefore, y  3 is a horizontal asymptote.
13
Put all this information into a chart:
f x 
   x  2
x  2
Does Not
Exist
2 x 
x-intercepts:
 0, 0
f x 
f x 
+
+
Does Not
Exist
Does Not
Exist
+
-
y-intercept:
 0, 0
Characteristics of the Graph
Decreasing, Concave Up
Vertical Asymptote at x  2
Increasing, Concave Down
Horizontal Asymptote:
y3
We can now plot the curve.
14
Example 5
Sketch the graph of f x  
x 2  2x  4
.
x2
Step 1 Test the Function.
Find intercepts. A fraction can only equal zero when the numerator is zero. So, to find the x-intercepts,
set the numerator equal to 0 and solve for x.
x-intercepts
y-intercept
x 2  2x  4
0
x2
x 2  2x  4  0
x
x
  2  
 2
21
2
 414 
2

0  20  4
4
f 0 

 2
0  2
2
2   12
2
x-intercepts:
None
y-intercept:
0,  2
Next, look for asymptotes. The denominator is undefined at x  2 , so there is a vertical asymptote at
x  2 . Take the left-hand and right-hand limits of the function at x  2 :
lim
x 2
x 2  2x  4
 
x2
and
lim
x 2
x 2  2x  4
 
x2
This means that the graph falls infinitely to the left of x  2 and rises infinitely to the right of x  2 .
Step 2 Test the First Derivative
Find critical points by taking the derivative of the function.
f x  
x  22 x  2  x 2  2 x  41  xx  4
x  22
x  22
(The algebra is shown at the end of the problem.)
Set the derivative equal to zero and solve for x:
x x  4 
 x  2 2
0

x  0, x  4
The derivative is undefined at x  2 .
15
f
Inc
Dec
Dec
Inc
____________ 0 ____________ 2 ___________ 4 ______________
f
+++++++
---------
---------
++++++++
We will use the First Derivative Test this time to check for relative extrema.
Since f 0  0 and f 4  0 , the x  0 and x  4 are critical numbers.
f has a relative maximum at x  0 because f x  changes from positive to negative at x  0 .
f has a relative minimum at x  4 because f x  changes from negative to positive at x  4 .
The relative maximum is f 0 
02  20  4  4  2 .
0  2
2
The relative minimum is f 4 
42  24  4  16  8  4  12  6
4  2
2
2
The coordinates of the extrema are:
Relative Maximum 0,  2
Relative Minimum 4, 6
Step 3 Test the Second Derivative.
Now take the second derivative to find any points of inflection. Use f x  
f x  
x 2  4x
 x  2 2
.
x  22 2 x  4  x 2  4 x 2x  2  8
x  24
x  23
(The algebra is shown at the end of the problem.)
Set the derivative equal to zero and solve for x:
8
x  23
0
There are no values of x where this expression is zero.
When x  2 , the expression is positive and so the graph is concave up.
When x  2 , the expression is negative and so the graph is concave down.
16
f
Concave Down
Concave Up
___________________ 2 ____________________
f 
---------------
+++++++++++
Step 4 Test the End Behavior.
Since the degree of the numerator is one greater than the degree of the denominator, the function will
not have a horizontal asymptote; instead, the function has an oblique (or slant) asymptote. From
algebra, we know that to find the slant asymptote, we should divide (either long division or synthetic
division) and ignore the remainder. The result of the division gives the equation for the oblique (slant)
asymptote.
Long Division
Synthetic Division
x
x2
x  2x  4
2

 x 2  2x

0 4
2 1 2
4
2
0
0
4
1
x 2  2x  4
4
Division gives
. Ignoring the remainder, we have the slant asymptote as y  x .
 x
x2
x2
As x increases infinitely to the left and infinitely to the right, the graph of the function will approach the
graph of the slant asymptote, y  x
17
Put all this information into a chart:
f x 
  x  0
x0
-2
0 x2
x2
Does Not
Exist
2 x4
x4
6
4 x
x-intercepts:
None
f x 
f x 
+
-
Increasing, Concave Down
0
-
Relative Maximum at 0, 2
-
-
Decreasing, Concave Down
Does Not
Exist
Does Not
Exist
Vertical Asymptote at x  2
-
+
Decreasing, Concave Up
0
+
Relative Minimum at 4, 6
+
+
Increasing, Concave Up
y-intercept:
0,  2
Characteristics of the Graph
Slant Asymptote:
yx
We can now plot the curve.
18
Here’s the algebra for example 6:
First Derivative:
f x  
x  22 x  2  x 2  2 x  41
x  22

2x2  2x  4x  4  x2  2x  4
x  22
x2  4x
x  22
x x  4 

x  22

Second Derivative:
x  22 2 x  4  x 2  4 x 2x  2
 x  2 4
x 2  4 x  42 x  4  x 2  4 x 2 x  4

 x  2 4
2 x  4x 2  4 x  4  x 2  4 x 

 x  2 4

2 x  4 x 2  4 x  4  x 2  4 x 

 x  2 4
2 x  44

 x  2 4
42 x  4 

 x  2 4
8 x  2 

 x  2 4
f  x  

8
x  23
19
Homework – Curve Sketching.
1 – 11. Sketch the graph of each function using the techniques described in this lesson.
1.
f x   x 4  4 x 3  16 x
3.
f x  
5.
f  x   3x 2 / 3  x
7.
f x  
9.
f  x   3x 4  4 x 3
11.
f x  
x4
x4
x 2  6 x  12
x4
2.
f x   2  x  x 3
4.
f x  
6.
f x   x 4  8x 3  18x 2  16 x  5
8.
f x  
10.
f  x    x  1
x2
x2  3
x2
x
5
x2  4
x2 1
20