3.03 Graphing polynomial functions

State the least possible degree of the polynomial.

To determine the degree of the polynomial, we need to find the zeros and their multiplicities. The sum of the multiplicities of each zero will be the degree of the polynomial.

The real zeros are the x-intercepts of the graph. These are at x=-3, 0, 1. Now, we need to look at how the graph behaves at each intercept to determine the multiplicity of each zero.

At x=-3, the graph crosses the x-axis with a point of inflection, so this zero has multiplicity 3. At x=0, the graph crosses the x-axis with no point of inflection, so this zero has multiplicity 1. At x=1, the graph is tangent to the x-axis, so this zero has multiplicity 2.

Adding the multiplicities of each zero, we find the polynomial has degree of at least 3+1+2=6.

The graph behaves in similar ways for zeros with even multiplicities or zeros with odd multiplicities greater than 1. For this reason, we cannot always assume that the multiplicity is 2 when the graph is tangent to the x-axis at a zero.

This is why the directions specify finding the least possible degree of the polynomial.

Determine whether the leading coefficient is positive or negative.

In part (a), we determined that the degree of the polynomial was even. When the degree is even, the end behavior will tend toward the same direction. The leading coefficient tells us whether both ends tend toward positive or negative infinity.

As x\to-\infty, f\left(x\right)\to\infty and as x\to\infty, f\left(x\right)\to\infty. Since both ends of the graph rise toward positive infinity, the leading coefficient is positive.

When both ends of the graph tend toward the same direction, the degree of the polynomial is even. When the ends of the graph tend toward opposite directions, the degree of the polynomial is odd.

State the equation of the least degree polynomial for the displayed graph.

In part (a), we found the zeros at x=-3,0,1, and their multiplicities were 3, 1, and 2 respectively. We can use this information with the factor theorem to write the factors of the function. Then, we need to use another point to find the leading coefficient.

In this case, we cannot use the y-intercept to determine the leading coefficient because it is one of the roots. Instead, we can use the point \left(-1,-8\right).

From the zeros and their multiplicities, we get the expression \left(x+3\right)^3\left(x-0\right)\left(x-1\right)^2 which can be simplified to x\left(x+3\right)^3\left(x-1\right)^2

Next, we need to use the point \left(-1,-8\right) to find the leading coefficient.

Therefore, the equation of this polynomial is f\left(x\right)=\dfrac{1}{4}x\left(x+3\right)^3\left(x-1\right)^2.

We can check this answer by graphing with technology. When you graph the equation, it would be helpful to adjust the window of your device to show the same domain and range as shown in the graph given.

Determine all the zeros of f\left(x\right) and their multiplicities.

Since the degree of the polynomial is 3, this polynomial has at most 2 turning points and a total of 3 complex roots. To find those roots, we can first try to factor the polynomial by grouping. However, that strategy will not work in this case.

When factoring does not work, we can try to use the rational roots theorem to find the possible rational roots. This theorem can only be used if the coefficients are integers. All the coefficients of this polynomial are integers, so we can use this theorem.

The leading coefficient is 2 and the constant term is -9. By the rational roots theorem, the possible rational roots will all be factors of 9 or half a factor of 9. So, the possible rational zeros of the polynomial are: \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, and \pm \frac{9}{2}.

First, we need to find one root by substituting the possible rational roots. By the remainder theorem, x=a is a root if f\left(a\right)=0. We can start by testing x=1:

\begin{aligned}f\left(1\right)&=2\left(1\right)^3-13\left(1\right)^2+24\left(1\right)-9\\&=2-13+24-9\\&=4\end{aligned}

Since 4\neq 0, x=1 is not a root. Next, we'll try x=-1:

\begin{aligned}f\left(-1\right)&=2\left(-1\right)^3-13\left(-1\right)^2+24\left(-1\right)-9\\&=-2-13-24-9\\&=-48\end{aligned}

Since -48\neq 0, x=-1 is not a root. Testing x=3 next:

\begin{aligned}f\left(3\right)&=2\left(3\right)^3-13\left(3\right)^2+24\left(3\right)-9\\&=2\left(27\right)-13\left(9\right)+24\left(3\right)-9\\&=54-117+72-9\\&=0\end{aligned}

So, x=3 is a root and \left(x-3\right) is a factor by the factor theorem.

Next, we can find \dfrac{f\left(x\right)}{x-3} to find the quadratic factor using synthetic or polynomial long division.

Finally, we can factor the quadratic to fully factor the original cubic f\left(x\right)=2x^3-13x^2+24x-9.

From here, we can see that x=3 is a zero with multiplicity 2, and x=\dfrac{1}{2} is a zero with multiplicity 1.

Notice how the leading coefficient in standard form is not written in front of the factors in factored form. Instead, the 2 is a coefficient inside a linear factor. When determining the leading coefficient of a polynomial, we should take into account any coefficients in front of the factors and within the factors.

Sketch the graph of f\left(x\right).

We will need to determine the end behavior using the leading coefficient and degree of the polynomial, find the y-intercept, and use the multiplicity of the x-intercepts to sketch a graph of this function.

The degree of the function is odd, so the end behavior of each side of the graph will go in opposite directions. The leading coefficient is positive, so we know that as {x\to\infty} we will have y\to\infty. This means that as {x\to -\infty} we will have y\to-\infty.

To find the y-intercept, we need to substitute x=0 into the function.

The y-intercept is at \left(0,-9\right).

The function will have x-intercepts at \left(3,0\right) and \left(\dfrac{1}{2},0\right). At x=\dfrac{1}{2}, the graph will cross the axis since the multiplicity is 1. After that, the function will have a turning point and will decrease to the next intercept at x=3. At x=3, the graph will touch the axis and turn back around since the multiplicity is even.

Now we have enough information to sketch the function.

Due to the x-intercepts and their multiplicities, there must be a relative maximum somewhere between the zeros. Eventually, we will have more mathematical tools to determine the exact positions of local turning points.

For now, the intercepts, their multiplicities, and the end behavior allow us to understand the function's basic shape which is enough for us to make a decent sketch.

• Rational coefficients

• Leading coefficient of -3

• Degree of 3

• Zeros include x=1+\sqrt{5} and x=2

Since the degree of the polynomial is 3, this polynomial has at most 2 turning points and a total of 3 complex roots. The zeros of the function that were given and their multiplicities are x=2 with multiplicity 1 and x=1+\sqrt{5} with multiplicity 1.

Since the polynomial has rational coefficients, we can use the irrational conjugate theorem to find the remaining zero.

By the irrational conjugate theorem, if x=1+\sqrt{5} is a zero, then x=1-\sqrt{5} must also be a zero. From these three zeros, we get the expression (x-2)\left[x-\left(1+\sqrt{5}\right)\right]\left[x-\left(1-\sqrt{5}\right)\right]

By the fundamental theorem of algebra, there are no additional factors as the product of the factors has degree 3. Since the leading coefficient is -3, f\left(x\right)=-3\left(x-2\right)\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right) is the polynomial function.

We can now find the y-intercept using the equation:

The y-intercept is \left(0,-24\right).

We will now determine the end behavior of f\left(x\right)=-3\left(x-2\right)\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right).

The degree of the function is 3 which is odd and the leading coefficient is -3 which is negative.

So as x\to\infty, we have f(x)\to-\infty and as x\to-\infty, we have f\left(x\right)\to\infty.

Sketching the graph gives us:

Notice that we estimated where 1-\sqrt{5} and 1+\sqrt{5} will be on the axis. If you cannot use a calculator to estimate the decimal, then you can use what you know about other square roots to estimate the decimal.

The \sqrt{5} lies between \sqrt{4} and \sqrt{9} which are both perfect squares.\begin{aligned}\sqrt{4}<&\sqrt{5}<\sqrt{9}\\2<&\sqrt{5}<3\end{aligned}

This means \sqrt{5} is somewhere between 2 and 3. Adding 1 to each expression, \begin{aligned}1+2<&1+\sqrt{5}<1+3\\3<&1+\sqrt{5}<4\end{aligned} we find that 1+\sqrt{5} lies somewhere between 3 and 4.

To estimate where 1-\sqrt{5} lies, we need to multiply by -1 which flips the inequality signs, then add 1.\begin{aligned}-2>&-\sqrt{5}>-3\\1-2>&1-\sqrt{5}<1-3\\-1>&1-\sqrt{5}>-2\end{aligned} This means 1-\sqrt{5} lies somewhere between -1 and -2.

• Increasing intervals: \left(-\infty,-2\right)\cup \left(0,2\right)

• Decreasing intervals: \left(-2,0\right)\cup \left(2,\infty\right)

• Only 2 roots, both real, each with degree 2

• Relative minimum at \left(0,-8\right)

Recall that increasing and decreasing intervals are given in terms of x. From this information, we know the x-values of the turning points, but we do not know the y-values. Picturing these increasing and decreasing intervals, the graph will go up, then down, then up, then down again, forming an M-shape.

The next piece of information given says there are only 2 roots and both are real, so we know there are 2 x-intercepts. Since they both have a multiplicity of 2, these will be tangent to the axis. This means the function must turn at the roots, so the x-intercepts are also the relative maxima.

The last thing we are told is the relative minimum is the y-intercept. This accounts for the last turning point, so we can now draw our sketch.

According to the information given, this function is even because it is symmetric to the y-axis. The function also has an even degree since the end behavior tends toward the same direction.

Not all even degree polynomials will also be even functions. If this function had x-intercepts at the origin and \left(4,0\right) for example, it would still have an even degree but it would no longer be an even function.

For a polynomial to be an even function, all terms must have a degree that is even or 0.