math 31 lessons chapter 4: max / min chapter 5: sketching 3. sketching polynomials
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MATH 31 LESSONS
Chapter 4: Max / Min
Chapter 5: Sketching
3. Sketching Polynomials
Steps for Sketching Polynomial Functions
Step 1. Degree
State the degree of f (x) and identify the shape.
When the degree is even, then:
if a > 1, it opens up
Deg = 2
Deg = 4
When the degree is even, then:
if a > 1, it opens up
if a < 1, it opens down
Deg = 2
Deg = 4
Deg = 2
Deg = 4
When the degree is odd, then:
if a > 1, it rises to the right
Deg = 3Deg = 5
When the degree is odd, then:
if a > 1, it rises to the right
if a < 1, it falls to the right
Deg = 3Deg = 5
Step 2. Intercepts
y-intercept(s)
Let x = 0, and then solve for y
x-intercept(s)
Let y = 0, and then solve for x
Note:
To find the x-intercept:
Factor completely and find the zeros
- for complex functions, you may need to use
the factor theorem and long division
To find the x-intercept:
For second degree factors of the form Ax2 + Bx + C =
0,
you can use the quadratic formula to solve.
i.e.
A
ACBBx
2
42
To find the x-intercept:
If you can’t factor or use the quadratic formula,
then use Newton’s method for finding roots.
i.e.
Take a first guess x1.
Then, to find x2:
1
112 xf
xfxx
Step 3. First Derivative Test
Differentiate and state the critical values
i.e. When f (x) = 0
Differentiate and state the critical values
i.e. When f (x) = 0
Use the interval test to show where the function is
increasing and decreasing
Differentiate and state the critical values
i.e. When f (x) = 0
Use the interval test to show where the function is
increasing and decreasing
Identify local (and absolute) max / mins
- substitute into the original function to get the y-values
Step 4. Sketch the Function
Place all intercepts and critical values on the grid
Using your knowledge of where the function is increasing
and decreasing, connect the dots
Extend the arms on either side to infinity
- recall that polynomial functions are continuous and
have a domain x
Ex. 1 Sketch the following function:
Try this example on your own first.Then, check out the solution.
xxy 483
Degree:
This function is of degree 3 and a > 0.
Thus, this function will rise to the right.
xxy 483
Intercepts:
y-intercept: (x = 0)
So, there is a y-intercept at (0, 0).
00480 3 y
xxy 483
x-intercepts: (y = 0)
xxy 483
0483 xx
0482 xx
0x 482 x
3448 x
So, there are x-intercepts at (0, 0) and (6.93, 0)
First Derivative Test:
xxxf 483
483 2 xxf
163 2 x
443 xx
xxxf 483
483 2 xxf
163 2 x
443 xx
0443when0 xxxf
4x Find CV’s
Interval test:
4f
-4
Sketch a number line, using the CV’s as boundaries
443 xxxf
x < -4:
Since f > 0, it is increasing.
e.g. x = -5
454535 f
0913
443 xxxf
4f
-4
() ()
-4 < x < 4:
Since f < 0, it is increasing.
e.g. x = 0
404030 f
0443
4f
-4
() () (+) () 443 xxxf
x > 4:
Since f > 0, it is increasing.
e.g. x = 5
454535 f
0193
4f
-4
() () (+) () (+) (+)
443 xxxf
Find the y-values by subbing them into the original function:
Local Min
Local Max
4f
-4
() () (+) () (+) (+)
xxxf 483
12844844atmaxLocal 3 f
12844844atminLocal 3 f
Sketch:
-6.93 6.93
128
-128
y
x
First, put the intercepts and the CV’s on the graph
-4 4
y
Then, use the interval test to connect the dots.
4f
-4
-6.93 6.93
128
-128
x-4 4
Ex. 2 Sketch the following function:
Try this example on your own first.Then, check out the solution.
45 24 xxy
Degree:
This function is of degree 4 and a > 0.
Thus, this function will open up.
45 24 xxy
Intercepts:
y-intercept: (x = 0)
So, there is a y-intercept at (0, 4).
45 24 xxy
44050 24 y
x-intercepts: (y = 0)
So, there are x-intercepts at (1, 0) and (2, 0)
45 24 xxy
045 24 xx
041 22 xx
02211 xxxx
2,1 x
First Derivative Test:
45 24 xxxf
xxxf 104 3
522 2 xx
0522when0 2 xxxf
0522when0 2 xxxf
02 x
0x
052 2 x
2
52 x
58.12
5x
Interval test:
Sketch a number line, using the CV’s as boundaries
522 2 xxxf
0f
-1.58 1.58
x < -1.58:
Since f < 0, it is decreasing.
e.g. x = -3
532323 2 f
01332
522 2 xxxf
0f
-1.58
() (+)
1.58
-1.58 < x < 0:
Since f > 0, it is increasing.
e.g. x = -1
512121 2 f
0312
522 2 xxxf
0f
-1.58
() (+) () ()
1.58
0 < x < 1.58:
Since f < 0, it is decreasing.
e.g. x = 1
512121 2 f
0312
522 2 xxxf
0f
-1.58
() (+) () () (+) ()
1.58
x > 1.58:
Since f > 0, it is increasing.
e.g. x = 3
532323 2 f
01332
522 2 xxxf
0f
-1.58
() (+) () () (+) (+) (+) ()
1.58
Find the y-values by subbing them into the original function:
25.2458.1558.158.1atminsLocal 24 f
Local Min
Local Max
0f
-1.58
() (+) () () (+) (+) (+) ()
1.58Local Min
45 24 xxy
440500atmaxLocal 24 f
Sketch:
-3
4
-4
y
x
First, put the intercepts and the CV’s on the graph
-1-2 1 32
(-1.58, -2.25) (1.58, 2.25)
Then, use the interval test to connect the dots.
0f
-1.58 1.58
-3
4
-4
y
x-1-2 1 32
(-1.58, -2.25) (1.58, 2.25)