Chapter 3.1 Polynomial Functions

In this chapter, we will focus on the principle of rigidity.

Rigidity

A rigid collection of objects is a collection in which fewer properties than one would expect uniquely determine an object in the collection.

An important principle in mathematics is to determine the way in which a collection is rigid and use the rigidity to specify certain elements of the collection. In this section, we focus our attention to polynomials and identify the ways in which it forms a rigid collection.

Quadratic Functions and Optimization

Many things in the world are described by quadratic functions. For example:

The height, as a function of time, of an object that falls a small distance without wind resistance;
The square of the distance, as a function of time, between two objects that have constant velocities;
The area, as a function of the ratio of the side lengths, of a rectangular yard of a fixed perimeter.

As such, we focus on quadratic functions. First, we define what a quadratic polynomial is.

Quadratic Polynomial

A quadratic polynomial, \(f\), is a function with the property that there are real numbers \(a\), \(b\), and \(c\) so that \(a\) is nonzero and \[f(x) = ax^2 + bx + c.\]

Here are some sketches of some quadratic polynomials

Notice that they all look roughly the same.

We will soon see that any quadratic polynomial will look like one of the two functions sketched below.

In this form, we can notice a couple of things.

The quadratic polynomial \(f\) that is given by the 1st graph will have a positive leading coefficient and the quadratic polynomial \(g\) that is given by the 2nd graph will have a negative leading coefficient.

The quadratic polynomial \(f\) will have exactly one local minimum, which will be also a global minimum, and the quadratic polynomial \(g\) will have exactly one local maximum, which will be a global maximum.

A quadratic polynomial \(f\) that is given for some real numbers \(a\), \(b\), and \(c\) by \[f(x) = ax^2 + bx + c,\] will have zeros at \(r_1\) and \(r_2\), where \[r_1 = \frac{-b-\sqrt{b^2 - 4ac}}{2a}\quad\text{and}\quad r_2 = \frac{-b+\sqrt{b^2 - 4ac}}{2a}.\]

The term inside of the square root is important for identifying the number of zeros, so we will give it a name.

Discriminant

The discriminant of \(f\), \(\Delta(f)\), is the quantity \[\Delta(f) = b^2 - 4ac.\]

There will be

two zeros if \(\Delta(f)\) is positive,
one if \(\Delta(f)\) is zero, and
no zeros if \(\Delta(f)\) is negative.

The formula that determines the zeros of a quadratic polynomial, the quadratic formula, follows from the transformation properties of quadratic polynomials.

But first, let’s make sure we understand how to use the discriminant.

Example 1

Take \(f\) to be given by \[f(x) = 3x^2 - x + c.\] Find values for \(c\) so that \(f\) has two zeros, one zero, and no zeros. For the first two cases, determine the zeros of \(f\).

Here \(a=3,b=-1,c=c\).

The discriminant of \(f\) is \(\Delta(f)=(-1)^2-4(3)c=1-12c\).

The function has two roots if \(\Delta(f)>0\), meaning \(1-12c>0\). Solve for \(c\) to get \(\frac{1}{12}>c\).
The function has one root if \(\Delta(f)=0\), meaning \(1-12c=0\). Solve for \(c\) to get \(\frac{1}{12}=c\).
The function has no root if \(\Delta(f)<0\), meaning \(1-12c<0\). Solve for \(c\) to get \(\frac{1}{12}<c\).

Quadratic polynomials form a rigid collection of functions in that they are completely determined by a translation and a scaling of the \(y\)-axis.

More precisely, for any quadratic polynomial \(f\) there are real numbers \(A\), \(h\), and \(k\) such that \[f(x) = A(x-h)^2 + k.\]

Moreover, a quadratic polynomial is uniquely determined by 3 non-colinear points.

In this example, we will determine how to transform a quadratic polynomial so it is of the form \(A(x-h)^2+k\).

Example 2

Define \(f\) by \[f(x) = 2x^2 -x + 5.\] Find \(A\), \(h\), and \(k\) so that \[f(x) = A(x-h)^2 + k.\]

Expand out \(A(x-h)^2 + k\):

\[ \begin{align*} A(x-h)^2+k\\ &=A(x-h)(x-h)+k\\ &=A(x^2-2hx+h^2)+k\\ &=Ax^2-2Ahx+Ah^2+k\\ &=[A]x^2+[-2Ah]x+[Ah^2+k]. \end{align*} \]

Thus \([2]x^2 +[-1]x + [5]=[A]x^2+[-2Ah]x+[Ah^2+k]\) if and only if the coefficients match:

\[ \begin{cases} 2&=A\\ -1&=-2Ah\\ 5&=Ah^2+k \end{cases} \]

The first equation says \(A=2\). Plug that in to the second equation to get \(-1=-2\cdot 2h\). Solve for \(h\) to get \(h=\frac{1}{4}\).

Plug in \(A\) and \(h\) into third equation to get \(5=(2)\cdot(\tfrac{1}{4})^2+k\). Solve for \(k\) to get \(k=\tfrac{39}{8}\).

Therefore, \(2x^2-x+5=2(x-\tfrac{1}{4})^2+\tfrac{39}{8}\)

As we can see, finding the right coefficients is based on the key property that the only way for two quadratic polynomials to be equal for all real numbers is for their coefficients to match.

Using this idea, we will derive a way to rewrite a general quadratic polynomial into the form \(A(x-h)^2+k\).

Example 3

Take \(a\) to be nonzero and define \(f\) by \[f(x) = ax^2 + bx + c.\] Find \(A\), \(h\), and \(k\) so that \[f(x) = A(x-h)^2 + k.\]

Expand out \(A(x-h)^2 + k\):

\[ \begin{align*} A(x-h)^2+k\\ &=A(x-h)(x-h)+k\\ &=A(x^2-2hx+h^2)+k\\ &=Ax^2-2Ahx+Ah^2+k\\ &=[A]x^2+[-2Ah]x+[Ah^2+k]. \end{align*} \]

Thus \([a]x^2 +[b]x + [c]=[A]x^2+[-2Ah]x+[-Ah^2+k]\) if and only if the coefficients match:

\[ \begin{cases} a&=A\\ b&=-2Ah\\ c&=Ah^2+k \end{cases} \]

Solve for \(A, h\) and \(k\) to get

\[ \begin{cases} a&=A\\ \frac{b}{-2a}&=h\\ c-a(\tfrac{b}{-2a})^2&=k \end{cases} \]

Thus \(ax^2+bx+c=a\left(x-\frac{b}{-2a}\right)^2+c-a\left(\tfrac{b}{-2a}\right)^2\)

Once again, we want to emphasis that that we can solve this problem by matching coefficients. This is because polynomials form a rigid collection: Polynomials are uniquely determined by their coefficients.

Writing a quadratic polynomial \(f\) in the form \[f(x) = A(x-h)^2 +k\] allows to quickly recognize somethings. One of which is a reflection property quadratic polynomials have.

Axis of Symmetry and Vertex

The axis of symmetry of \(f\) is the line with the property that the reflection of \(f\) across \(L\) is equal to \(f\).

The vertex of \(f\) is the point on \(f\) that lies on the axis of symmetry.

Another way to view is as follows:

Take \(g\) to be the quadratic polynomial given by \[g(x) = x^2\quad \text{so that}\quad f = \langle h,k\rangle + Y_A(g).\]

From this transformational perspective,

The vertex of \(g\) is the the point \((0,0)\), and so the vertex of \(f\) is the point \((h,k)\).

Understand this idea with the next example.

Example 4

Take \(f\) to be the function that is given by \[f(x) = (x-2)^2 - 1.\]

Determine the axis of symmetry and vertex of \(f\).
Find all roots (\(x\)-intercepts) of \(f\).
Sketch \(f\).

axis of symmetry is \(x=2\) and vertex is at \((2,-1)\)
roots are at \(x\) in which \(f(x)=0\) meaning \((x-2)^2 - 1=0\). We get \[\begin{align*}(x-2)^2-1&=0\\(x-2)^2&=1\\ (x-2)&=\pm\sqrt{1}\\ x-2&=\pm 1\\ x&=2\pm 1\end{align*}.\] Thus \(x=3\) or \(x=1\).
The vertex is at \((2,-1)\) and the roots are at \(x=3\), \(x=1\). The function opens up since \(a=1>0\). The graph is this:

In this next example, use the equation of the quadratic polynomial to recognize what its graph will look like.

Example 5

Take \(f_1\), \(f_2\), \(f_3\), and \(f_4\) to be given by the equations \[f_1(x) = -3(x+2)^2 +1, \quad f_2(x) = -\tfrac{1}{2}(x+1)^2 - 1,\] \[f_3(x) = 3(x+1)^2 +2,\quad f_4(x) = \tfrac{1}{2}(x-3)^2 - 4.\] Match the sketches of the functions with the functions.

\(f_2\)

\(f_4\)

\(f_1\)

\(f_3\)

In this next example, we will use the rigidity of quadratic polynomials to derive many features, one of which is the quadratic formula that gives the roots of the quadratic polynomial.

Example 6

Take \(a\) to be nonzero and take \(f\) to be the quadratic polynomial that is given by \[f(x) = ax^2 + bx + c.\]

Determine the vertex of \(f\).
Find a condition that is sufficient to guarantee that \(f\) has roots.
Given that \(f\) satisfies this condition, use transformations to derive the quadratic formula and find the roots of \(f\).

In a previous example, \(ax^2+bx+c=a\left(x-\frac{b}{-2a}\right)^2+c-a\left(\tfrac{b}{-2a}\right)^2\) so the vertex is at \(\left(-\tfrac{b}{2a},c-a\left(\tfrac{b}{-2a}\right)^2\right)\) or \(\left(-\tfrac{b}{2a},c-\tfrac{b^2}{4a}\right)\)
The discriminant is \(\Delta(f)=b^2-4ac\); thus \(b^2-4ac\geq 0\) for there to be a root.
Solve for \(x\):

\[ \begin{align*} 0&=a\left(x-\frac{b}{-2a}\right)^2+c-a\left(\tfrac{b}{-2a}\right)^2\\ 0&=a\left(x+\frac{b}{2a}\right)^2+c-\tfrac{b^2}{4a}\\ \tfrac{b^2}{4a}-c&=a\left(x+\frac{b}{2a}\right)^2\\ \tfrac{b^2}{4a}-\tfrac{4ac}{4a}&=a\left(x+\frac{b}{2a}\right)^2\\ \tfrac{b^2-4ac}{4a}&=a\left(x+\frac{b}{2a}\right)^2\\ \tfrac{b^2-4ac}{4a^2}&=\left(x+\frac{b}{2a}\right)^2\\ \pm\sqrt{\tfrac{b^2-4ac}{4a^2}}&=x+\tfrac{b}{2a}\\ \pm\tfrac{\sqrt{b^2-4ac}}{2a}&=x+\tfrac{b}{2a}\\ \pm\tfrac{\sqrt{b^2-4ac}}{2a}-\tfrac{b}{2a}&=x\\ \tfrac{-b\pm\sqrt{b^2-4ac}}{2a}&=x\\ \end{align*} \]

This is the quadratic formula.

Let’s focus our attention the extremal values of quadratic polynomials.

Take \(f\) and \(g\) to be the quadratic polynomials given by \[g(x) = x^2\quad \text{and}\quad f(x) = A(x-h)^2 +k.\]

Use the principle of transformation to study \(f\).

The vertex of \(g\) is \((0,0)\) and so the vertex of \(f\) is \((h, k)\) because \[f = \langle h,k\rangle + Y_A(g).\]

If \(A\) is positive (resp. negative), then \(f\) has a unique minimum (resp. maximum).

The \(x\)-coordinate of the vertex of \(f\) is the unique extreme of \(f\).

Note that optimizing a function means finding the local and global extrema of the function.

An optimization problem is a problem that involves optimizing a function.

Practice optimizing a function with this next example.

Example 7

Find and classify the extremal points of \(f\), where \[f(x) = 2x^2 -x + 5.\]

Rewrite into the form \(A(x-h)^2+k\) to get \(2x^2-x+5=2(x-\tfrac{1}{4})^2+\tfrac{39}{8}.\)

Thus \((\tfrac{1}{4},\tfrac{39}{8})\) is the vertex and that is minimum point since \(A=2>0\).

In this example, we will revisit one the questions we solved in Chapter 2, but solve it a different way. You should be comfortable solving with the methods we used before and the methods we use in this next example.

Example 8

Take \(L\) to be the line that is given by the equation \[y = 3x+1.\] Without appealing to geometry, find the point on \(L\) that is closest to the point \((1,5)\).

Closest point \(p=(x,3x+1)\) on \(L\) means that the distance is minimized: \[ \begin{align*} \|(x,3x+1)-(1,5)\|^2&=\|(x-1,3x-4)\|^2\\ &=(x-1)^2+(3x-4)^2\\ &=x^2-2x+1+9x^2-24x+16\\ &=10x^2-26x+17 \end{align*} \]

The minimum is the vertex. Work it out and get \(\left(\frac{13}{10},\frac{1}{10}\right)\). The \(x\)-coordinate is \(\frac{13}{10}\). Plug that in to \(L\) to get the \(y\) coordinate. So the point on \(L\) that is closest to \((1,5)\) is \((\tfrac{13}{10},\tfrac{49}{10})\)

Do you see some “advantages” and “disadvantages” in the methods we used in Chapter 2 compared to how we did in using our knowledge from Chapter 3?

The Factor Theorem

Like integers, polynomials can be decomposed in many different ways. First, lets remember how we can rewrite integers.

Quotient, Remainder, Dividend, Divisor

Given any integer \(M\) and any non-zero integer \(N\), there is a unique pair of integers \(q\) and \(r\) with \[0\leq |r| < |N|\] such that \[M = qN + r.\]

Refer to \(q\) as the quotient (of \(M\) and \(N\)) and \(r\) as the remainder.

Refer to \(M\) as the dividend and \(N\) as the divisor.

You may remember this concept when you learned about how to divide integers.

Refresh your memory or learn how to do this with this example.

Example 9

Write \[25 = q\cdot 7 + r,\] where \(r\) is an integer and \(|r|\) is less than \(7\). What is the dividend, divisor, quotient, and remainder?

Here \(q=3\) so that \(3\cdot 7=21\). Therefore, \(r=4\).

Thus \(25=3\cdot 7+4\).

Now, let’s discuss how we can do a similar decomposition with polynomials.

Take \(a\) and \(b\) to be polynomial functions.

If the degree of \(a\) is less than the degree of \(b\), then \(\frac{a}{b}\) is a proper fraction.

The quotient of two polynomials uniquely decomposes as a sum of a polynomial function and a proper fraction.

The collection of possible decompositions of a quotient of polynomials is rigid: A decomposition is uniquely determined if the fractional part is proper.

This rigidity implies a relationship between the roots and the factors of a polynomial.

For any polynomials \(a\) and \(b\) with \[{\rm deg}(b) \leq {\rm deg}(a),\] there is a unique pair of polynomials \(q\) and \(r\) so that \[{\rm deg}(r) < {\rm deg}(b) \quad \text{and} \quad a = qb + r.\]

If \(r\) is the zero polynomial, then \(b\) divides \(a\).

To determine the above decomposition, match coefficients.

Given that \({\rm deg}(r) < {\rm deg}(q)\), here is an example of how to match coefficients to find \(q\) and \(r\) so that \[3x^5 + 2x^2 +x - 1 = q(x)(x^2 +x-1) + r(x).\]

Write \[q(x) = ax^3 + bx^2 + cx + d\quad \text{and} \quad r(x) = ex + f.\]

Determine constants \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) so that \[q(x)(x^2 +x-1) + r(x) = Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F .\]

Solve the system of equalities obtained from the equality \[Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F = 3x^5 + 2x^2 +x - 1.\]

Let’s practice with this example.

Example 10

Find polynomials \(q\) and \(r\) so that \(\frac{r(x)}{x+2}\) is a proper fraction and \[x^2 + 3x + 1 = q(x)(x + 2) + r(x).\]

For the degrees to work out, \(q(x)=ax+b\) and \(r=c\).

Then \[ \begin{align*} q(x)(x+2)+r(x)&=(ax+b)(x+2)+c\\ &=(ax^2+2ax+bx+2b)+c\\ &=ax^2+(2a+b)x+2b+c. \end{align*} \]

Therefore \([1]x^2+[3]x+[1]=[a]x^2+[2a+b+c]x+[2b+d]\) if and only if \[ \begin{cases} 1&=a\\ 3&=2a+b\\ 1&=2b+c \end{cases} \]

Solve to get \(a=1\), \(b=1\), and \(c=-1\).

Thus \(x^2+3x+1=(x+1)(x+2)-1.\)

Here is another example.

Example 11

Find polynomials \(q\) and \(r\) so that \(\frac{r(x)}{x^2 + 1}\) is a proper fraction and \[3x^4 + x^2 + x - 5 = q(x)(x^2 + 1) + r(x).\]

For the degrees to work out, \(q(x)=ax^2+bx+c\) and \(r=dx+e\).

Then \[ \begin{align*} q(x)(x^2+1)+r(x)&=(ax^2+bx+c)(x^2+1)+dx+e\\ &=(a x^4 + a x^2 + b x^3 + b x + c x^2 + c)+dx+e\\ &=ax^4+bx^3+(a+c)x^2+(b+d)x+c+e. \end{align*} \]

Therefore \([3]x^4+[0]x^3+[1]x^2+[1]x+[-5]=[a]x^4+[b]x^3+[a+c]x^2+[b+d]x+[c+e].\) if and only if \[ \begin{cases} 3&=a\\ 0&=b\\ 1&=a+c\\ 1&=b+d\\ -5&=c+e \end{cases} \]

Solve to get \(a=3\), \(b=0\), \(c=-2\), \(d=1\), and \(e=-3\)

Thus \(3x^4 + x^2 + x - 5=(3x^2-2)(x^2+1)+x-3.\)

One important thing to note when using this decomposition is that if \(f\), \(q\), and \(r\) are polynomials, \(\frac{r(x)}{x-a}\) is a proper fraction, and \[f(x) = q(x) \cdot (x-a) + r(x),\] then \[f(a) = r(a).\]

So, we have the following theorem.

Factor Theorem

The Factor Theorem: The monomial \((x-a)\) divides \(f(x)\) if and only if \[f(a) = 0.\]

Why? The polynomial \(r(x)\) is a constant polynomial, and so it must be the zero polynomial.

Here is an example of understanding the reminder.

Example 11

What is the remainder of \(x^3 + x^2 + x +1\) when divided by \(x-1\)?

Take \(f(x)=x^3 + x^2 + x +1\). Here \((x-1)\) means \(a=1\). So \(f(1)\) will be the reminder when we divide by \((x-1)\). So the remainder was \(f(1)=4\).

In this example, we use the factor theorem to factor the following polynomial.

Example 12

Write \(5x^3 + 20x^2 + 5x - 30\) as a product of linear factors given that \(1, -2, -3\) are roots (zeros).

Since \(x=1\) is a root, then there is a polynomial \(q\) so that \(5x^3 + 20x^2 + 5x - 30=q(x)(x-1)\). For the degrees to match, \(q(x)=ax^2+bx+c\). So

Then \[ \begin{align*} q(x)(x-1)&=(ax^2+bx+c)(x-1)\\ &=(a x^3 - a x^2 + b x^2 - b x + c x - c)\\ &=ax^3+(-a+b)x^2+(-b+c)x-c. \end{align*} \]

Therefore \([5]x^3 + [20]x^2 + [5]x +[- 30]=[a]x^3+[-a+b]x^2+[-b+c]x+[-c].\) if and only if \[ \begin{cases} 5&=a\\ 20&=-a+b\\ 5&=-b+c\\ -30&=-c\\ \end{cases} \]

Solve to get \(a=5\), \(b=25\) and \(c=30\). So \(5x^3+20x^2+5x-30=(x-1)(5x^2+25x+30)\) or \(5(x-1)(x^2+5x+6)\). Continue the process by factoring the quadratic polynomial. Therefore

\(5x^3+20x^2+5x-30=5(x-1)(x+2)(x+3).\)

The zeros of a polynomial play an important role in understanding its graph. Before we can articulate this, we need a word to quantify an important part of a zero.

Order of a Zero

A real number \(a\) is a zero of order \(n\) for a polynomial function \(f\) if \((x-a)^n\) divides \(f\) but \((x-a)^{n+1}\) does not.

Determining the zeros of a polynomial and the orders of the zeros helps us sketch the polynomial. Practice finding the zero and order of the zeros with this example.

Example 13

Take \(f\) to be the polynomial function that is given by \[f(x) = x(x-1)^2(x+3)^5.\] List the zeros of \(f\) together with their orders.

The zeros are \(x=0\) with order 1, \(x=1\) with order 2, and \(x=-3\) with order 5.

Sketching Polynomials

To sketch a polynomial, we need to understand what its graph looks like near zeros, but also what its graph looks like “far away”.

Here is specifically what we mean.

Asymptotic Behavior

Polynomial functions are dominated by their leading term whenever \(|x|\) is large.

This behavior is called asymptotic behavior.

A polynomial \(P\) is asymptotically equal to \(ax^n\) if \(ax^n\) is the leading term of \(P\).

The polynomial \(P\) will “look like’’ \[y = ax^n.\]

Finding the asymptotic behavior is the same as finding the leading term. Finding the leading term means writing out the polynomial in expanded form. However, we do not need to expand out the entire polynomial! Pay attention to what we do in this next example.

Example 14

Determine the leading term of the polynomial \(f\) that is given by \[f(x) = (x+200)^8(3-2x)^3(5x+1)(x-7)^{20}.\] Describe the asymptotic behavior of \(f\).

The leading term is \(x^8\cdot (-2x)^3\cdot (5x)\cdot x^{20}=-40x^{32}.\) Therefore the asymptotic behavior looks like \(-40x^{32}\).

As we see in the above example, we only multiplied the leading terms from each product. This comes from the rigidity of polynomials!

Next, we focus on the zeros.

The zeros of a polynomial \(P\) are the \(x\)-intercepts of \(P\).

The Factor Theorem restricts the possible number of \(x\)-intercepts of a polynomial of a given degree.

For example, any odd degree polynomial, \(P\), must have at least one \(x\)-intercept and so there must be at least one real number \(a\) so that \((x-a)\) divides \(P\). However, an even polynomial need not have an \(x\)-intercept.

Moreover, the sum of the degrees of all zeros of a polynomial \(P\) cannot exceed the degree of \(P\).

Use this next example to understand all these facts to determine the possible decompositions of a polynomial with certain features.

Example 15

Take \(f\) to be a degree 6 polynomial and \(g\) to be a degree 5 polynomial. How many distinct zeros must \(f\) and \(g\) have? How many distinct zeros can \(f\) and \(g\) have?

The polynomial \(f\) can have at most \(6\) distinct zeros while \(g\) can have at most \(5\) distinct zeros. \(f\) can have \(0,1,2,3,4,5\) or \(6\) zeros while \(g\) can have \(1,2,3,4\), or \(5\) zeros.

In this next example use the asymptotic information and pointwise information to determine what the polynomial may look like.

Example 16

Find all possible degree 3 polynomials \(f\) that are asymptotically equal to \(5x^3\) and that satisfy the equality \[f(2) = f(-3) = 0.\]

The equalities \(f(2)=0\) and \(f(-3)=0\) mean that \(f\) has \(x=2\) and \(x=-3\) as the zeros, meaning its factored form must contain \((x-2)\) and \((x+3)\). Since \(f\) is asymptotically equal to \(5x^3\), its degree should be \(3\) and leading coefficient should be \(5x^3\). So here are some options

\(5(x-2)^2(x+3)\)
\(5(x-2)(x+3)^2\)

Now, let’s focus our attention how the zero of a polynomial and its order affects its shape near the zero. First, here is a definition of what we mean by near the zero.

Local Behavior

If \(a\) is an order \(n\) zero of the polynomial \(P\), then up to a real number \(A\) and for \(x\) in a sufficiently small open interval containing \(a\), \(P\) will look like a solution to the equation \[y = A(x-a)^n.\]

This local behavior allows us to sketch any polynomial of the form \[f(x) = a(x-a_1)^{n_1}(x-a_2)^{n_2}\cdots(x-a_m)^{n_m}.\]

The rigidity of polynomials allows us to combine the local behavior with asymptotic behavior, the global behavior, to sketch any polynomial that is in factored form.

Here is one example of how we can do that.