Algebra 2 by Richard Wright
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Objectives:
SDA NAD Content Standards (2018): AII.5.1, AII.5.3, AII.6.3, AII.7.1
Quadratic functions are useful for measuring things like the motion of objects with constant acceleration or the shape of the cables on the Golden Gate bridge. However, some situtations require more complicated models such as the power output of a wind turbine which is modeled by the cubic P = 0.5Aρcfvw3.
A polynomial in one variable is a function that has one variable with exponents and all the exponents are positive. For example, 2x4 + 3x3 − 2x2 is a polynomial in one variable, but 3x−1 is not. The degree of a polynomial is the highest exponent of the variable. A polynomial is usually written so that the terms are in descending order of the exponent.
What is the degree of 3x15 − 2x12 + 5x2 − 7?
Solution
The degree is the highest exponent of the variable which is 15.
The degree of the polynomial influences the shape of its graph, so polynomials are classified by their degree.
| Degree | Type | Example | Graph |
|---|---|---|---|
| 0 | Constant | y = 1 |  |
| 1 | Linear | y = x + 1 |  |
| 2 | Quadratic | y = x2 + 2x + 1 |  |
| 3 | Cubic | y = x3 + 2x2 + x + 1 |  |
| 4 | Quartic | y = 0.5x4 − 4x2 + 4 |  |
Notice that, except for the constant, the ends of each graph go either up or down. When the degree was 1 or 3, one end went up and the other end went down. When the degree was 2 or 4, both ends went up. Remember multiplying a function by −1 will cause it to reflect over the x-axis. That would cause the ends of the graph to go in the opposite direction. All this can be summed up the following table about end behavior.
The end behavior of a polynomial function is as is in the table with the ends either rising up towards positive infinity or falling towards negative infinity.
| Positive Leading Coefficient | Negative Leading Coefficient | |
|---|---|---|
| Even Degree |  |  |
| Odd Degree |  |  |
What is the end behavior of y = −2x3 + 2x2 − 5x + 10?
Solution
The degree of the function is 3 which is odd. The leading coefficient which is the coefficient of the variable with the highest exponent is −2 which is negative. Odd degree and negative leading coefficient is the bottom right cell in the table.
We say the graph rises to the left and falls to the right because it goes up on the left and down on the right.
To graph a polynomial function,
Graph y = x3 − x2 − 2x.
Solution
Start by making a table of values.
| x | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| y | −8 | 0 | 0 | −2 | 0 | 12 |
Plot the points and draw a smooth curve through the points.
The degree is 3 which is odd, and the leading coefficient is 1 which is positive. An odd degree and leading coefficient means that the end behavior is falls to the left and rises to the right just like in the graph.
A turning point is a point on the graph of a polynomial function where the graph turns from going up to going down or vise versa. Turning points are also called local extrema. A local minimum is the lowest point in an area of the graph, and a local maximum is the highest point in the area of a graph.
Other important points on a graph are the x-intercepts which are the points where the graph intersects the x-axis. These are the points where the polynomial function equals zero. If the polynomial is written in factored form such as (x – k1)(x – k2)…, then the x-intercepts are the k's.
Use the graph to identify the turning points and x-intercepts.
Solution
The local maximum is (0, 0) because that is the highest point in the area. The local minimum is (2, −4) because that is the lowest point in the area. The x-intercepts are (0, 0) and (3, 0) because those are the points where the graph intersects the x-axis.
Graph \(f\left(x\right)=\frac{1}{4}x^4-2x^2\) and estimate the turning points and x-intercepts.
Solution
The table of values is
| x | −4 | −3 | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| y | 32 | 2.25 | −4 | −1.75 | 0 | −1.75 | −4 | 2.25 | 32 |
The local minimums are (−2, −4) and (2, −4) because they are both the lowest point in their area. The local maximum is (0, 0) because it is the highest point in its area. The x-intercepts are (−2.8, 0), (0, 0), and (2.8, 0) because those are all the places the graph intersects the x-axis.
Graph \(f\left(x\right)=\frac{1}{2}\left(x-3\right)\left(x-1\right)\left(x+2\right)\) and estimate the turning points and x-intercepts.
Solution
The table of values is
| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|
| y | −12 | 0 | 4 | 3 | 0 | −2 | 0 | 9 |
The local minimum is about (2.1, −2.0) because that is the lowest point in the area. The local maximum is about (−0.8, 4.1) because it is the highest point in its area. The x-intercepts are (−2, 0), (1, 0), and (3, 0) because those are all the places the graph intersects the x-axis.
Note: When the function is written in factored form (x − k), then the x-intercepts are all the k's. In this case the factors are (x – 3)(x – 1)(x + 2) so the x-intercepts are 3, 1, −2.
158 #1, 3, 7, 19, 21, 23, 25, 29, 31; 210 #1, 3, 7, 23, 25, 27, Mixed Review = 20
Mixed Review
