2.4: Graphs of Polynomial Functions
- Refresher Topics
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Factoring Techniques
- Factoring by Grouping
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Factoring Techniques
- Foundational Topics
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Geometry
- Common Volumes
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Geometry
In this section, you will:
- Recognize characteristics of graphs of polynomial functions.
- Use factoring to find zeros of polynomial functions.
- Identify zeros and their multiplicities.
- Determine end behavior.
- Understand the relationship between degree and turning points.
- Graph polynomial functions.
- Use the Intermediate Value Theorem.
- Graph the absolute value of a polynomial function.
- Solve a polynomial inequality (using a graph).
The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table \( \PageIndex{ 1 } \).
| Year | 2006 | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 | 2013 |
| Revenues | 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\[R(t)=-0.037 t^4 +1.414 t^3 -19.777 t^2 +118.696t-205.332, \nonumber \]where \(R\) represents the revenue in millions of dollars and \(t\) represents the year, with \(t=6\) 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.
Recognizing Characteristics of Graphs of Polynomial Functions
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 (again, as mention in the previous section, this is not technically true; however, we will use this flawed definition of continuity until we reach Calculus). Figure \( \PageIndex{ 1 } \) shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.
Figure \( \PageIndex{ 1 } \)
Which of the graphs in Figure \( \PageIndex{ 2 } \) represents a polynomial function?
Figure \( \PageIndex{ 2 } \)
- Solution
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The graphs of \(f\) and \(h\) are graphs of polynomial functions. They are smooth and continuous.
The graphs of \(g\) and \(k\) are graphs of functions that are not polynomials. The graph of function \(g\) has a sharp corner. The graph of function \(k\) 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.
Using Factoring to Find Zeros of Polynomial Functions
Recall that if \(f\) is a polynomial function, the values of \(x\) for which \(f( x )=0\) are called zeros of \(f\). 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 \( x \)-intercepts because at the \( x \)-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 in this section:
- The polynomial can be factored using known methods: greatest common factor and trinomial factoring.
- The polynomial is given in factored form.
- Technology is used to determine the intercepts.
Find the \( x \)-intercepts of \(f(x)= x^6 -3 x^4 +2 x^2 \).
- Solution
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We can attempt to factor this polynomial to find solutions for \(f( x )=0\).\[ \begin{array}{rrclcl} & x^6 -3 x^4 +2 x^2 & = & 0 & & \\[6pt] \implies & x^2(x^4 -3 x^2 +2) & = & 0 & \quad & \left( \text{factoring out the GCF} \right) \\[6pt] \implies & x^2(u^2 -3 u +2) & = & 0 & \quad & \left( \text{substituting }u = x^2\text{ in the trinomial} \right) \\[6pt] \implies & x^2(u - 2)(u - 1) & = & 0 & \quad & \left( \text{factoring the trinomial} \right) \\[6pt] \implies & x^2(x^2 - 2)(x^2 - 1) & = & 0 & \quad & \left( \text{resubstituting }u = x^2 \right) \\[6pt] \implies & x^2(x^2 - 2)(x - 1)(x + 1) & = & 0 & \quad & \left( \text{factoring a Difference of Squares} \right) \\[6pt] \end{array} \nonumber \]Thus,\[ \begin{array}{rccccccc} & x^2 = 0 & \text{ or } & x^2 - 2 = 0 & \text{ or } & x - 1 = 0 & \text{ or } & x + 1 = 0 \\[6pt] \implies & x = 0 & & x^2 = 2 & & x = 1 & & x = -1 \\[6pt] \implies & x = 0 & & x = \pm \sqrt{2} & & x = 1 & & x = -1 \\[6pt] \end{array} \nonumber \]This gives us five \( x \)-intercepts: \((0,0)\), \( (1,0) \), \( (-1,0) \), \((\sqrt{ 2 } ,0)\), and \((-\sqrt{ 2 } ,0)\) (see Figure \( \PageIndex{ 3 } \)). We can see that this is an even function.
Figure \( \PageIndex{ 3 } \)
Find the \( x \)-intercepts of \(f(x)= x^3 -5 x^2 -x+5\).
- Solution
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Find solutions for \(f(x)=0\) by factoring.\[ \begin{array}{rrclcl} & x^3 - 5x^2 - x + 5 & = & 0 & & \\[6pt] \implies & x^2(x - 5) - 1 (x - 5) & = & 0 & \quad & \left( \text{factoring by grouping} \right) \\[6pt] \implies & (x - 5)(x^2 - 1) & = & 0 & \quad & \left( \text{factor out the GCF} \right) \\[6pt] \implies & (x - 5)(x - 1)(x + 1) & = & 0 & \quad & \left( \text{factoring the Difference of Squares} \right) \\[6pt] \end{array} \nonumber \]Hence, \( x = 5 \), or \( x = 1 \), or \( x = -1 \). There are three \( x \)-intercepts: \((-1,0)\), \((1,0)\), and \(( 5,0 )\) (see Figure \( \PageIndex{ 4 } \)).
Figure \( \PageIndex{ 4 } \)
Find the \( y \)- and \( x \)-intercepts of \(g(x)= (x-2)^2 (2x+3)\).
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The \( y \)-intercept can be found by evaluating \(g( 0 )\).\[g(0) = (0-2)^2 (2(0)+3) = 12. \nonumber \]So the \( y \)-intercept is \((0,12)\).
The \( x \)-intercepts can be found by solving \(g( x )=0\).\[ (x - 2)^2 (2x + 3) = 0 \implies x = 2 \text{ or } x = -\dfrac{3}{2}. \nonumber \]So the \( x \)-intercepts are \((2,0)\) and \(\left( -\frac{3}{2}, 0 \right)\).
We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial. For example, Figure \( \PageIndex{ 5 } \) shows that our answers for Example \( \PageIndex{ 4 } \) are correct.
Figure \( \PageIndex{ 5 } \)
Find the \( x \)-intercepts of \(h(x)= x^3 +4 x^2 +x-6\).
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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 \( \PageIndex{ 6 } \), it appears that there are \( x \)-intercepts at \(x=-3\), \(-2\), and \(1\).
Figure \( \PageIndex{ 6 } \)We can check whether these are correct by substituting these values for \(x\) and verifying that\[h(-3)=h(-2)=h(1)=0. \nonumber \]Since \(h(x)= x^3 +4 x^2 +x-6\), we have:\[ \begin{array}{rcl} h(-3) & = & (-3)^3 + 4(-3)^2 + (-3) - 6 = -27 + 36 - 3 - 6 = 0 \\[6pt] h(-2) & = & (-2)^3 + 4(-2)^2 + (-2) - 6 = -8 + 16 - 2 - 6 = 0 \\[6pt] h(1) & = & (1)^3 + 4(1)^2 + (1) - 6 = 1 + 4 + 1 - 6 = 0 \\[6pt] \end{array} \nonumber \]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.\[h(x)= x^3 +4 x^2 +x-6 =(x+3)(x+2)(x-1). \nonumber \]
Find the \( y \)- and \( x \)-intercepts of the function \(f(x)= x^4 -19 x^2 +30x\).
Identifying Zeros and Their Multiplicities
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\[f(x)=(x+3) (x-2)^2 (x+1)^3. \nonumber \]Notice in Figure \( \PageIndex{ 7 } \) that the behavior of the function at each of the \( x \)-intercepts is different.
Figure \( \PageIndex{ 7 } \): Identifying the behavior of the graph at an \( x \)-intercept by examining the multiplicity of the zero.
The \( x \)-intercept at \(x = -3\) is the solution of equation \((x+3)=0\). The graph passes directly through the \( x \)-intercept at \(x=-3\). 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 at \(x = 2\) is the repeated solution of equation \((x-2)^2 = 0\). 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.\[(x-2)^2 =(x-2) (x-2). \nonumber \]The factor is repeated, that is, the factor \(( x-2 )\) 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, \(x=2\), has multiplicity 2 because the factor \(( x-2 )\) occurs twice.
The \( x \)-intercept at \(x = -1\) is the repeated solution of factor \((x+1)^3 = 0\). 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 \(f( x )= x^3 \). 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 \( \PageIndex{ 8 } \) for examples of graphs of polynomial functions with multiplicity \( 1 \), \( 2 \), and \( 3 \).
Figure \( \PageIndex{ 8 } \)
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 \((x-h)^p \) (and if this is the greatest such factor of \( (x - p) \)), the behavior near the \( x \)-intercept \(h\) is determined by the power \(p\). We say that \(x=h\) is a zero of multiplicity \(p\) .
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 of the factors for the polynomial is the degree of the polynomial function.
It is important to make sure that the higher the multiplicity of the zero, the flatter the graph of the polynomial appears near the zero.
Use the graph of the function of degree 6 in Figure \( \PageIndex{ 9 } \) to identify the zeros of the function and their possible multiplicities.
Figure \( \PageIndex{ 9 } \)
- Solution
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The polynomial function is of degree \(n\). The sum of the multiplicities must be \(n\).
Starting from the left, the first zero occurs at \(x=-3\). The graph touches the \( x \)-axis, so the multiplicity of the zero must be even. The zero of \(-3\) could have multiplicity \(2\), \( 4 \), or \( 6 \) (remember, we were told this polynomial has degree 6, so the sum of the multiplicities of the zeros must be 6).
The next zero occurs at \(x=-1\). The graph looks almost linear at this point, so this is likely a 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. From the shape of the curve near to \( x = 4 \), we know the multiplicity cannot be \( 1 \) (the graph is not locally linear near \( x = 4 \)). Hence, \( x = 4 \) must be a zero of multiplicity either \( 3 \) or \( 5 \).
Now that we have an idea of the possible multiplicities, let's build a table and use a little deductive logic to identify the exact multiplicities.\[ \begin{array}{cc} \textbf{Zero} & \textbf{Possible Multiplicities} \\[6pt] \hline -3 & 2, 4, 6 \\[6pt] -1 & 1 \\[6pt] 4 & 3, 5 \\[6pt] \end{array} \nonumber \]Since the sum of the multiplicities must be 6, and we have three zeros, the maximum multiplicity for any given zero is 3. This cleans up our table quite a bit:\[ \begin{array}{cc} \textbf{Zero} & \textbf{Possible Multiplicities} \\[6pt] \hline -3 & 2 \\[6pt] -1 & 1 \\[6pt] 4 & 3 \\[6pt] \end{array} \nonumber \]Thus, we have identified the zeros are their exact multiplicities.
Use the graph of the function of degree 7 in Figure \( \PageIndex{ 10 } \) to identify the zeros of the function and their multiplicities.
Figure \( \PageIndex{ 10 } \)
Determining End Behavior
As we have already learned, the behavior of a graph of a polynomial function of the form\[f(x)= a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \nonumber \]will either ultimately rise or fall as \(x\) increases without bound and will either rise or fall as \(x\) 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 negative 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, \( a_n x^n \), is an even power function, as \(x\) increases or decreases without bound, \(f(x)\) increases without bound. When the leading term is an odd power function, as \(x\) decreases without bound, \(f(x)\) also decreases without bound; as \(x\) increases without bound, \(f(x)\) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure \( \PageIndex{ 11 } \) summarizes all four cases.
Figure \( \PageIndex{ 11 } \)
Understanding the Relationship Between Degree and Turning Points
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 \(f(x)= x^4 - x^3 -4 x^2 +4x\) in Figure \( \PageIndex{ 12 } \). The graph has three turning points.
Figure \( \PageIndex{ 12 } \)
This function is a \(4^{\text{th}}\) degree polynomial function and has 3 turning points. This should be familiar because, as mentioned in the previous section, the maximum number of turning points of a polynomial function is always one less than the degree of the function.
Find the maximum number of turning points of each polynomial function.
- \(f(x) = -x^3 + 4 x^5 - 3 x^2 +1\)
- \(f(x) = -(x-1)^2 (1 + 2x^2)\)
- Solutions
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- 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 \(5-1=4\).
- First, identify the leading term of the polynomial function if the function were expanded. Looking at the given function, \(f(x) = -(x-1)^2 (1 + 2x^2)\), we can see that the factor \( (x - 1)^2 \) will contribute \( x^2 \) (if we multiplied it out) and the factor \( 1 + 2x^2 \) contributes \( 2x^2 \). Therefore, the leading term is\[ a_n x^n = - (x^2) (2x^2) = -2 x^4. \nonumber \]Therefore, this polynomial function is of degree 4. As such, the maximum number of turning points is \(4-1=3\).
Graphing Polynomial Functions
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.
- Find the intercepts.
- Check for symmetry. If the function is an even function, its graph is symmetric about the \( y \)-axis. That is, \(f(-x)=f( x )\). If a function is an odd function, its graph is symmetric about the origin. That is, \(f( -x )=-f( x )\).
- Use the multiplicities of the zeros to determine the behavior of the polynomial at the \( x \)-intercepts.
- Determine the end behavior by examining the leading term.
- Use the end behavior and the behavior at the intercepts to sketch a graph.
- Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
- Optionally, use technology to check the graph.
Sketch a graph of \(f(x) = -2(x+3)^2 (x-5)\).
- Solution
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This graph has two \( x \)-intercepts. At \(x=-3\), the factor is squared, indicating a multiplicity of 2. The graph will bounce at this \( x \)-intercept. At \(x=5\), the function has a multiplicity of one, indicating the graph will cross through the axis like a line at this intercept.
The \( y \)-intercept is found by evaluating \(f(0)\).\[f(0) = -2(0+3)^2(0-5) = -2 \cdot 9 \cdot (-5) = 90. \nonumber \]The \( y \)- intercept is \((0,90)\).
Additionally, we can see the leading term, if this polynomial were multiplied out, would be \(-2 x^3 \), 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 \( \PageIndex{ 13 } \).
Figure \( \PageIndex{ 13 } \)To sketch this, we consider that:
- As \(x \to - \infty\), \(f(x) \to \infty \). So we know the graph starts in the second quadrant and is decreasing toward the \( x \)-axis.
- Since \(f( -x ) = -2 (-x+3)^2 (-x–5)\) is not equal to \(\pm f( x )\), the graph does not display symmetry.
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At \((-3,0)\), the graph bounces off of the \( x \)-axis, so the function must start increasing.
At \((0,90)\), the graph crosses the \( y \)-axis at the \( y \)-intercept. See Figure \( \PageIndex{ 14 } \).
Figure \( \PageIndex{ 14 } \)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 \((5,0)\). See Figure \( \PageIndex{ 15 } \).
Figure \( \PageIndex{ 15 } \)As \(x \to \infty\), the function \(f(x) \to - \infty \), 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 \( \PageIndex{ 16 } \), and verify that the resulting graph looks like our sketch in Figure \( \PageIndex{ 15 } \).
Figure \( \PageIndex{ 16 } \): The complete graph of the polynomial function \(f(x) = -2(x+3)^2 (x-5)\)
Sketch a graph of \(f(x)= \frac{1}{4} x (x-1)^4 (x+3)^3 \).
Using the Intermediate Value Theorem
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 \(f\) whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers \(a\) and \(b\) in the domain of \(f\), if \(a < b\) and \(f( a ) \neq f( b )\), then the function \(f\) takes on every value between \(f( a )\) and \(f( b )\). 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 \(f\) at \(x=a\) lies above the \( x \)-axis and another point at \(x=b\) lies below the \( x \)-axis, there must exist a third point between \(x=a\) and \(x=b\) where the graph crosses the \( x \)-axis. Call this point \(( c,f( c ) )\). This means that we are assured there is a solution \(c\) where \(f( c )=0\).
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 \( x \)-axis. Figure \( \PageIndex{ 17 } \) shows that there is a zero between \(a\) and \(b\).
Figure \( \PageIndex{ 17 } \): Using the Intermediate Value Theorem to show there exists a zero.
Let \(f\) be a polynomial function. If \(f( a )\) and \(f( b )\) have opposite signs, then there exists at least one value \(c\) between \(a\) and \(b\) for which \(f( c )=0\).
To be honest, the Intermediate Value Theorem applies to any continuous function (not just polynomials); however, that extra bit of power to the theorem is reserved for Calculus (where the Intermediate Value Theorem becomes incredibly useful).
Show that the function \(f(x)= x^3 -5 x^2 + 3x +6\) has at least two real zeros between \(x=1\) and \(x=4\).
- Solution
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As a start, evaluate \(f(x)\) at the integer values \(x=1\), \(2\), \(3\), and \(4\). See Table \( \PageIndex{ 2 } \).
Table \( \PageIndex{ 2 } \) \(x\) 1 2 3 4 \(f(x)\) 5 0 –3 2 We see that one zero occurs at \(x=2\). Also, since \(f(3)\) is negative and \(f(4)\) is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4. Thus, we have shown that there are at least two real zeros between \(x=1\) and \(x=4\).
We can verify this result by looking at the graph of the function in Figure \( \PageIndex{ 18 } \).
Figure \( \PageIndex{ 18 } \)
Show that the function \(f(x)=7 x^5 -9 x^4 - x^2\) has at least one real zero between \(x=1\) and \(x=2\).
Writing Formulas for Polynomial Functions
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 \(p\) has horizontal intercepts at \(x= x_1, x_2, \ldots ,x_n \), then the polynomial can be written in the factored form:\[ f(x) = a(x- x_1)^{p_1} (x - x_2)^{p_2} \cdots (x - x_n)^{p_n}, \nonumber \]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.
Write a formula for the polynomial function shown in Figure \( \PageIndex{ 19 } \).
Figure \( \PageIndex{ 19 } \)
- Solution
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This graph has three \( x \)-intercepts: \(x=-3\), \(2\), 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\[f(x) = a(x+3) (x-2)^2 (x-5). \nonumber \]To determine the stretch factor, we utilize another point on the graph. We will use the \( y \)-intercept \((0,–2)\), to solve for \(a\).\[ \begin{array}{rrclcl} & f(0) & = & a(0 + 3)(0 - 2)^2 (0 - 5) & \quad & \left( \text{substituting} \right) \\[6pt] \implies & -2 & = & a(0 + 3)(0 - 2)^2 (0 - 5) & \quad & \left( \text{substituting} \right) \\[6pt] \implies & -2 & = & -60a & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & \dfrac{1}{30} & = & a & \quad & \left( \text{dividing both sides by }-60 \right) \\[6pt] \end{array} \nonumber \]The graphed polynomial appears to represent the function \(f(x)= \frac{1}{30} (x+3) (x-2)^2 (x-5)\).
Given the graph shown in Figure \( \PageIndex{ 20 } \), write a formula for the function shown.
Figure \( \PageIndex{ 20 } \)
Using Local and Global Extrema
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 \(x=a\) (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 \(x=a\). If a function has a local maximum at \(a\), then \(f(a) \geq f(x)\) for all \(x\) in an open interval around \(x=a\). If a function has a local minimum at \(a\), then \(f(a) \leq f(x)\) for all \(x\) in an open interval around \(x=a\).
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 \(a\), then \(f(a) \geq f(x)\) for all \(x\). If a function has a global minimum at \(a\), then \(f(a) \leq f(x)\) for all \(x\).
We can see the difference between local and global extrema in Figure \( \PageIndex{ 21 } \).
Figure \( \PageIndex{ 21 } \)
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, \(f( x )=x\) 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 then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box.
- Solution
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We will start this problem by drawing a picture like that in Figure \( \PageIndex{ 22 } \), labeling the width of the cut-out squares with a variable, \(w\).
Figure \( \PageIndex{ 22 } \)Notice that after a square is cut out from each end, it leaves a \(( 14 - 2w )\) cm by \(( 20 - 2w )\) cm rectangle for the base of the box, and the box will be \(w\) cm tall. This gives the volume\[V(w) = (20-2w)(14-2w)w = 280w - 68w^2 + 4w^3. \nonumber \]Notice, since the factors are \(w\), \(20–2w\), and \(14–2w\), the three zeros are \( 0 \), \( 10 \), and \( 7 \), 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 \(w\) may take on are greater than zero or less than 7. This means we will restrict the domain of this function to \(0<w<7\). Using technology to sketch the graph of \(V( w )\) on this reasonable domain, we get a graph like that in Figure \( \PageIndex{ 23 } \). We can use this graph to estimate the maximum value for the volume, restricted to values for \(w\) that are reasonable for this problem - values from 0 to 7.
Figure \( \PageIndex{ 23 } \)From this graph, we turn our focus to only the portion on the reasonable domain, \([ 0,7 ]\). 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 \( \PageIndex{ 24 } \).
Figure \( \PageIndex{ 24 } \)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 \([-1,4]\) of the function \(f(x) = -0.2(x-2)^3 (x+1)^2 (x-4)\).
Revisiting Graphs of Absolute Values of Functions
While you will not be graphing the absolute value of polynomial functions in Calculus, you will be graphing the absolute value of other functions. Therefore, it's a good idea to keep that skill fresh (it's a simple skill once you know how to graph the non-absolute value function).
Graph\[ f(x) = \left| -0.9(x + 2)^3(x - 1)^6(x - 5)^5\right|. \nonumber \]
- Solution
-
As we have done in previous sections, we initially concern ourselves with the
non-absolute value
function\[ f(x) = -0.9(x + 2)^3 (x - 1)^6(x - 5)^5. \nonumber \]We list the roots and their multiplicities, as well as the behavior of the graph near that root:\[ \begin{array}{c|c|c}
\textbf{Root} & \textbf{Multiplicity} & \text{Behavior} \\[6pt]
\hline -2 & 3 & \text{like }x^3 \\[6pt]
1 & 6 & \text{like }x^6 \text{ (a very flat parabola)} \\[6pt]
5 & 5 & \text{like }x^5 \text{ (a flat cubic)} \\[6pt]
\end{array} \nonumber \]Next, we note that the degree of the polynomial is \( 3 + 6 + 5 = 14 \), so the end behavior is like that of \( x^2 \) except the leading coefficient is negative. Therefore, the end behavior is more like \( -x^2 \). That is, as \( x \to -\infty \), \( f(x) \to -\infty \), and as \( x \to \infty \), \( f(x) \to -\infty \). Reading from left to right, the graph of \( f \) should- start low
- increase until it crosses the \( x \)-axis at \( x = -2 \) like a cubic
- climb for a bit, but then turn to fall back to the \( x \)-axis to "bounce" like a very flat parabola at \( x = 1 \)
- climb for a bit, but then turn to fall back to the \( x \)-axis to cross at \( x = 5 \) like a flat cubic
- it then continues to fall forever.
To graph the absolute value of this function, we take all the function outputs (\( y \)-values) that are negative and we force them to be positive. In effect, this keeps the part of the graph from Figure \( \PageIndex{ 25 } \) that is already above the \( x \)-axis in place; however, all the parts that are below the \( x \)-axis get reflected about the \( y \)-axis to become positive (above the \( x \)-axis). See Figure \( \PageIndex{ 26 } \).
Figure \( \PageIndex{ 25 } \)
Figure \( \PageIndex{ 26 } \)
Solving Polynomial Inequalities
We finish this section with a topic that might not seem related to graphing polynomial functions - solving polynomial inequalities.
There are those who insist on students solving inequalities using sign charts, and then there are those who want to reinforce a student's graphing skills. If you cannot tell, I am in the latter camp. To solve any inequality, you could move all terms to one side of the inequality, graph the resulting expression on the non-zero side, and use that graph to solve the inequality. The key to being able to do this is know how to graph that non-zero expression.
Let's see this in an example.
Solve\[ 3x^3 -4x \geq 6x^2 - 8. \nonumber \]
- Solution
-
Let's start by getting all terms to one side.\[ \begin{array}{rrcl}
& 3x^3 -4x & \geq & 6x^2 - 8 \\[6pt]
\implies & 3x^3 - 6x^2 - 4x + 8 & \geq & 0 \\[6pt]
\end{array} \nonumber \]Now we want to graph the expression on the left side. That is, let's graph\[ g(x) = 3x^3 - 6x^2 - 4x + 8. \nonumber \]Since this is a polynomial, we would like to find the roots and their multiplicities. Let's try factoring by grouping:\[ \begin{array}{rcl}
g(x) & = & 3x^3 - 6x^2 - 4x + 8 \\[6pt]
& = & 3x^2(x - 2) - 4(x -2)\\[6pt]
& = & (x - 2)(3x^2 - 4)\\[6pt]
\end{array} \nonumber \]Hence, this polynomial has the following roots (and multiplicities):\[ \begin{array}{c|c}
\textbf{Root} & \textbf{Multiplicity} \\[6pt]
\hline 2 & 1 \\[6pt]
\dfrac{2}{\sqrt{3}} & 1 \\[6pt]
-\dfrac{2}{\sqrt{3}} & 1 \\[6pt]
\end{array} \nonumber \]Since \( g(x) \) is a cubic, we know it "starts low and ends high." A quick sketch yields the graph in Figure \( \PageIndex{ 27 } \).
Remember, we are being asked to solve \( 3x^3 - 6x^2 - 4x + 8 \geq 0 \). This is equivalent to asking, "When is the graph of \( g(x) = 3x^3 - 6x^2 - 4x + 8 \) at or above the \( x \)-axis?" The answer to this question is simple once you have the graph. We can see from Figure \( \PageIndex{ 27 } \), that \( g(x) \geq 0 \) when\[ x \in \left[ -\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}} \right] \cup \left[ 2, \infty \right). \nonumber \]
Figure \( \PageIndex{ 27 } \)
Access the following online resource for additional instruction and practice with graphing polynomial functions.