How to find the degree of a polynomial graph calculator

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Find the degree, leading coefficient, and leading term of a polynomial step by step

The calculator will find the degree, leading coefficient, and leading term of the given polynomial function.

Your Input

Find the degree, the leading coefficient, and the leading term of $$$p{\left(x \right)} = 5 x^{7} + 2 x^{5} - 4 x^{3} + x^{2} + 15$$$.

Solution

The degree of a polynomial is the highest of the degrees of the polynomial's individual terms. In our case, the degree is $$$7$$$.

The leading term is the term with the highest degree. In our case, the leading term is $$$5 x^{7}$$$.

The leading coefficient is the coefficient of the leading term. In our case, the leading coefficient is $$$5$$$.

Answer

Degree: $$$7$$$A.

Leading coefficient: $$$5$$$A.

Leading term: $$$5 x^{7}$$$A.

Let's see how to graph the polynomial function P(x) = x³ - x.

First of all, we use Omni's polynomial graphing calculator to do the work for us. There, we begin by telling what type of a function we have. In our case, it's a cubic polynomial, so we choose 3 under "Polynomial degree." That'll show a symbolic representation of such an expression underneath and corresponding variable fields farther down. Looking back at our polynomial function example, we input:

a₃ = 1, a₂ = 0, a₁ = -1, and a₀ = 0.

(Note how we have a₂ = 0 and a₀ = 0 since P(x) has no terms with x² or with no x at all. Also, we input a₃ = 1 and a₁ = -1 even though there are no numbers in the corresponding places in P(x). That's because, by convention, we don't write 1-s in front of variables. However, observe that we needed to remember about the minus in a₁.)

The moment we input the last coefficient, Omni's polynomial graphing calculator will draw the graph, as well as find the zeros of the polynomial together with its critical points, extrema, and inflection points. Let us also mention that in case you'd like to see some other section of the graph than the one presented, you may go into the advanced mode and input a custom interval.

Now let's try to describe the graph ourselves. First of all, we need to find the zeros of the polynomial, so we solve the equation P(x) = 0:

x³ - x = 0

x * (x² - 1) = 0

x * (x - 1) * (x + 1) = 0

We obtained a product which is equal to 0, which means one of its factors must be zero. In other words, we have x = 0, x - 1 = 0, or x + 1 = 0, which gives us three solutions: x = 0, x = 1, and x = -1.

Next, we look for critical points. For that, we compute the derivative P'(x) according to the formula from the above section:

P'(x) = (x³ - x)' = 3x² - 1.

Now, we solve the equation P'(x) = 0:

3x² - 1 = 0

3x² = 1

x² = ⅓

which gives us two solutions (i.e., critical points): -√3/3 ≈ -0.577 and √3/3 ≈ 0.577.

Lastly, we draw the graph. Note that the leading coefficient of the polynomial is positive (i.e., equal to 1), and it's in front of an odd power of x (i.e., x³). That means the end behavior of the polynomial function is as follows:

• P(x) goes to plus infinity when x goes to plus infinity; and
• P(x) goes to minus infinity when x goes to minus infinity.

In between, the graph must touch the vertical axis in points -1, 0, and 1, and flatten in -0.577 and 0.577. That takes us to the conclusion that P(x) has a local maximum in x = -0.577 and a local minimum in x = 0.577; it has no inflection points.

All in all, the graph of P(x) = x³ - x looks like this:

Make sure to experiment with the polynomial graphing calculator to see how different coefficients affect the zeros and the bumps. Also, check out other algebraic tools on the website that can help in other polynomial-related problems.

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