4.1: An Introduction to The Graphs of Polynomial Functions
- Page ID
- 99730
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Focus: Polynomial Functions and Their Graphs
In this chapter, we will explore describing relationships with larger polynomial functions like \(f(x)=x^3+x^2 \) or \( g(x) = x^4+2x^3+4x^2+2\).
A polynomial is function that can be written as a sum of whole number powers of \(x\) such as
\[f(x)=a_{n} x^{n} +\cdots + a_{2} x^{2} +a_{1} x + a_{0} \nonumber \]
To distinguish between different size polynomials and their graphs, we need to develop some new terminology to describe different polynomials.
The Degree of a Polynomial Function
The degree of the polynomial is the highest power of the variable that occurs in the polynomial.
Example \(\PageIndex{1}\)
Identify the degree of each polynomials.
- \(f(x)=3+2x^{2} -4x^{3}\)
- \(g(t)=5t^{5} -2t^{3} +7t\)
- \(h(p)=6p-p^{3} -2\)
Solution
- For the function \(f(x)\), the degree is 3 since the term with the largest power is \(-4x^{3} \)
- For \(g(t)\), the degree is 5 since the term with the largest power is \(5t^{5}\).
- For \(h(p)\), the degree is 3 since the term with the largest power is \(-p^{3}\).
Notice that a linear function, such as \(f(x)=3x+1\) is a polynomial function of degree one. A quadratic function, such as \(g(x)=x^2+3x+1\) is a polynomial function of degree two. Our goal for this chapter is describing relationships with polynomial functions of degree three or larger.
With quadratic functions in chapter three, we identified three key features of the graph of a parabola in order to help us analyze the relationship the quadratic functions were describing. The key features of a quadratic function are
- The vertex
- The zeroes
- The end behavior
The vertex and end behavior allow us to determine where the parabola is increasing or decreasing as well as identify where the maximum or minimum value occurs. The real zeros of a parabola are the x-intercepts which help us determine where the function is positive or negative. With a larger polynomial, we seek to make a similar analysis. On the graph of the larger degree polynomial \(f(x) = x^3 - x^2\) below,
the polynomial changes sign from positive to negative (or vice versus) at the real zeros (or x-intercepts) at \(x=0\) and \(x=1\). The graph changes direction at \(x = 0\) and \(x \approx 0.67 \). Unlike a quadratic function, this third degree polynomial changes direction twice. Whereas a quadratic function has only one extrema, this polynomial has two relative extrema at the turning points where the graph changes direction. At the turning point \(x=0\), a relative maximum occurs. By a relative maximum, we mean that in a small interval containing \(x=0\), the function has its largest value in the interval at \(x=0\). At the turning point \(x \approx 0.67\), a relative minimum occurs. By a relative minimum, we mean that in a small interval containing \(x\approx 0.67\), the function has its smallest value in the interval at \(x\approx 0.67\). The relative maximums and minimums of a function are typically referred to as the relative extrema.
The function \(y=f(x)\) has a relative maximum at x = a if \(f(a) \ge f(x)\) for any \(x\) in a small interval containing \(x=a\).
The function \(y=f(x)\) has a relative minimum at x = a if \(f(a) \le f(x)\) for any \(x\) in a small interval containing \(x=a\).
Notice the end behavior of the larger polynomial \(f(x)\) above: it opens up on the right (as \(x \rightarrow + \infty , y \rightarrow + \infty \)) and down on the left (as \(x \rightarrow - \infty , y \rightarrow - \infty \)). These three features of the graph of a polynomial function will help us analyze the relationship it is describing.
The key features of the graph of a polynomial function are
- The zeros
- The relative extrema
- The end behavior
In Calculus, we will explore techniques to locate the relative extrema. At this point in Precalculus, we will focus on identifying the zeros and the end behavior. We can use technology, such as a graphing calculator or graphing app, to approximate relative extrema on the graphs of polynomials at this point.
Focus: The Zeros of a Polynomial Function
As with quadratic functions, the zeros of polynomials are the values of \(x\) for which the function value \(f(x)=0\). Real zeros are x-intercepts (or intercepts on the horizontal axis) on the graph. The zeros are a significant feature of a polynomial function since it will only change sign from positive to negative (or vice versus) at a real zero. Many of the techniques we used for solving polynomial equations in algebra will help us find the zeros of some polynomials.
Example \(\PageIndex{2}\)
Find the zeros of the polynomial functions.
- \(f(x)=x^3-x^2-6x\)
- \(f(x)=x^4 +x^3 +x^2\)
Solution
- The zeros occur when \(y=f(x)=0\). So \[x^3-x^2-6x=0\nonumber \] Factoring out a common factor of \(x\), then factoring the trinomial we have \[x(x^2-x-6)=0\nonumber \] \[x(x-3)(x+2)=0\nonumber \] The linear factors correspond to three distinct zeros \[x=0, x=3, \text{and } x=2\nonumber \]
- Letting \(y=f(x)=0\), we have \[x^4 +x^3 +x^2=0\nonumber \] Factoring out a common factor of \(x^2\) \[x^2(x^2+x+1)=0\nonumber \] Since the trinomial in not easily factorable into linear factors, we can set each factor to zero \[x^2=x \cdot x = 0 \text{ and } x^2+x+1=0\nonumber \] In the first equation, we have two factors of \(x\) that correspond to the zero \(x=0\). In the second equation, we must use the quadratic equation with \(a=1\), \(b=1\), and \(c=1\) since the equation is not easily factored into linear factors. \[x= \dfrac{-1 \pm \sqrt{1^2-4(1)(1)}}{2(1)}\nonumber \] \[x= \dfrac{-1 \pm \sqrt{-3}}{2} = \dfrac{-1 \pm i \sqrt{3}}{2} = \dfrac{-1}{2} \pm i \dfrac{\sqrt{3}}{2}\nonumber \]
Notice in the first polynomials in the previous example, we had three distinct real zeros that corresponding to a single linear factor each. Whereas in the second polynomial, we had a repeated real zero \(x=0\) that corresponds to two of the same linear factor as well as two complex zeros. In general, we could have a distinct real zero(s), a repeated zero(s), or complex conjugate zeros with a larger polynomial function. With quadratic functions, we had either two real zeros or two complex zeros. We need some language to describe the number of time a zero of a polynomial is repeated. We call the number of linear factors \((x-k) \) that corresponding to a zero \(x=k\) the multiplicity off the zero.
A zero \(x=k\) of a polynomial \(y=f(x)\) has multiplicity \(n\) if there are \(n\) factors of the corresponding factor \( (x-k) \) in the factorization of the polynomial.
Let \(f(x)=x^5+4x^4+4x^3\). Find the zeros and their multiplicity.
Solution
Letting \(y=f(x)=0\), we have \[x^5+4x^4+4x^3=0\nonumber \] Factoring our a common factor of \(x^3\), then factoring the trinomial we have \[x^3(x^2+4x+4)=0\nonumber \]\[x^3(x+2)(x+2)=0\nonumber \]These factors correspond to the zeros \(x=0\) with multiplicity three and \(x=-2\) with multiplicity two.
Let \(f(x)=x^6-6x^5+9x^4\). Find the zeros and their multiplicity.
- Answer
-
\(f(x)\) has zeros \(x=0\) with multiplicity four and \(x=3\) with multiplicity two.
Notice when finding the zeros and their multiplicities in each of the preceding examples and exercise, the polynomials had a common factor that when factored out of the polynomial, left a quadratic factor. When we set the quadratic factor to zero, we could solve with the quadratic formula or by factoring. With some larger polynomials, such as \(f(x)=x^3-13x+12\), there is no common factor. We can't use the quadratic formula since it is a cubic not a quadratic function. None of the techniques of solving polynomial equations from algebra help us find the zeros in this case. We will develop more tools in the next section to find the zeros of polynomials in this case. As with quadratic functions, real zeros are x-intercepts one the graph. However, we need to determine the effect of the multiplicity of a zero on the graph of a polynomial near that zero.
Building on what we did with quadratic functions, we can use the zeros of a polynomial and their multiplicities to find the formula for a polynomial function.
Find a formula for a polynomial function with zeros \(x=1\) with multiplicity one and \(x=-3\) with multiplicity two that passes through the point (0,4).
Solution
Since the polynomial has a zero \(x=1\) with multiplicity one, there is a single factor of \( (x-1) \). Since the polynomial has a zero \(x=-3\) with multiplicity two, there are two factors of \( (x+3) \). However, the polynomial \(y=f(x)=(x-1)(x+3)^2 \) passes through the point (0,-9) since \(f(0) = -9\). As with quadratic functions, the polynomial can also have a constant stretch factor \(a\). So \[f(x)=a(x-1)(x+3)^2 \nonumber\] To find \(a\), we can substitute the point \(x=0\) and \(y=4\) into the polynomial \[a(0-1)(0+3)^2=4 \nonumber \] Then solving for \(a\), we have \[-9a=4 \nonumber \] \[a=\dfrac{-4}{9} \nonumber \] Therefore, the formula for the polynomial is \(f(x) =\dfrac{-4}{9}(x-1)(x+3)^2\)
For our polynomial in example 4, we could leave our function in factored form as \(f(x) =\dfrac{-4}{9}(x-1)(x+3)^2\) or multiply the factors out as \(f(x) =\dfrac{-4}{9}x^3+\dfrac{20}{9}x^2-\dfrac{-4}{3}x+4\). Depending on what we want to do with the formula for the polynomial, either form is useful.
Find a formula for a polynomial function with zeros \(x=0\) with multiplicity three and \(x=2i\) with multiplicity one that passes through the point (1,2).
Solution
Since the polynomial has a zero \(x=0\) with multiplicity three, the polynomial has three factors of \(x\). Since the polynomial has the complex zero \(x=2i\) of multiplicity one, the complex conjugate \(x=-2i\) must also be a zero since complex zeros of polynomials appear in conjugate pairs. These complex zeros correspond to the factors \( (x-2i) \) and \( (x+2i) \) in the polynomial. Therefore, the polynomial has the formula \[f(x)=ax^3(x-2i)(x+2i) \nonumber\] Simplifying by multiplying out the complex factors, we have \[f(x)=ax^3(x^2-2ix+2ix-4i^2) \nonumber\]\[f(x)=ax^3(x^2+4) \nonumber\] Substituting the given point \(x=1\) and \(y=2\) into the polynomial to find the stretch factor \(a\) then solving, we have \[2=a1^3(1^2+4) \nonumber\] \[2=5a \nonumber\] \[a=\dfrac{2}{5} \nonumber\. Therefore, the polynomial has a formula \(f(x)=\dfrac{2}{5}x^3(x^2+4) \)
Find a formula for a polynomial function with zeros \(x=1\) with multiplicity one, \(x=-2\) with multiplicity one, and \(x=3\) with multiplicity two that passes through the point (0,5).
- Answer
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\(f(x)=\dfrac{-5}{18}(x-1)(x+2)(x-3)^2 \)
In example 4.1.2, the third degree polynomial \(f(x)=x^3-x^2-6x\) had three corresponding linear factors \(f(x)=x(x-3)(x+2)\). Each of these three linear factors corresponded to three distinct zeros \(x=0\), \(x=3\), and \(x=-2\) each with multiplicity one. In example 4.1.3, the fifth degree polynomial \(f(x)=x^5+4x^4+4x^3\) had five corresponding linear factors \( f(x)=x^3(x+2)^2 \). The factor \(x\) was repeated three times and the factor \( (x+2) \) was repeated two times. These corresponded to the zeros \(x=0\) with multiplicity three and the zero \(x=-2\) with multiplicity two. Conversely, in example 4.1.3 we were given two distinct zeros of a polynomial: \(x=1\) with multiplicity one and \(x=-3\) with multiplicity two. The zero \(x=1\) with multiplicity one corresponds to a single linear factor \( (x-1) \) in the polynomial. The zero \(x=-3\) with multiplicity two corresponds to two factors of the linear factor \( (x+3) \) in the polynomial. So we could write the polynomial as \(f(x)=a(x-1)(x+3)^2 \). Multiplying the three linear factors, we have the third degree polynomial \(f(x)=a(x^3+5x^2+3x-9) \) where \(a\) is the constant stretch factor. If we have five zeros in a polynomial (some may be repeated zeros with multiplicity two or more), these correspond to five linear factors in the polynomial. When we multiply these linear factors, we have a polynomial of degree five. This demonstrates the relationship between the degree of the polynomial, the number of linear factors, and the number of distinct zeros.
A polynomial of degree \(n\) will have at most \(n\) distinct real zeros.
Important Topics of this Section
- The Graph of a Polynomial Function
- The Zeros of a Polynomial Function
- The Multiplicity of a Zero
- Degree of a polynomial
- The Fundamental Theorem of Algebra