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Polynomial Functions

Definition

A polynomial of degree n is a function of the form \[f(x) = a_nx^n + a_{n-1}x^{n-1} + \dotso + a_2x^2+a_1x+a_0\] where the a's are the coeffiecients.

For example, $f(x) = 4x^6 - 3x^2 +2$ is a polynomial of degree 6 as 6 is the highest power of $x$.

A polynomial of degree 1 is called linear. A polynomial of degree 2 is called a quadratic. A polynomial of degree 3 is called a cubic. A polynomial of degree 4 is called a quartic.

Turning Points

A turning point of a function is a point where the graph of the function changes from sloping downwards to upwards, or vice versa. The gradient changes from positive to negative or negative to positive.

A polynomial of degree $n$ can have up to $(n-1)$ turning points. The number of turning points can be found by differentiating the function and setting the derivative equal to zero which will then give the $x$ coordinates of any turning points. The number of solutions found corresponds to the number of turning points. These $x$ values can then be substituted into the original equation to determine the coordinates (i.e. both the $x$ and $y$ values) of the turning points.

Worked Examples

Example 1

State the degree of the following polynomials:

a) $f(x) = x^2 - x + 4$ b) $f(x) = x^5 - x^4 + 3x^2 -1$ c) $f(x) = 6x$ d) $f(x) = 2x^8 - 3$

Solution

a) This polynomial has degree $2$.

b) This polynomial has degree $5$.

c) This polynomial has degree $1$.

d) This polynomial has degree $8$.

Example 2

Find all the turning points of the function $f(x) = x^3 +8x^2 -12x + 4$.

Solution

Differentiate the function.

\[f'(x) = 3x^2+16x-12\]

Set the function equal to zero.

\[f'(x) = 3x^2+16x-12 =0\]

Solve for $x$.

\begin{align} 3x^2 +16x-12 &=0\\ (3x-2)(x+6) &=0\\ x = \frac{2}{3}, \; x=-6 \end{align}

There are two possible solutions so $f(x)$ has two turning points, when $x=\dfrac{2}{3}$ and $x=-6$.

To find the coordinates of these turning points, substitute these values for $x$ into the original equation.

\[f\left(\dfrac{2}{3}\right) = \left(\dfrac{2}{3}\right)^3 +8\left(\dfrac{2}{3}\right)^2 -12\left(\dfrac{2}{3}\right) + 4 = -\dfrac{4}{27}\]

\[f(-6) = (-6)^3 +8(-6)^2 -12(-6) + 4 = 148\]

So the coordinates of the turning points are $\left(\dfrac{2}{3},-\dfrac{4}{27}\right)$ and $(-6, 148)$.

Video Examples

Example 1

Prof. Robin Johnson sketches the graph of $y=3x^2+6x$.

Workbooks

These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples.

External Resources

Whiteboard maths

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