
Every polynomial function is continuous. This means that graphing polynomial functions won’t have any edges or holes. Zeros are important because they are the points where the graph will intersect our touches the x- axis.
If $ x_0$ is the root of the polynomial f(x) with multiplicity k then:
- If the multiplicity k is odd, the graph will cross the x-axis.
- If the multiplicity k is even, the graph will only touch the x- axis.
- If k > 1 the graph will flatten at $ x_0$.
There is just one more thing you should pay attention to the leading coefficient. When increasing x the function value increases also, in negative or positive way. This means that the ends of our graph will either decrease or increase without bound.
The leading coefficient test $ f(x) = a_n x^n + a_{n – 1} x^{n – 1} + … + a_1 x + a_0$
- If $ a > 0$ and n is even both ends of the graph will increase.
- If $ a > 0$ and n is odd then the graph will increase at the right end and decrease at the left end.
- If $ a < 0$ and n is even both ends of the graph will decrease.
- If $ a < 0$ and n is odd the graph will decrease at the right end and increase at the left end.
What is the process for graphing polynomial functions?
- You should determine all the zeros and their multiplicity. Also you should determine where the graph intercepts the y – axis.
- Use the leading coefficient test so you know how your graph will act at the ends.
- If you want to be more precise, you can always plot more points. The more points you find, the better your sketch will be.
First let’s observe this on the basic polynomials.
Example 1. Graph $ f(x) = x^2 + 2x$
Zeros of the function f(x) are 0 and -2, and zeros of the function $ g(x)$ are 0 and 2.
First let’s focus on the function f(x). The leading coefficient is positive and the leading exponent is even number. By the leading coefficient test, both ends of the graph will increase, which we know is true.
This graph will intersect the y – axis for f(0). This means that the graph will cut the y – axis in (0, 0).
Example 2. Graph $ x^3 + 8$
Zeros of this function are $ -2, 1 + i\sqrt{3}, 1 – i\sqrt{3}$. The only real root is -2. Graph will intersect y – axis in (0, 8).
The leading coefficient is a positive number and the leading exponent is odd, this means that the graph will decrease at the right end and increase at the left end.
Example 3. Graph $ f(x) = x^4 – 4x^2 + x – 1$
Real roots are $ x_1 \approx -2,1625$, $ x_2 \approx 1,9366$.
The graph will increase at the right end and decrease at the left end. Since there are 3 sign changes, the graph will change its course exactly three times.
If the function was set as $ f(x) = – x^4 + 4x^2 – x + 1$ its graph would look like this: