Figure \(\PageIndex{5c}\): The complete graph of the polynomial function \(f(x)=−2(x+3)^2(x−5)\).
Example \(\PageIndex{6}\)
Sketch a graph of \(f(x)=x^2(x^2−1)(x^2−2)\). State the end behaviour, the \(y\)-intercept, and \(x\)-intercepts and their multiplicity.
Step 1. The leading term is the product of the high order terms of each factor: \( (x^2)(x^2)(x^2) = x^6\). The leading term is positive so the curve rises on the right. The degree of the leading term is even, so both ends of the graph go in the same direction (up). \(\qquad \nwarrow \dots \nearrow \)
Step 2. The \(y\)-intercept can be found by evaluating \(f(0)\). So \(f(0)=0^2(0^2-1)(0^2-2)=(0)(-1)(-2)=0 \).
The \(x\)-intercepts can be found by solving \(f(x)=0\). Set each factor equal to zero.
\[\begin{align*} x^2&=0 & & & (x^2−1)&=0 & & & (x^2−2)&=0 \\ x^2&=0 & &\text{ or } & x^2&=1 & &\text{ or } & x^2&=2 \\ x&=0, \:x=0 &&& x&={\pm}1 &&& x&={\pm}\sqrt{2} \end{align*}\] .
This gives us five \(x\)-intercepts: \( (0,0)\), \((1,0)\), \((−1,0)\), \((\sqrt{2},0)\), and \((−\sqrt{2},0)\). The \(x\)-intercepts \((1,0)\), \((−1,0)\), \((\sqrt{2},0)\), and \((−\sqrt{2},0)\) all have odd multiplicity of 1, so the graph will cross the \(x\)-axis at those intercepts. The \(x\)-intercept \((0,0)\) has even multiplicity of 2, so the graph will stay on the same side of the \(x\)-axis at 2. (The graph is said to be tangent to the x- axis at 2 or to "bounce" off the \(x\)-axis at 2).
Step 3. The polynomial is an even function because \(f(-x)=f(x)\), so the graph is symmetric about the y-axis. The graph appears below.
Example \(\PageIndex{7}\)
Sketch a graph of the polynomial function \(f(x)=x^4−4x^2−45\). State the end behaviour, the \(y\)-intercept, and \(x\)-intercepts and their multiplicity.
Step 1. The leading term is \(x^4\). The leading term is positive so the curve rises on the right. The degree of the leading term is even, so both ends of the graph go in the same direction (up). \(\qquad \nwarrow \dots \nearrow \)
Step 2. The \(y\)-intercept occurs when the input is zero.
\[ \begin{align*} f(0) &=(0)^4−4(0)^2−45 =−45 \end{align*}\]
The \(y\)-intercept is \((0,−45)\).
The \(x\)-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial.
\[\begin{align*} f(x)&=x^4−4x^2−45 \\ &=(x^2−9)(x^2+5) \\ &=(x−3)(x+3)(x^2+5) \end{align*}\]
\[0=(x−3)(x+3)(x^2+5) \nonumber \]
\( \begin{array}{ccccc} x−3=0 & \text{or} & x+3=0 &\text{or} & x^2+5=0 \\ x=3 & \text{or} & x=−3 &\text{or} &\text{(no real solution)} \end{array} \)
The \(x\)-intercepts \((3,0)\) and \((–3,0)\) all have odd multiplicity of 1, so the graph will cross the \(x\)-axis at those intercepts.
Step 3. The polynomial is an even function because \(f(-x)=f(x)\), so the graph is symmetric about the y-axis. The graph appears below. The imaginary zeros are not \(x\)-intercepts, but the graph below shows they do contribute to "wiggles" (truning points) in the graph of the function.
Sketch a graph of \(f(x)=\dfrac{1}{6}(x−1)^3(x+3)(x+2)\).
Definition: Turning Points
A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).
A polynomial of degree \(n\) will have, at most, \(n\) \(x\)-intercepts and \(n−1\) turning points.
The degree of a polynomial function helps us to determine the number of \(x\)-intercepts and the number of turning points. A polynomial function of \(n\) th degree is the product of \(n\) factors, so it will have at most \(n\) roots or zeros. When the zeros are real numbers, they appear on the graph as \(x\)-intercepts. The graph of the polynomial function of degree \(n\) can have at most \(n–1\) turning points. This means the graph has at most one fewer turning points than the degree of the polynomial or one fewer than the number of factors.
The maximum number of turning points of a polynomial function is always one less than the degree of the function.
Example \(\PageIndex{9}\): Find the Maximum Number of Turning Points of a Polynomial Function
Find the maximum number of turning points of each polynomial function.
a. \(f(x)=−x^3+4x^5−3x^2+1\)
First, rewrite the polynomial function in descending order: \(f(x)=4x^5−x^3−3x^2+1\)
Identify the degree of the polynomial function. This polynomial function is of degree 5.
The maximum number of turning points is \(5−1=4\).
b. \(f(x)=−(x−1)^2(1+2x^2)\)
First, identify the leading term of the polynomial function if the function were expanded: multiply the leading terms in each factor together.
\( \begin{array}{rl} f(x) & =−(x−1)^2(1+2x^2)\\ &= {\color{Cerulean}{-1}} ( {\color{Cerulean}{x}}-1)^{ {\color{Cerulean}{2}} }(1+{\color{Cerulean}{2x^2}})\\ \text{High order term} &= {\color{Cerulean}{-1}} ( {\color{Cerulean}{x}})^{ {\color{Cerulean}{2}} }({\color{Cerulean}{2x^2}})\\ &= -2x^4\\ \end{array} \)
Then, identify the degree of the polynomial function. This polynomial function is of degree 4.
The maximum number of turning points is \(4−1=3\).
Example \(\PageIndex{10}\): Find the Maximum Number of Intercepts and Turning Points of a Polynomial
Without graphing the function, determine the maximum number of \(x\)-intercepts and turning points for \(f(x)=−3x^{10}+4x^7−x^4+2x^3\).
The polynomial has a degree of \(n\)= 10 , so there are at most 10 \(x\)-intercepts and at most 9 turning points.
Without graphing the function, determine the maximum number of \(x\)-intercepts and turning points for \(f(x)=108−13x^9−8x^4+14x^{12}+2x^3\)
There are at most 12 \(x\)-intercepts and at most 11 turning points.
Example \(\PageIndex{12}\): Drawing Conclusions about a Polynomial Function from the Factors
Given the function \(f(x)=−4x(x+3)(x−4)\), determine the \(y\)-intercept and the number, location and multiplicity of \(x\)-intercepts, and the maximum number of turning points.
The \(y\)-intercept is found by evaluating \(f(0)\).
\[\begin{align*} f(0)&=−4(0)(0+3)(0−4)=0 \end{align*}\]
The \(y\)-intercept is \((0,0)\).
The \(x\)-intercepts are found by determining the zeros of the function.
\( \begin{array}{ccc} &0=-4x(x+3)(x-4) \\ x=0 & \text{or} \quad x+3=0 \quad\text{or} & x-4=0 \\ x=0 & \text{or} \quad x=−3 \quad\text{or} & x=4 \end{array} \)
The three \(x\)-intercepts \((0,0)\),\((–3,0)\), and \((4,0)\) all have odd multiplicity of 1.
The degree is 3 so the graph has at most 2 turning points.
Given the function \(f(x)=0.2(x−2)(x+1)(x−5)\), determine the local behavior.
The function is a 3rd degree polynomial with three \(x\)-intercepts \((2,0)\), \((−1,0)\), and \((5,0)\) all have multiplicity of 1, the \(y\)-intercept is \((0,2)\), and the graph has at most 2 turning points.
Example \(\PageIndex{14}\): Drawing Conclusions about a Polynomial Function from the Graph
What can we conclude about the degree of the polynomial and the leading coefficient represented by the graph shown below based on its intercepts and turning points?
|
Conclusion: the degree of the polynomial is even and at least 4. |
What can we conclude about the polynomial represented by the graph shown below based on its intercepts and turning points?
A polynomial function
The end behavior indicates an odd-degree polynomial function (ends in opposite direction), with a negative leading coefficient (falls right). There are 3 \(x\)-intercepts each with odd multiplicity, and 2 turning points, so the degree is odd and at least 3.
Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an \(x\)-intercept where each factor is equal to zero, we can form a function that will pass through a set of \(x\)-intercepts by introducing a corresponding set of factors.
Note: Factored Form of Polynomials
If a polynomial of lowest degree \(p\) has horizontal intercepts at \(x=x_1,x_2,…,x_n\), then the polynomial can be written in the factored form: \(f(x)=a(x−x_1)^{p_1}(x−x_2)^{p_2}⋯(x−x_n)^{p_n}\) where the powers \(p_i\) on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor \(a\) can be determined given a value of the function other than the \(x\)-intercept.
Example \(\PageIndex{16}\): Writing a Formula for a Polynomial Function from the Graph
Construct the factored form of a possible equation for each graph given below.
(a) |
Looking at the graph of this function, as shown in Figure \(\PageIndex{16}\), it appears that there are \(x\)-intercepts at \(x=−3,−2, \text{ and }1\). Each \(x\)-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form. \( h(x)=a(x+3)(x+2)(x−1) \) The stretch factor \(a\) can be found by using another point on the graph, like the \(y\)-intercept, \((0,-6)\). \[\begin{align*} f(0)&=a(0+3)(0+2)(0−1) \\ −6&=a(-6) \\ a&=1 \end{align*}\] Thus \( h(x)=(x+3)(x+2)(x−1). \) |
(b) |
This graph has three \(x\)-intercepts: \(x=−3,\;2,\text{ and }5\). The \(y\)-intercept is located at \((0,2).\) At \(x=−3\) and \( x=5\), the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At \(x=2\), the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us \[f(x)=a(x+3)(x−2)^2(x−5)\] To determine the stretch factor, we utilize another point on the graph. We will use the \(y\)-intercept \((0,–2)\), to solve for \(a\). \[\begin{align*} f(0)&=a(0+3)(0−2)^2(0−5) \\ −2&=a(0+3)(0−2)^2(0−5) \\ −2&=−60a \\ a&=\dfrac{1}{30} \end{align*}\] The graphed polynomial appears to represent the function \(f(x)=\dfrac{1}{30}(x+3)(x−2)^2(x−5)\). |
Given the graph shown in Figure \(\PageIndex{21}\), write a formula for the function shown.
\(f(x)=−\frac{1}{8}(x−2)^3(x+1)^2(x−4)\)
Write a formula for a polynomial of degree 5, with zeros of multiplicity 2 at \(x\) = 3 and \(x\) = 1, a zero of multiplicity 1 at \(x\) = -3, and vertical intercept at (0, 9)
\(f(x) = \dfrac{1}{3} (x - 1)^2 (x - 3)^2 (x + 3)\)
multiplicity the number of times a given factor appears in the factored form of the equation of a polynomial; if a polynomial contains a factor of the form \((x−h)^p\), \(x=h\) is a zero of multiplicity \(p\).
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus .
Learning objectives.
In this section, you will:
The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 1 .
2006 | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 | 2013 | |
52.4 | 52.8 | 51.2 | 49.5 | 48.6 | 48.6 | 48.7 | 47.1 |
The revenue can be modeled by the polynomial function
where R R represents the revenue in millions of dollars and t t represents the year, with t = 6 t = 6 corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general.
Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.
Which of the graphs in Figure 2 represents a polynomial function?
The graphs of f f and h h are graphs of polynomial functions. They are smooth and continuous .
The graphs of g g and k k are graphs of functions that are not polynomials. The graph of function g g has a sharp corner. The graph of function k k is not continuous.
Do all polynomial functions have as their domain all real numbers?
Yes. Any real number is a valid input for a polynomial function.
Recall that if f f is a polynomial function, the values of x x for which f ( x ) = 0 f ( x ) = 0 are called zeros of f . f . If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros .
We can use this method to find x - x - intercepts because at the x - x - intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases:
Given a polynomial function f , f , find the x -intercepts by factoring.
Find the x -intercepts of f ( x ) = x 6 − 3 x 4 + 2 x 2 . f ( x ) = x 6 − 3 x 4 + 2 x 2 .
We can attempt to factor this polynomial to find solutions for f ( x ) = 0. f ( x ) = 0.
This gives us five x -intercepts: ( 0 , 0 ) , ( 1 , 0 ) , ( −1 , 0 ) , ( 2 , 0 ) , ( 0 , 0 ) , ( 1 , 0 ) , ( −1 , 0 ) , ( 2 , 0 ) , and ( − 2 , 0 ) . ( − 2 , 0 ) . See Figure 3 . We can see that this is an even function because it is symmetric about the y -axis.
Find the x -intercepts of f ( x ) = x 3 − 5 x 2 − x + 5. f ( x ) = x 3 − 5 x 2 − x + 5.
Find solutions for f ( x ) = 0 f ( x ) = 0 by factoring.
There are three x -intercepts: ( −1 , 0 ) , ( 1 , 0 ) , ( −1 , 0 ) , ( 1 , 0 ) , and ( 5 , 0 ) . ( 5 , 0 ) . See Figure 4 .
Find the y - and x -intercepts of g ( x ) = ( x − 2 ) 2 ( 2 x + 3 ) . g ( x ) = ( x − 2 ) 2 ( 2 x + 3 ) .
The y -intercept can be found by evaluating g ( 0 ) . g ( 0 ) .
So the y -intercept is ( 0 , 12 ) . ( 0 , 12 ) .
The x -intercepts can be found by solving g ( x ) = 0. g ( x ) = 0.
So the x -intercepts are ( 2 , 0 ) ( 2 , 0 ) and ( − 3 2 , 0 ) . ( − 3 2 , 0 ) .
We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5 .
Find the x -intercepts of h ( x ) = x 3 + 4 x 2 + x − 6. h ( x ) = x 3 + 4 x 2 + x − 6.
This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.
Looking at the graph of this function, as shown in Figure 6 , it appears that there are x -intercepts at x = −3 , −2 , x = −3 , −2 , and 1. 1.
We can check whether these are correct by substituting these values for x x and verifying that
Since h ( x ) = x 3 + 4 x 2 + x − 6 , h ( x ) = x 3 + 4 x 2 + x − 6 , we have:
Each x -intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.
Find the y - and x -intercepts of the function f ( x ) = x 4 − 19 x 2 + 30 x . f ( x ) = x 4 − 19 x 2 + 30 x .
Graphs behave differently at various x -intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and "bounce" off.
Suppose, for example, we graph the function shown.
Notice in Figure 7 that the behavior of the function at each of the x -intercepts is different.
The x -intercept x = −3 x = −3 is the solution of equation ( x + 3 ) = 0. ( x + 3 ) = 0. The graph passes directly through the x -intercept at x = −3. x = −3. The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.
The x -intercept x = 2 x = 2 is the repeated solution of equation ( x − 2 ) 2 = 0. ( x − 2 ) 2 = 0. The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept.
The factor is repeated, that is, the factor ( x − 2 ) ( x − 2 ) appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity . The zero associated with this factor, x = 2 , x = 2 , has multiplicity 2 because the factor ( x − 2 ) ( x − 2 ) occurs twice.
The x -intercept x = − 1 x = − 1 is the repeated solution of factor ( x + 1 ) 3 = 0. ( x + 1 ) 3 = 0. The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit function f ( x ) = x 3 . f ( x ) = x 3 . We call this a triple zero, or a zero with multiplicity 3.
For zeros with even multiplicities, the graphs touch or are tangent to the x -axis. For zeros with odd multiplicities, the graphs cross or intersect the x -axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3.
For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x -axis.
For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x -axis.
If a polynomial contains a factor of the form ( x − h ) p , ( x − h ) p , the behavior near the x - x - intercept h h is determined by the power p . p . We say that x = h x = h is a zero of multiplicity p . p .
The graph of a polynomial function will touch the x -axis at zeros with even multiplicities. The graph will cross the x -axis at zeros with odd multiplicities.
The sum of the multiplicities is the degree of the polynomial function.
Given a graph of a polynomial function of degree n , n , identify the zeros and their multiplicities.
Use the graph of the function of degree 6 in Figure 9 to identify the zeros of the function and their possible multiplicities.
The polynomial function is of degree 6. The sum of the multiplicities must be 6.
Starting from the left, the first zero occurs at x = −3. x = −3. The graph touches the x -axis, so the multiplicity of the zero must be even. The zero of −3 −3 most likely has multiplicity 2. 2.
The next zero occurs at x = −1. x = −1. The graph looks almost linear at this point. This is a single zero of multiplicity 1.
The last zero occurs at x = 4. x = 4. The graph crosses the x -axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is 6.
Use the graph of the function of degree 9 in Figure 10 to identify the zeros of the function and their multiplicities.
As we have already learned, the behavior of a graph of a polynomial function of the form
will either ultimately rise or fall as x x increases without bound and will either rise or fall as x x decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say –100 or –1,000.
Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, a n x n , a n x n , is an even power function, as x x increases or decreases without bound, f ( x ) f ( x ) increases without bound. When the leading term is an odd power function, as x x decreases without bound, f ( x ) f ( x ) also decreases without bound; as x x increases without bound, f ( x ) f ( x ) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure 11 summarizes all four cases.
In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function f ( x ) = x 4 − x 3 − 4 x 2 + 4 x f ( x ) = x 4 − x 3 − 4 x 2 + 4 x in Figure 12 . The graph has three turning points.
This function f f is a 4 th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function.
A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).
A polynomial of degree n n will have at most n − 1 n − 1 turning points.
Find the maximum number of turning points of each polynomial function.
First, rewrite the polynomial function in descending order: f ( x ) = 4 x 5 − x 3 − 3 x 2 + 1 f ( x ) = 4 x 5 − x 3 − 3 x 2 + 1
Identify the degree of the polynomial function. This polynomial function is of degree 5.
The maximum number of turning points is 5 − 1 = 4. 5 − 1 = 4.
First, identify the leading term of the polynomial function if the function were expanded.
Then, identify the degree of the polynomial function. This polynomial function is of degree 4.
The maximum number of turning points is 4 − 1 = 3. 4 − 1 = 3.
We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.
Given a polynomial function, sketch the graph.
Sketch a graph of f ( x ) = −2 ( x + 3 ) 2 ( x − 5 ) . f ( x ) = −2 ( x + 3 ) 2 ( x − 5 ) .
This graph has two x -intercepts. At x = −3 , x = −3 , the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x -intercept. At x = 5 , x = 5 , the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.
The y -intercept is found by evaluating f ( 0 ) . f ( 0 ) .
The y -intercept is ( 0 , 90 ) . ( 0 , 90 ) .
Additionally, we can see the leading term, if this polynomial were multiplied out, would be − 2 x 3 , − 2 x 3 , so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure 13 .
To sketch this, we consider that:
At ( 0 , 90 ) , ( 0 , 90 ) , the graph crosses the y -axis at the y -intercept. See Figure 14 .
Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at ( 5 , 0 ) . ( 5 , 0 ) . See Figure 15 .
As x → ∞ x → ∞ the function f ( x ) → −∞ , f ( x ) → −∞ , so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.
Using technology, we can create the graph for the polynomial function, shown in Figure 16 , and verify that the resulting graph looks like our sketch in Figure 15 .
Sketch a graph of f ( x ) = 1 4 x ( x − 1 ) 4 ( x + 3 ) 3 . f ( x ) = 1 4 x ( x − 1 ) 4 ( x + 3 ) 3 .
In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x -axis, we can confirm that there is a zero between them. Consider a polynomial function f f whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers a a and b b in the domain of f , f , if a < b a < b and f ( a ) ≠ f ( b ) , f ( a ) ≠ f ( b ) , then the function f f takes on every value between f ( a ) f ( a ) and f ( b ) . f ( b ) . (While the theorem is intuitive, the proof is actually quite complicated and requires higher mathematics.) We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function f f at x = a x = a lies above the x - x - axis and another point at x = b x = b lies below the x - x - axis, there must exist a third point between x = a x = a and x = b x = b where the graph crosses the x - x - axis. Call this point ( c , f ( c ) ) . ( c , f ( c ) ) . This means that we are assured there is a solution c c where f ( c ) = 0. f ( c ) = 0.
In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x - x - axis. Figure 17 shows that there is a zero between a a and b . b .
Let f f be a polynomial function. The Intermediate Value Theorem states that if f ( a ) f ( a ) and f ( b ) f ( b ) have opposite signs, then there exists at least one value c c between a a and b b for which f ( c ) = 0. f ( c ) = 0.
Show that the function f ( x ) = x 3 − 5 x 2 + 3 x + 6 f ( x ) = x 3 − 5 x 2 + 3 x + 6 has at least two real zeros between x = 1 x = 1 and x = 4. x = 4.
As a start, evaluate f ( x ) f ( x ) at the integer values x = 1 , 2 , 3 , x = 1 , 2 , 3 , and 4. 4. See Table 2 .
1 | 2 | 3 | 4 | |
5 | 0 | –3 | 2 |
We see that one zero occurs at x = 2. x = 2. Also, since f ( 3 ) f ( 3 ) is negative and f ( 4 ) f ( 4 ) is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.
We have shown that there are at least two real zeros between x = 1 x = 1 and x = 4. x = 4.
We can also see on the graph of the function in Figure 18 that there are two real zeros between x = 1 x = 1 and x = 4. x = 4.
Show that the function f ( x ) = 7 x 5 − 9 x 4 − x 2 f ( x ) = 7 x 5 − 9 x 4 − x 2 has at least one real zero between x = 1 x = 1 and x = 2. x = 2.
Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x -intercept where each factor is equal to zero, we can form a function that will pass through a set of x -intercepts by introducing a corresponding set of factors.
If a polynomial of lowest degree p p has horizontal intercepts at x = x 1 , x 2 , … , x n , x = x 1 , x 2 , … , x n , then the polynomial can be written in the factored form: f ( x ) = a ( x − x 1 ) p 1 ( x − x 2 ) p 2 ⋯ ( x − x n ) p n f ( x ) = a ( x − x 1 ) p 1 ( x − x 2 ) p 2 ⋯ ( x − x n ) p n where the powers p i p i on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor a a can be determined given a value of the function other than the x -intercept.
Given a graph of a polynomial function, write a formula for the function.
Write a formula for the polynomial function shown in Figure 19 .
This graph has three x -intercepts: x = −3 , 2 , x = −3 , 2 , and 5. 5. The y -intercept is located at ( 0 , 2 ) . ( 0 , 2 ) . At x = −3 x = −3 and x = 5 , x = 5 , the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At x = 2 , x = 2 , the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us
To determine the stretch factor, we utilize another point on the graph. We will use the y - y - intercept ( 0 , – 2 ) , ( 0 , – 2 ) , to solve for a . a .
The graphed polynomial appears to represent the function f ( x ) = 1 30 ( x + 3 ) ( x − 2 ) 2 ( x − 5 ) . f ( x ) = 1 30 ( x + 3 ) ( x − 2 ) 2 ( x − 5 ) .
Given the graph shown in Figure 20 , write a formula for the function shown.
With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.
Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum . These are also referred to as the absolute maximum and absolute minimum values of the function.
A local maximum or local minimum at x = a x = a (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around x = a . x = a . If a function has a local maximum at a , a , then f ( a ) ≥ f ( x ) f ( a ) ≥ f ( x ) for all x x in an open interval around x = a . x = a . If a function has a local minimum at a , a , then f ( a ) ≤ f ( x ) f ( a ) ≤ f ( x ) for all x x in an open interval around x = a . x = a .
A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at a , a , then f ( a ) ≥ f ( x ) f ( a ) ≥ f ( x ) for all x . x . If a function has a global minimum at a , a , then f ( a ) ≤ f ( x ) f ( a ) ≤ f ( x ) for all x . x .
We can see the difference between local and global extrema in Figure 21 .
Do all polynomial functions have a global minimum or maximum?
No. Only polynomial functions of even degree have a global minimum or maximum. For example, f ( x ) = x f ( x ) = x has neither a global maximum nor a global minimum.
An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic and then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box.
We will start this problem by drawing a picture like that in Figure 22 , labeling the width of the cut-out squares with a variable, w . w .
Notice that after a square is cut out from each end, it leaves a ( 14 − 2 w ) ( 14 − 2 w ) cm by ( 20 − 2 w ) ( 20 − 2 w ) cm rectangle for the base of the box, and the box will be w w cm tall. This gives the volume
Notice, since the factors are w , w , 20 – 2 w 20 – 2 w and 14 – 2 w , 14 – 2 w , the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values w w may take on are greater than zero or less than 7. This means we will restrict the domain of this function to 0 < w < 7. 0 < w < 7. Using technology to sketch the graph of V ( w ) V ( w ) on this reasonable domain, we get a graph like that in Figure 23 . We can use this graph to estimate the maximum value for the volume, restricted to values for w w that are reasonable for this problem—values from 0 to 7.
From this graph, we turn our focus to only the portion on the reasonable domain, [ 0 , 7 ] . [ 0 , 7 ] . We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure 24 .
From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side.
Use technology to find the maximum and minimum values on the interval [ −1 , 4 ] [ −1 , 4 ] of the function f ( x ) = − 0.2 ( x − 2 ) 3 ( x + 1 ) 2 ( x − 4 ) . f ( x ) = − 0.2 ( x − 2 ) 3 ( x + 1 ) 2 ( x − 4 ) .
Access the following online resource for additional instruction and practice with graphing polynomial functions.
What is the difference between an x - x - intercept and a zero of a polynomial function f ? f ?
If a polynomial function of degree n n has n n distinct zeros, what do you know about the graph of the function?
Explain how the Intermediate Value Theorem can assist us in finding a zero of a function.
Explain how the factored form of the polynomial helps us in graphing it.
If the graph of a polynomial just touches the x -axis and then changes direction, what can we conclude about the factored form of the polynomial?
For the following exercises, find the x - x - or t -intercepts of the polynomial functions.
C ( t ) = 2 ( t − 4 ) ( t + 1 ) ( t − 6 ) C ( t ) = 2 ( t − 4 ) ( t + 1 ) ( t − 6 )
C ( t ) = 3 ( t + 2 ) ( t − 3 ) ( t + 5 ) C ( t ) = 3 ( t + 2 ) ( t − 3 ) ( t + 5 )
C ( t ) = 4 t ( t − 2 ) 2 ( t + 1 ) C ( t ) = 4 t ( t − 2 ) 2 ( t + 1 )
C ( t ) = 2 t ( t − 3 ) ( t + 1 ) 2 C ( t ) = 2 t ( t − 3 ) ( t + 1 ) 2
C ( t ) = 2 t 4 − 8 t 3 + 6 t 2 C ( t ) = 2 t 4 − 8 t 3 + 6 t 2
C ( t ) = 4 t 4 + 12 t 3 − 40 t 2 C ( t ) = 4 t 4 + 12 t 3 − 40 t 2
f ( x ) = x 4 − x 2 f ( x ) = x 4 − x 2
f ( x ) = x 3 + x 2 − 20 x f ( x ) = x 3 + x 2 − 20 x
f ( x ) = x 3 + 6 x 2 − 7 x f ( x ) = x 3 + 6 x 2 − 7 x
f ( x ) = x 3 + x 2 − 4 x − 4 f ( x ) = x 3 + x 2 − 4 x − 4
f ( x ) = x 3 + 2 x 2 − 9 x − 18 f ( x ) = x 3 + 2 x 2 − 9 x − 18
f ( x ) = 2 x 3 − x 2 − 8 x + 4 f ( x ) = 2 x 3 − x 2 − 8 x + 4
f ( x ) = x 6 − 7 x 3 − 8 f ( x ) = x 6 − 7 x 3 − 8
f ( x ) = 2 x 4 + 6 x 2 − 8 f ( x ) = 2 x 4 + 6 x 2 − 8
f ( x ) = x 3 − 3 x 2 − x + 3 f ( x ) = x 3 − 3 x 2 − x + 3
f ( x ) = x 6 − 2 x 4 − 3 x 2 f ( x ) = x 6 − 2 x 4 − 3 x 2
f ( x ) = x 6 − 3 x 4 − 4 x 2 f ( x ) = x 6 − 3 x 4 − 4 x 2
f ( x ) = x 5 − 5 x 3 + 4 x f ( x ) = x 5 − 5 x 3 + 4 x
For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval.
f ( x ) = x 3 − 9 x , f ( x ) = x 3 − 9 x , between x = −4 x = −4 and x = −2. x = −2.
f ( x ) = x 3 − 9 x , f ( x ) = x 3 − 9 x , between x = 2 x = 2 and x = 4. x = 4.
f ( x ) = x 5 − 2 x , f ( x ) = x 5 − 2 x , between x = 1 x = 1 and x = 2. x = 2.
f ( x ) = − x 4 + 4 , f ( x ) = − x 4 + 4 , between x = 1 x = 1 and x = 3 x = 3 .
f ( x ) = −2 x 3 − x , f ( x ) = −2 x 3 − x , between x = –1 x = –1 and x = 1. x = 1.
f ( x ) = x 3 − 100 x + 2 , f ( x ) = x 3 − 100 x + 2 , between x = 0.01 x = 0.01 and x = 0.1 x = 0.1
For the following exercises, find the zeros and give the multiplicity of each.
f ( x ) = ( x + 2 ) 3 ( x − 3 ) 2 f ( x ) = ( x + 2 ) 3 ( x − 3 ) 2
f ( x ) = x 2 ( 2 x + 3 ) 5 ( x − 4 ) 2 f ( x ) = x 2 ( 2 x + 3 ) 5 ( x − 4 ) 2
f ( x ) = x 3 ( x − 1 ) 3 ( x + 2 ) f ( x ) = x 3 ( x − 1 ) 3 ( x + 2 )
f ( x ) = x 2 ( x 2 + 4 x + 4 ) f ( x ) = x 2 ( x 2 + 4 x + 4 )
f ( x ) = ( 2 x + 1 ) 3 ( 9 x 2 − 6 x + 1 ) f ( x ) = ( 2 x + 1 ) 3 ( 9 x 2 − 6 x + 1 )
f ( x ) = ( 3 x + 2 ) 5 ( x 2 − 10 x + 25 ) f ( x ) = ( 3 x + 2 ) 5 ( x 2 − 10 x + 25 )
f ( x ) = x ( 4 x 2 − 12 x + 9 ) ( x 2 + 8 x + 16 ) f ( x ) = x ( 4 x 2 − 12 x + 9 ) ( x 2 + 8 x + 16 )
f ( x ) = x 6 − x 5 − 2 x 4 f ( x ) = x 6 − x 5 − 2 x 4
f ( x ) = 3 x 4 + 6 x 3 + 3 x 2 f ( x ) = 3 x 4 + 6 x 3 + 3 x 2
f ( x ) = 4 x 5 − 12 x 4 + 9 x 3 f ( x ) = 4 x 5 − 12 x 4 + 9 x 3
f ( x ) = 2 x 4 ( x 3 − 4 x 2 + 4 x ) f ( x ) = 2 x 4 ( x 3 − 4 x 2 + 4 x )
f ( x ) = 4 x 4 ( 9 x 4 − 12 x 3 + 4 x 2 ) f ( x ) = 4 x 4 ( 9 x 4 − 12 x 3 + 4 x 2 )
For the following exercises, graph the polynomial functions. Note x - x - and y - y - intercepts, multiplicity, and end behavior.
f ( x ) = ( x + 3 ) 2 ( x − 2 ) f ( x ) = ( x + 3 ) 2 ( x − 2 )
g ( x ) = ( x + 4 ) ( x − 1 ) 2 g ( x ) = ( x + 4 ) ( x − 1 ) 2
h ( x ) = ( x − 1 ) 3 ( x + 3 ) 2 h ( x ) = ( x − 1 ) 3 ( x + 3 ) 2
k ( x ) = ( x − 3 ) 3 ( x − 2 ) 2 k ( x ) = ( x − 3 ) 3 ( x − 2 ) 2
m ( x ) = − 2 x ( x − 1 ) ( x + 3 ) m ( x ) = − 2 x ( x − 1 ) ( x + 3 )
n ( x ) = − 3 x ( x + 2 ) ( x − 4 ) n ( x ) = − 3 x ( x + 2 ) ( x − 4 )
For the following exercises, use the graphs to write the formula for a polynomial function of least degree.
For the following exercises, use the graph to identify zeros and multiplicity.
For the following exercises, use the given information about the polynomial graph to write the equation.
Degree 3. Zeros at x = –2, x = –2, x = 1, x = 1, and x = 3. x = 3. y -intercept at ( 0 , –4 ) . ( 0 , –4 ) .
Degree 3. Zeros at x = –5, x = –5, x = –2 , x = –2 , and x = 1. x = 1. y -intercept at ( 0 , 6 ) ( 0 , 6 )
Degree 5. Roots of multiplicity 2 at x = 3 x = 3 and x = 1 x = 1 , and a root of multiplicity 1 at x = –3. x = –3. y -intercept at ( 0 , 9 ) ( 0 , 9 )
Degree 4. Root of multiplicity 2 at x = 4, x = 4, and a roots of multiplicity 1 at x = 1 x = 1 and x = –2. x = –2. y -intercept at ( 0 , – 3 ) . ( 0 , – 3 ) .
Degree 5. Double zero at x = 1 , x = 1 , and triple zero at x = 3. x = 3. Passes through the point ( 2 , 15 ) . ( 2 , 15 ) .
Degree 3. Zeros at x = 4 , x = 4 , x = 3 , x = 3 , and x = 2. x = 2. y -intercept at ( 0 , −24 ) . ( 0 , −24 ) .
Degree 3. Zeros at x = −3 , x = −3 , x = −2 x = −2 and x = 1. x = 1. y -intercept at ( 0 , 12 ) . ( 0 , 12 ) .
Degree 5. Roots of multiplicity 2 at x = −3 x = −3 and x = 2 x = 2 and a root of multiplicity 1 at x = −2. x = −2.
y -intercept at ( 0 , 4 ) . ( 0 , 4 ) .
Degree 4. Roots of multiplicity 2 at x = 1 2 x = 1 2 and roots of multiplicity 1 at x = 6 x = 6 and x = −2. x = −2.
y -intercept at ( 0, 18 ) . ( 0, 18 ) .
Double zero at x = −3 x = −3 and triple zero at x = 0. x = 0. Passes through the point ( 1 , 32 ) . ( 1 , 32 ) .
For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum.
f ( x ) = x 3 − x − 1 f ( x ) = x 3 − x − 1
f ( x ) = 2 x 3 − 3 x − 1 f ( x ) = 2 x 3 − 3 x − 1
f ( x ) = x 4 + x f ( x ) = x 4 + x
f ( x ) = − x 4 + 3 x − 2 f ( x ) = − x 4 + 3 x − 2
f ( x ) = x 4 − x 3 + 1 f ( x ) = x 4 − x 3 + 1
For the following exercises, use the graphs to write a polynomial function of least degree.
For the following exercises, write the polynomial function that models the given situation.
A rectangle has a length of 10 units and a width of 8 units. Squares of x x by x x units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a polynomial function in terms of x . x .
Consider the same rectangle of the preceding problem. Squares of 2 x 2 x by 2 x 2 x units are cut out of each corner. Express the volume of the box as a polynomial in terms of x . x .
A square has sides of 12 units. Squares x + 1 x + 1 by x + 1 x + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms of x . x .
A cylinder has a radius of x + 2 x + 2 units and a height of 3 units greater. Express the volume of the cylinder as a polynomial function.
A right circular cone has a radius of 3 x + 6 3 x + 6 and a height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone is V = 1 3 π r 2 h V = 1 3 π r 2 h for radius r r and height h . h .
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This Polynomial and Rational Functions Unit Bundle includes guided notes, homework assignments, four quizzes, a study guide and a unit test that cover the following topics:
• Graphing Power Functions
• Graphing Power Functions with Negative Exponents
• Graphing Power Functions with Rational Exponents
• Graphing Polynomial Functions
• End Behavior and Turning Points of Polynomial Functions
• Zeros, Linear Factors, and Multiplicity
• Effects of Multiplicity on the Graph of a Polynomial Function
• Dividing Polynomials using Long Division
• Dividing Polynomials using Synthetic Division
• Factoring Polynomials using Division
• Remainder Theorem
• Factor Theorem
• Rational Zero Theorem
• Irrational Zeros
• Descartes' Rule of Signs
• Complex Zeros
• Fundamental Theorem of Algebra
• Using Zeros to Write Polynomial Functions
• Graphing Rational Functions
• Vertical, Horizontal, and Slant (Oblique) Asymptotes
• Solving Nonlinear Inequalities (Polynomial and Rational)
ADDITIONAL COMPONENTS INCLUDED:
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Unit 6 – Trigonometric Identities and Equations
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Unit 8 – Vectors
Unit 9 – Conic Sections Unit 10 – Systems of Equations and Matrices Unit 11 – Sequences and Series
Unit 12 – Introduction to Calculus
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© All Things Algebra (Gina Wilson), 2012-present
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IMAGES
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Video discussing on how to find the zeros of polynomial functions and what the multiplicity tells you.
6. f(x) = -8P-2012 Zero Multiplicity Effect Zero Multiplicity Effect . 7. A polynomial function has a zeros at -1, 2, and 7 (all multiplicity 1). Write a function in standard form that could represent this function. [/A . 8. A polynomial function has a zeros at 5/2 (multiplicity 2), 3 (multiplicity 1), and 0 (multiplicity 4). Write a function ...
Final answers should positive only. 1. Subtract -3n2 from -7n2 C-1nÒ-G3nZ) two monomials with a product of 3. (—4a3b2)2 • (3a2b) 15? y -6x7y (3xy)2 8. The side length of a cu can be repræented by the expression 215. If the side length is doubled, write an expression to represent the new volume of the cube. = in sbndard form if need4 then ...
Final answer: In polynomial functions, zeros are the values which make the function equals to zero. Their multiplicity, depicted by the exponent, influences the graph. An example is given for question 1 where , its zeros are x=0, x=1 and x=-5 with a multiplicity of 1 each, and for question 7, the function that represents zeros at -1, 2, 7 is ...
3a. Dividing using Long division - fraction answers 3b. Example with synthetic division 3.3 4a. Rational Roots Test 4b. Find integer roots of polynomial 4c. Find rational roots of polynomial (synthetic division then factor) 4d. Descartes Rule of Signs 4e. Find polynomial given fractional roots 4a. Find rational zeros using TI-84 calculator 4b.
The zero of −3 has multiplicity 2. The next zero occurs at \(x=−1\). The graph looks almost linear at this point. This is a single zero of multiplicity 1. The last zero occurs at \(x=4\). The graph crosses the \(x\)-axis, so the multiplicity of the zero must be odd. Since the graph is flat around this zero, the multiplicity is likely 3 ...
Unit 3 Polynomials ANSWER KEY Introduction to Polynomials; Polynomial Graphs and Key Features Polynomial Vocabulary Review ... (x-3)factored equation: (x+2)(x+1)(x-2)(x-3)(x-4) WATCH OUT! Multiplicities of Zeros If c is a zero of the function P and the corresponding factor (x − c) occurs exactly m times in the factorization of P then we say ...
Graphing Polynomials; End Behavior; Zeros & Multiplicity . Day 4. Synthetic Substitution; Fundamental Theorem of Algebra. Day 5 . Applying the Remainder and Factor Theorem . Day 6 . Finding All Rational Zeros . Day 7 . ... Handouts and homework keys are posted under assignments. You may also email Ms. Huelsman at [email protected] with ...
The zero of −3 −3 most likely has multiplicity 2. 2. The next zero occurs at x = −1. x = −1. The graph looks almost linear at this point. This is a single zero of multiplicity 1. The last zero occurs at x = 4. x = 4. The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and ...
Verify that the zeros -3, 4, and 6 are the only solutions to the equation y = 0 for each option Step 3 Identify any option that does not have the zeros -3, 4, and 6 as the only solutions, which would mean it cannot be the correct equation for the polynomial function
Step 1. Provided monomials are − 3 n 2 and 7 n 2 . Provided statement is, subtract − 3 n 2 from 7 n 2 . Name: Treonna Brown Unit 5: Polynomial Functions Date: 1018/20 Bell: Homework 1: Monomials & Polynomials Directions: Simplify the monomials below. Final answers should contain positive exponents only.
Learning Objectives: SWBAT. Determine the real zeros of a polynomial algebraically and graphically • Determine the multiplicity of a zero of a polynomial. Describe how the zero's multiplicity affects the graph of the zero. Lesson 1.9 - Zeros and Multiplicity. Example 2 - Find all real zeros of the following polynomial algebraically (FACTOR IT!)
This curriculum includes 975+ pages of instructional materials (warm-ups, notes, homework, quizzes, unit tests, review materials, a midterm exam, a final exam, and many other extras) for Algebra 2. ... • Zeros of a Polynomial Function, Multiplicity, Effect of Multiplicity on a Graph ... Answer Key. Included. Teaching Duration. N/A.
b) 5 real zeros at approximately -2.6 (multiplicity 2), -1, and 2.6 (multiplicity 2) c) 0 imaginary zeros; a) Leading coefficient: 1; Degree: 4 b) 4 real zeros at approximately -1 and 3 (multiplicity 3) c) 0 imaginary zeros [-2, -1] [-1, 3] [-2, 4] [-2, 2] The Bounded Roots Theorem is based on continuous functions, and this rational function is ...
Interactive worksheet: Homework 3: Zeros and Multiplicity. Algebra online exercise. EN . English Español Catal à Français Deutsch ... Complete the data to send the answers to your teacher . Finish . Share on social networks ×. Twitter. Facebook. Google Classroom. Pinterest. Whatsapp ...
Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. ... calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Mathway. Visit Mathway on the web. Start 7-day free trial on the app ...
Description. This Polynomial and Rational Functions Unit Bundle includes guided notes, homework assignments, four quizzes, a study guide and a unit test that cover the following topics: • Graphing Power Functions. • Graphing Power Functions with Negative Exponents. • Graphing Power Functions with Rational Exponents.
Question: Given the graph of the following degree 3 polynomial function, find all of the zeros and their multiplicities. Given the graph of the following degree 3 polynomial function, find all of the zeros and their multiplicities. There are 2 steps to solve this one.
Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: The graph of a polynomial function has zeros of 3 (multiplicity 2), and -7 (multiplicity 1). What is a function, in standard form, that could represent this function? The graph of a polynomial function has zeros of 3 (multiplicity 2), and -7 ...
Algebra questions and answers; Find the zeros and state their multiplicity. Enter your answer in as a whole number or as a fraction.f(x)=3x(x-2)(x+3)(x2-1)with a multiplicity ofwith a multiplicity ofwith a multiplicity ofwith a multiplicity ofwith a multiplicity of
Refer our service to your friend and receive 10% from every order. Homework 3 Zeros And Multiplicity Answer Key. Level: Master's, University, College, PHD, High School, Undergraduate. Rating:
Find the zeros of the function and state the multiplicities.m = x 5 - 3x 3Select one: Your solution's ready to go! Enhanced with AI, our expert help has broken down your problem into an easy-to-learn solution you can count on.
Find the zeros and give the multiplicity of each. f(x)=3x^(2)(2x+5)^(3)(5x-7) Your solution's ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.