Families of Polynomials
A Family of Polynomials is a collection of equations based on shared characteristics like degrees and roots.
We can start writing a function in factored form given these characteristics; however, we will need additional information to identify the coefficient, \(a\), in order to determine the exact function.
Degrees and Families
- Linear Family: Degree 1, e.g., \( f(x) = ax + b \)
- Quadratic Family: Degree 2, e.g., \( f(x) = ax^2 + bx + c \)
- Cubic Family: Degree 3, e.g., \( f(x) = ax^3 + bx^2 + cx + d \)
x-Intercepts and Factored Form
The roots or x-intercepts of a polynomial can be shown in its factored form:
Where \( r_1, r_2, ..., r_n \) are the roots, and 'a' is the leading coefficient.
Leading Coefficient
The leading coefficient '\(a\)' affects:
- The width of the graph
- The end behavior of the graph with respect to infinity
For instance, cubic function with roots at \(-2\), \(1\), and \(3\) is expressed as:
An additional point on the graph would determine the specific value of '\(a\)'.
Example
Write the exact family member of a quadratic function with roots at \(-5\) and \(2\) that passes through \((8, 6) ).
We can start with the equation for the family of polynomials with these roots:
Next, we can substitute \((8, 39)\) into the equation:
\(39 = a(8+5)(8-2)\)
\(39 = a(13)(6)\)
\(78a = 39\)
\(a = \cfrac{1}{2}\)
Finally, we can plug the \(a\)-value into the original equation to determine the exact family member:
Therefore, we can determine the equation for the exact member of the family is \(\boldsymbol{f(x) = \cfrac{1}{2}(x+5)(x-2)}\).
Recall that if you are given a set of conditions you may have many answers (or a family) of equations that will satisfy them. There will be one unique equation only if enough information is provided.
- Determine the exact member of the polynomial family that satisfies the conditions
- Draw a sketch of the function
A cubic polynomial with a double root at \(4\), a root at \(8\), and a \(y\)-intercept of \(6\).
i. We can start with the equation for the family of polynomials with these roots:
Since this function has a \(y-\)-intercept of \(6\), we can substitute \((0, 6)\) into the equation:
Next, we can solve for \(k\):
\(6 = k(-4)^2(-8)\)
\(6 = k(16)(-8)\)
\(-128k = 6\)
\(k = -\cfrac{3}{64}\)
Finally, we can plug the \(k\)-value into the original equation to determine the exact member of the polynomial family:
Therefore, we ccan determine the exact member of the polynomial family is \(\boldsymbol{f(x) = -\cfrac{3}{64}(x- 4)^2(x-8)}\).
ii. We can sketch a graph of the function as such:
A polynomial of degree \(6\), with zeros at \(-2\), \(4\) (order \(2\)), \(8\) (order \(3\)) that goes through the point \((-4, 80)\).
We can start with the equation for the family of polynomials with these roots:
Next, we can substitute \((-4, 80)\) into the equation:
Then, we can solve for \(k\):
\(80 = k(-2)(-8)^2(-12)^3\)
\(80 = k(-2)(64)(-1728)\)
\(221184k = 80\)
\(k = \cfrac{5}{13824}\)
Finally, we can plug the \(k\)-value into the original equation to determine the exact member of the polynomial family:
Therefore, the equation for the exact member of the family is \(\bolsymbol{f(x) = \cfrac{5}{13824}(x+2)(x-4)^2(x-8)^3}\).
ii. We can sketch a graph of the function as such:
| X | Y |
|---|---|
| 1 | 4 |
| 2 | 15 |
| 3 | 30 |
| 4 | 49 |
| 5 | 72 |
| 6 | 99 |
In order to determine the degree of the function, we can calculate the differences to identify the pattern.
We can start by calculating the first differences:
First Differences (ΔY): 11, 15, 19, 23, 27
Since the First Differences aren't constant, we can move on to calculating Second Differences:
Second Differences (Δ²Y): 4, 4, 4, 4
Since the Second Differences are constant, we can determine that the function's degree is \(2\) and is therefore quadratic.
Next, we can determine the coefficient, \(a\) by dividing the constant second difference, \(4\), by the factorial of the degree, \(2\):
\(a = \cfrac{\text{4}}{2!}\)
\(a = \cfrac{\text{4}}{(2)(1)}\)
\(a = \cfrac{4}{2} = 2\)
We can express the function in general form as:
Then, we can help determine \(b\) and \(c\) using points from the table.
First, we can evaluate \(f(1)\):
\(f(1) = 2(1)^2 + b(1) + c\)
\(4 = 2 + b + c\)
\(2 = b + c\)
Next, we can evaluate \(f(2)\):
\(f(2) = 2(2)^2 + b(2) + c\)
\(15 = 8 + 2b + c\)
\(7 = 2b + c\)
We can use the Elimination Method to solve for \(b\):
\(2 = b + c\)
\(7 = 2b + c\)
\((7-2) = (2b - b) + (c - c)\)
\(b = 5\)
We can express the updated general formula as:
Finally, we can solve \(c\) by substituting one of the points from the table:
\(f(3) = 2(3)^2 + 5(3) + c\)
\(30 = 18 + 15 + c\)
\(30 = 33 + c\)
\(c = -3\)
Finally, we can determine the final equation by plugging in the appropriate values:
Therefore, we can determine the final equation is \(\boldsymbol{f(x) = 2x^2 + 5x -3}\).
| X | Y |
|---|---|
| 1 | 0 |
| 2 | 1 |
| 3 | 10 |
| 4 | 33 |
| 5 | 76 |
| 6 | 145 |
| 7 | 246 |
In order to determine the degree of the function, we can calculate the differences to identify the pattern.
We can start by calculating the first differences:
First Differences (ΔY): 1, 9, 23, 43, 69, 101
Since the First Differences aren't constant, we can move on to calculating Second Differences:
Second Differences (Δ²Y): 8, 14, 20, 26, 32
Since the Second Differences aren't constant, we can move on to calculating Third Differences:
Third Differences (Δ³Y): 6, 6, 6, 6
We can determine that the degree is 3 since the third differences are constant.
Since we have determined that the function is cubic, we can express it in general form as:
Next, we can determine the coefficient, \(a\) by dividing the constant third difference, \(6\), by the factorial of the degree, \(3\):
We can determine the coefficient, \(a\) by dividing the constant third difference, \(6\), by the factorial of \(3\):
\(a = \cfrac{\text{6}}{3!}\)
\(a = \cfrac{6}{(3)(2)(1)}\)
\(a = \cfrac{6}{6} = 1\)
Next, we can determine \(b\), \(c\) and \(d\) using points from the table:
\(f(1) = 1(1)^3 + b(1)^2 + c(1) + d \)
\(0 = 1 + b + c + d \)
\(-1 = b + c + d \)
\(f(2) = 1(2)^3 + b(2)^2 + c(2) + d \)
\(1 = 8 + 4b + 2c + d \)
\(-7 = 4b + 2c + d \)
\(f(3) = 1(3)^3 + b(3)^2 + c(3) + d \)
\(10 = 27 + 9b + 3c + d \)
\(-17 = 9b + 3c + d \)
\(f(4) = 1(4)^3 + b(4)^2 + c(4) + d \)
\(33 = 64 + 16b + 4c + d\)
\(-31 = 16b + 4c + d\)
We can use the Elimination Method to solve for \(b\):
\(-1 = b + c + d \)
\(-7 = 4b + 2c + d \)
\(-17 = 9b + 3c + d \)
\(-31 = 16b + 4c + d \)
\((-7-(-1)) = (4b - b) + (2c - c) + (d-d)\)
\(-6 = 3b + c\)
\((-17-(-7)) = (9b - 4b) + (3c - 2c) + (d-d)\)
\(-10 = 5b + c\)
\((-31-(-17)) = (16b - 9b) + (4c - 3c) + (d-d)\)
\(-14 = 7b + c\)
We can continue using the Elimination Method for the equations we determined in the last step to solve for \(b\):
\(-6 = 3b + c\)
\(-10 = 5b + c\)
\((-10 -(-6)) = (5b - 3b) + (c-c)\)
\(-4 = 2b\)
\(b = -2\)
We can substitute the value of \(b\) into one of the equations above to solve for \(c\):
\(-14 = 7b + c\)
\(-14 = 7(-2) + c\)
\(-14 = -14 + c\)
\(c = 0\)
Finally, we can substitute \(b\) and \(c\) into one of the other equations above to solve for \(d\):
\(-1 = b + c + d\)
\(-1 = -2 + 0 + d\)
\(d = 1\)
Finally, we can determine the final equation by plugging in the appropriate values:
Therefore, we can determine the final equation is \(\boldsymbol{f(x) = x^3 - 2x^2 + 0x + 1}\).