College Algebra
Answer/Discussion to Practice
Problems
on Graphs of Polynomial Functions
Answer/Discussion
to 1a

|
| Use the Leading Coefficient
Test, described above, to find if the graph rises or falls to the
left and to the right.
Do you think that the graph rises or falls to the left and to the
right?
Since the degree of the polynomial, 4, is even and the leading
coefficient, -3, is negative, then the graph of the given polynomial falls
to the left and falls to the right. |
 |
*Setting the 1st factor = 0
*Solve for x
*x = 0 is a zero |
Since the exponent on this factor is 2, then the multiplicity for
the zero x = 0 is 2.
Since the multiplicity is 2, which is even, then the graph touches
the x-axis and turns around at the zero x
= 0. |
 |
*Setting the 2nd factor = 0
*Solve for x
*x = -3 is a zero |
Since the exponent on this factor is 2, then the multiplicity for
the zero x = -3 is 2.
Since the multiplicity is 2, which is even, then the graph touches
the x-axis and turns around at the zero x
= -3. |
 |
*Plug in 0 for x |
| The y-intercept is (0, 0). |
| Step 4: Determine if
there is any symmetry. |
 |
*Plug in -x for
x
|
| It is not symmetric about the y-axis. |
 |
*Plug in -x for
x
*Take the opposite of f(x)
|
| It is not symmetric about the origin. |
| Since the degree of the function is 4, then there is at most 4 -
1 = 3 turning points. |
| To get a more accurate curve, lets find some points that are in between
the points we found in steps 2 and 3: |
|
x
|
|
(x, y)
|
|
-2
|
|
(-2, -12)
|
|
-1
|
|
(-1, -12)
|
Answer/Discussion
to 1b

|
| Use the Leading Coefficient
Test, described above, to find if the graph rises or falls to the
left and to the right.
Do you think that the graph rises or falls to the left and to the
right?
Since the degree of the polynomial, 3, is odd and the leading coefficient,
1, is positive, the graph of the given polynomial falls to the left
and rises to the right. |
 |
*Factor out a GCF
|
 |
*Setting the 1st factor = 0
*Solve for x
*x = 0 is a zero |
Since the exponent on this factor is 2, then the multiplicity for
the zero x = 0 is 2.
Since the multiplicity is 2, which is even, then the graph touches
the x-axis and turns around at the zero x
= 0. |
 |
*Setting the 2nd factor = 0
*Solve for x
*x = 3 is a zero |
Since the exponent on this factor is 1, then the multiplicity for
the zero x = 3 is 1.
Since the multiplicity is 1, which is odd, then the graph crosses
the x-axis at the zero x
= 3. |
 |
*Plug in 0 for x |
| The y-intercept is (0, 0). |
| Step 4: Determine if
there is any symmetry. |
 |
*Plug in -x for
x
|
| It is not symmetric about the y-axis. |
 |
*Plug in -x for
x
*Take the opposite of f(x)
|
| It is not symmetric about the origin. |
| Since the degree of the function is 3, then there is at most 3 -
1 = 2 turning points. |
| To get a more accurate curve, lets find some points that are in between
the points we found in steps 2 and 3: |
|
x
|
|
(x, y)
|
|
1
|
|
(1, -2)
|
|
2
|
|
(2, -4)
|
All contents copyright (C) 2002 - 2008, WTAMU and Kim Seward. All rights reserved. Last revised on October 8, 2002 by Kim Seward. |