3.1 Polynomial Functions and Their Graphs
3.1.1 Learning Objectives
By the end of this section, you should be able to:
- identify polynomial functions and describe their degree and leading coefficient;
- predict end behaviour from the leading term;
- find zeros and determine their multiplicities;
- sketch polynomial graphs using intercepts, signs, and end behaviour; and
- relate degree to the possible number of turning points.
3.1.2 Polynomial Functions and Basic Shapes
A polynomial function has the form
\[ P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0, \]
where \(n\) is a non-negative integer, the coefficients are real numbers, and \(a_n\ne0\). The degree is \(n\), \(a_n\) is the leading coefficient, \(a_nx^n\) is the leading term, and \(a_0\) is the constant term.
Polynomial functions are defined and continuous for every real number. Their graphs have no breaks, holes, or sharp corners. Basic monomials such as \(x^2\), \(x^3\), and \(x^4\) provide useful parent graphs for understanding more complicated polynomials.
Example 3.1 Identifying Polynomial Features
For
\[ P(x)=-4x^5+2x^3-x+7, \]
identify the degree, leading coefficient, leading term, and constant term.
Solution
The greatest exponent is \(5\), so the degree is \(5\). The coefficient of \(x^5\) is \(-4\).
\[ \boxed{\text{Degree }5,\quad \text{leading coefficient }-4,\quad \text{leading term }-4x^5,\quad \text{constant }7} \]
Exercise 3.1 Identify the degree and leading coefficient of \(Q(x)=6-3x^2+8x^7\).
Show answer
Answer: In descending order, \(Q(x)=8x^7-3x^2+6\), so the degree is \(7\) and the leading coefficient is \(8\).
3.1.3 End Behaviour
The end behaviour of a polynomial describes what happens to \(P(x)\) as \(x\) becomes very large positively or negatively. Far from the origin, the leading term dominates the other terms.
- Even degree, positive leading coefficient: both ends rise.
- Even degree, negative leading coefficient: both ends fall.
- Odd degree, positive leading coefficient: the left end falls and the right end rises.
- Odd degree, negative leading coefficient: the left end rises and the right end falls.
Example 3.2 Predicting End Behaviour
Describe the end behaviour of \(P(x)=-2x^6+5x^3-1\).
Solution
The leading term is \(-2x^6\). Its degree is even and its leading coefficient is negative. Therefore, both ends fall:
\[ \boxed{P(x)\to-\infty\text{ as }x\to-\infty,\qquad P(x)\to-\infty\text{ as }x\to\infty} \]
Exercise 3.2 Describe the end behaviour of \(P(x)=3x^5-x^2+4\).
Show answer
Answer: \(\boxed{P(x)\to-\infty\text{ as }x\to-\infty,\quad P(x)\to\infty\text{ as }x\to\infty}\)
3.1.4 Zeros, Intercepts, and Multiplicity
A real number \(c\) is a zero of \(P\) if \(P(c)=0\). The corresponding \(x\)-intercept is \((c,0)\), and \(x-c\) is a factor of \(P(x)\).
If \((x-c)^m\) is a factor, then \(c\) has multiplicity \(m\):
- at a zero of odd multiplicity, the graph crosses the \(x\)-axis;
- at a zero of even multiplicity, the graph touches the axis and turns around;
- larger multiplicities make the graph flatter near the intercept.
Example 3.3 Interpreting Multiplicity
For
\[ P(x)=(x+2)^2(x-1)^3, \]
identify the zeros and describe the graph near each one.
Solution
The zero \(-2\) has multiplicity \(2\), so the graph touches the \(x\)-axis and turns. The zero \(1\) has multiplicity \(3\), so the graph crosses the axis and flattens near the intercept.
\[ \boxed{x=-2\text{ has multiplicity }2;\quad x=1\text{ has multiplicity }3} \]
Exercise 3.3 For \(Q(x)=(x-4)^4(x+1)\), state each zero, its multiplicity, and whether the graph crosses or touches the \(x\)-axis.
Show answer
Answer: \(x=4\) has multiplicity \(4\), so the graph touches and turns. \(x=-1\) has multiplicity \(1\), so the graph crosses.
3.1.5 Sketching Polynomial Graphs
To sketch a polynomial graph:
- determine the end behaviour;
- find the \(x\)- and \(y\)-intercepts;
- record the multiplicity of each zero;
- use test points to determine the sign between consecutive zeros; and
- draw a smooth, continuous curve consistent with all the information.
The Intermediate Value Theorem helps explain why sign changes matter. If a polynomial has values of opposite signs at two inputs, then it must have at least one zero between those inputs.
Example 3.4 Sketching from Factored Form
Describe the key features of
\[ P(x)=-(x+1)(x-2)^2. \]
Solution
The degree is \(3\) and the leading coefficient is negative, so the left end rises and the right end falls. The zeros are \(-1\), with multiplicity \(1\), and \(2\), with multiplicity \(2\). The graph crosses at \((-1,0)\) and touches at \((2,0)\).
The \(y\)-intercept is
\[ P(0)=-(1)(4)=-4. \]
\[ \boxed{\text{Cross at }(-1,0),\quad \text{touch at }(2,0),\quad y\text{-intercept }(0,-4)} \]
Exercise 3.4 Describe the end behaviour and intercept behaviour of \(P(x)=x(x-3)^2(x+2)\).
Show answer
Answer: Both ends rise. The graph crosses at \(x=-2\) and \(x=0\), and touches at \(x=3\).
3.1.6 Turning Points and Desmos
A local maximum is a point where a function changes from increasing to decreasing. A local minimum is a point where it changes from decreasing to increasing. Together, they are local extrema.
A polynomial of degree \(n\) can have at most \(n-1\) turning points. This is an upper bound, not a guarantee. Desmos can help estimate turning points and confirm a hand-drawn sketch.
Example 3.5 Estimating Turning Points
Use Desmos to investigate
\[ P(x)=x^3-3x. \]
Solution
The graph has odd degree with positive leading coefficient. Desmos shows a local maximum near \((-1,2)\) and a local minimum near \((1,-2)\). The degree is \(3\), so at most \(2\) turning points are possible.
\[ \boxed{\text{Local maximum }(-1,2),\quad \text{local minimum }(1,-2)} \]
Exercise 3.5 What is the greatest possible number of turning points of a degree-\(7\) polynomial?
Show answer
Answer: \(\boxed{6}\)
3.1.7 Conceptual Takeaways
- A polynomial’s leading term controls its end behaviour.
- Zeros connect algebraic factors to graphical \(x\)-intercepts.
- Multiplicity determines whether a graph crosses or touches the \(x\)-axis.
- Polynomial graphs are smooth and continuous.
- Degree limits the number of zeros and turning points a polynomial can have.
3.1.8 Skills You Should Be Able to Do
- Identify the degree, leading coefficient, and leading term of a polynomial.
- Predict end behaviour.
- Find zeros and multiplicities from factored form.
- Sketch a polynomial using intercepts, signs, and end behaviour.
- Use Desmos to verify a graph and estimate local extrema.
3.1.9 Practice Problems with Solutions
State the degree and leading coefficient of \(P(x)=5x^4-2x^7+9\).
Show Solution
In descending order, \(P(x)=-2x^7+5x^4+9\). Therefore,
\[ \boxed{\text{degree }7,\quad \text{leading coefficient }-2}. \]
Describe the end behaviour of \(P(x)=4x^8-3x+1\).
Show Solution
The leading term \(4x^8\) has even degree and a positive coefficient. Thus,
\[ \boxed{P(x)\to\infty\text{ as }x\to\pm\infty}. \]
Describe the end behaviour of \(Q(x)=-x^5+6x^2\).
Show Solution
The leading term is \(-x^5\), with odd degree and negative coefficient:
\[ \boxed{Q(x)\to\infty\text{ as }x\to-\infty,\quad Q(x)\to-\infty\text{ as }x\to\infty}. \]
Find the zeros and multiplicities of \(P(x)=3(x-2)^3(x+5)^2\).
Show Solution
Each factor is zero when
\[ x-2=0\quad\text{or}\quad x+5=0. \]
Therefore, \(\boxed{x=2\text{ has multiplicity }3;\quad x=-5\text{ has multiplicity }2}\).
Find the \(x\)- and \(y\)-intercepts of \(P(x)=x(x-4)(x+2)\).
Show Solution
The factors give zeros \(0\), \(4\), and \(-2\), so the \(x\)-intercepts are \((0,0)\), \((4,0)\), and \((-2,0)\). Since \(P(0)=0\), the \(y\)-intercept is also \((0,0)\).
\[ \boxed{x\text{-intercepts }(-2,0),(0,0),(4,0);\quad y\text{-intercept }(0,0)} \]
A polynomial has zeros \(-1\) and \(3\). The zero \(-1\) has multiplicity \(2\), and \(3\) has multiplicity \(1\). Write a least-degree polynomial with leading coefficient \(2\).
Show Solution
The required factors are \((x+1)^2\) and \((x-3)\). Multiplying by the specified leading coefficient gives
\[ \boxed{P(x)=2(x+1)^2(x-3)}. \]
Determine whether \(P(x)=x^4-5x^2+4\) has a zero between \(1\) and \(2\).
Show Solution
\[ P(1)=1-5+4=0. \]
The endpoint \(x=1\) is already a zero, but there is no zero strictly between \(1\) and \(2\), since
\[ P(x)=(x^2-1)(x^2-4) \]
has real zeros \(-2,-1,1,2\). Thus, \(\boxed{\text{no zero lies strictly between }1\text{ and }2}\).
Use Desmos to graph \(P(x)=x^4-4x^2\). State its zeros, end behaviour, and approximate local extrema.
Show Solution
Factoring gives
\[ P(x)=x^2(x-2)(x+2), \]
so the zeros are \(-2\), \(0\), and \(2\), with \(0\) having multiplicity \(2\). Both ends rise. Desmos shows a local maximum at \((0,0)\) and local minima at approximately \((-1.41,-4)\) and \((1.41,-4)\).
\[ \boxed{\text{Zeros }-2,0,2;\quad \text{local extrema }\approx(-1.41,-4),(0,0),(1.41,-4)} \]