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# Describing Graphs of Polynomial Functions

As has been seen, the basic characteristics of polynomial functions, zeros and end behavior, allow a sketch of the function's graph to be made. Using other characteristics, such as increasing and decreasing intervals and turning points, it's possible to give a much more detailed description of polynomial graphs.
Concept

## Increasing and Decreasing Intervals

A function is said to be increasing when, as the $x$-values increase (from left to right), the values of $f(x)$ increase. Conversely, a function is said to be decreasing when, as $x$ increases, $f(x)$ decreases. The graph below shows increasing intervals with green arrows and decreasing intervals with red arrows.

The function above contains two increasing intervals and one decreasing interval. To describe each, use the $x$-values. Commonly a point where a function has a relative maximum or a relative minimum is neither included in an increasing nor a decreasing interval.

\begin{aligned} \textbf{Increasing interval:} & \ \text{-} \infty < x < \text{-} 2 \ \text{and} \ 0 < x < + \infty \\[0.8em] \textbf{Decreasing interval:} & \qquad \qquad \ \text{-} 2 < x < 0 \\ \end{aligned}
Concept

## Relative Minimum and Maximum

For some functions, their graphs extend infinitely in the vertical direction. For these graphs, there are no highest or lowest points. However, it's possible for these functions to have relative minimum or relative maximum values. A relative minimum is the lowest point for a region of the graph. Similarly, a relative maximum is the highest point for a region of the graph.

Concept

## Turning Point

Relative minimums and maximums of polynomial functions are also called turning points. This is because a function's graph turns from increasing to decreasing, or vice versa, at these points.

The graph of a polynomial function of degree $n$ can have at most $n - 1$ turning points. The function shown is a $3$rd degree polynomial and has $2$ turning points.

Furthermore, a polynomial function with $n$ real zeros must have at least $n - 1$ turning points.
Concept

## Even and Odd Symmetry

If a function has a symmetry, it is either even or odd. The symmetry is even when the graph is symmetric with respect to the $y$-axis, and odd when it's symmetric about the origin.

Concept

### Even Symmetry

If a function has even symmetry, the following rule applies: $f(\text{-} x)=f(x).$ The rule comes from the fact that even symmetry is a reflection across the $y$-axis. Therefore, changing the sign of the $x$-value does not affect the function value.

The concept applies both ways. Hence, if the rule is true for the entire domain, the function has even symmetry.

Concept

### Odd Symmetry

Instead, if a function has odd symmetry, the rule it must follow is $f(\text{-} x)=\text{-} f(x).$ An odd symmetry means graphically that the graph is rotated $180^\circ$ about the origin. Therefore, changing the sign of the $x$-value also changes the sign of the function value.

If this rule is satisfied on the entire domain, the function has odd symmetry.
Exercise

Using their function rules, determine if each of the following functions has even or odd symmetry, or neither. $h(x) = x^5 + 2x \qquad g(x) = 3x^3 + 1$

Solution

We can determine if a function has even or odd symmetry, using its function rule and the following relationships.

\begin{aligned} \textbf{even} &: f(\text{-} x)=f(x)\\ \textbf{odd} &: f(\text{-} x)=\text{-} f(x) \end{aligned} To determine if any function has even or odd symmetry, $f(x), f(\text{-} x),$ and $\text{-} f(x)$ must be known. $h(x)$ and $g(x)$ are given. Thus, the following must be found. $h(\text{-} x), \quad \text{-} h(x), \quad g(\text{-} x), \quad \text{and} \quad \text{-} g(x).$ Focusing first on $h(x),$ we'll find $h(\text{-} x).$
$h(x)=x^5+2x$
$h(\text{-} x)=(\text{-} x)^5+2(\text{-} x)$
Simplify power and product
$h(\text{-} x)=\text{-} x^5-2x$
Next, we'll find $\text{-} h(x).$
$h(x)=x^5+2x$
$\text{-} h(x) = \text{-} \left(x^5+2x\right)$
$\text{-} h(x)= \text{-} x^5-2x$
We can summarize the needed relationships for $h(x)$ as follows.
$h(x)$ $h(\text{-} x)$ $\text{-} h(x)$
$x^5+2x$ $\text{-} x^5-2x$ $\text{-} x^5-2x$

It can be seen that $h(\text{-} x) = \text{-} h(x).$ Thus $h$ has odd symmetry. The needed relationships for $g$ can be found in the same way. This gives the following.

$g(x)$ $g(\text{-} x)$ $\text{-} g(x)$
$3x^3+1$ $\text{-} 3x^3+1$ $\text{-} 3x^3-1$

Notice that $g(\text{-} x) \neq g(x) \quad \text{and } \quad g(\text{-} x) \neq \text{-} g(x).$ Thus, $g$ has neither even nor odd symmetry.

info Show solution Show solution
Exercise

The figure below shows the graph of a polynomial function.

Use the graph to determine

• any zeros,
• the $y$-intercept,
• the coordinates of any turning points,
• any relative minimums or maximums,
• the end behavior, and
• any symmetries.
Solution

Let's determine the characteristics in order from the list.

Example

### Zeros

Zeros are the $x$-values where the value of the function is equal to $0.$ In other words, where the graph intersects the $x$-axis. In the figure, we can see that this occurs at approximately $x = \text{-}6,$ $x = 1,$ $x = 3,$ and $x = 6.5.$

Example

### $y$-intercept

A $y$-intercept is the $y$-value of a function where $x = 0$. In other words, where its graph intersects the $y$-axis. In the figure, we can see that this occurs at approximately $y = \text{-} 4.$

Example

### Turning points

Turning points are found where the graph turns from increasing to decreasing, or vice versa. Notice that the function has the following increasing and decreasing intervals. \begin{aligned} \textbf{decreasing:}\,\, \text{-} \infty & < x < \text{-}1 \\ \,\, 3 & < x < 5 \\ \textbf{increasing:}\,\, \text{-} 1 & < x < 2 \\ \,\, 5 & < x <+ \infty \\ \end{aligned}

Thus, the three turning points are $(\text{-}1,\text{-}5), \quad (5,\text{-}7), \quad \text{and} \quad (2,3).$ The rightmost minimum point is an absolute minimum point because it is where the function has its lowest value. The other minimum and maximum points are relative.

Example

### End Behavior

The end behavior shows the graph extending upward when $x$ approaches both positive and negative infinity.

Assuming that there are no additional turning points outside the graph, we can conclude that the degree of the function is even, as the directions are the same. The leading coefficient of the polynomial must be positive, as the right-end extends upward.

Example

### Symmetries

Finally, we must determine if the function has any symmetries. If we look at the graph, we can see that it is not symmetric with respect to the $y$-axis, and also not symmetric about the origin. Thus, the function has neither even nor odd symmetry.

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Exercise

Draw a graph of a polynomial function, $f(x),$ having these characteristics.

• $f(x)<0$ when $\text{-} 3 < x < 1$ and $x>3$
• $f(x)>0$ when $x < \text{-} 3$ and $1 < x < 3$
• $f(x)$ is decreasing when $x<\text{-} 1$ and $x>2$
• $f(x)$ is increasing when $\text{-} 1
Solution

To draw this graph, we'll make sense of the first two bullets then the last two. If $f(x)<0,$ the graph lies below the $x$-axis. Similarly, when $f(x)>0,$ the graph lies above the $x$-axis. Notice that, to go from below to above the $x$-axis, the graph must cross it, which happens at a zero. Thus, the zeros are: $(\text{-} 3,0),\quad (1,0),\quad \text{and} \quad (3,0).$ We'll plot these points in a coordinate plane.

Two of these points, $(\text{-} 3, 0)$ and $(3,0),$ lie in intervals where $f(x)$ is decreasing. The third point, $(1,0),$ is instead in an interval where the function is increasing. Using this, we can draw short sections of the function through the zeros.

When a function changes from increasing to decreasing, or vice versa, it does so at a turning point. From the given information, these changes occur at $x = \text{-} 1$ and $x = 2.$ We can extend the already-drawn sections to these $x$-values.

Note that, because we don't have the exact coordinates of the turning point, their position is approximate. The function is decreasing both when $x<\text{-} 1$ and when $x>2.$ Since we know that, we can now draw the rest of the graph.

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