Features of Polynomial Graphs

Just from the degree and the leading coefficient of a polynomial, you can say a lot about its graph: where its ends point, how many times it can cross the $x$-axis, and how many hills and valleys it can have. You can even find the degree from a table of values. This page collects these tools so you can picture a polynomial before you graph it.

Key ideas

Finite differences

Take a table of values with equally spaced $x$-values. Subtract each $y$-value from the next to get the first differences. Subtract those to get the second differences, and so on.

For a polynomial of degree $n$:

• the $n$th differences are all the same (constant), and
• with $x$-values $1$ apart, that constant equals $a \cdot n!$, where $a$ is the leading coefficient.

Recall $n! = n(n - 1)\cdots(2)(1)$, so $2! = 2$, $3! = 6$, $4! = 24$, $5! = 120$.

Degree	Constant differences	Value (steps of 1)
$1$ (linear)	first	$a$
$2$ (quadratic)	second	$2a$
$3$ (cubic)	third	$6a$
$4$ (quartic)	fourth	$24a$

If the $x$-values go up by a step $h$ instead of $1$, the constant is $a \cdot n! \cdot h^n$.

End behaviour

End behaviour describes what $y$ does as $x$ becomes very large ($x \to \infty$) or very negative ($x \to -\infty$). For large $|x|$, the leading term $a_n x^n$ is so much bigger than the others that it decides everything. So you only need two things: is the degree odd or even, and is the leading coefficient positive or negative?

The degree (odd or even) and the sign of the leading coefficient set the end behaviour.

Degree	Leading coefficient	As $x \to -\infty$	As $x \to \infty$	Quadrants
odd	positive	$y \to -\infty$	$y \to \infty$	Q3 to Q1
odd	negative	$y \to \infty$	$y \to -\infty$	Q2 to Q4
even	positive	$y \to \infty$	$y \to \infty$	Q2 to Q1
even	negative	$y \to -\infty$	$y \to -\infty$	Q3 to Q4

Odd degree: the ends go in opposite directions (like a line). Even degree: the ends go the same way (like a parabola).

Intercepts and turning points

A turning point is where the graph changes from rising to falling (a local maximum) or from falling to rising (a local minimum). For a polynomial of degree $n \ge 1$:

	Odd degree	Even degree
$x$-intercepts	at least $1$, at most $n$	from $0$ up to $n$
turning points	an even number, at most $n - 1$	an odd number, at most $n - 1$

Why must an odd-degree polynomial have an $x$-intercept? Its ends point in opposite directions, one below the $x$-axis and one above, and its graph has no breaks, so it has to cross somewhere.

So a quartic ($n = 4$) has $0$, $1$, $2$, $3$, or $4$ $x$-intercepts and either $1$ or $3$ turning points.

Domain and range

The domain of every polynomial function is $\{x \in \mathbb{R}\}$.

• Odd degree: one end goes up forever and the other goes down forever, so the range is $\{y \in \mathbb{R}\}$.
• Even degree: both ends go the same way, so there’s a lowest point (if $a \gt 0$) or a highest point (if $a \lt 0$). The range is $\{y \in \mathbb{R} \mid y \ge \text{minimum}\}$ or $\{y \in \mathbb{R} \mid y \le \text{maximum}\}$.

Worked examples

Example 1: Degree from a table

Use finite differences to find the degree and leading coefficient of the polynomial function in the table.

$x$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$-3$	$1$	$-1$	$3$	$25$	$77$

Solution. The $x$-values go up by $1$. Find the differences:

Differences
first	$4$	$-2$	$4$	$22$	$52$
second	$-6$	$6$	$18$	$30$
third	$12$	$12$	$12$

The third differences are constant, so the degree is $3$. The constant is $6a$:

$$6a = 12 \quad\Rightarrow\quad a = 2$$

The leading coefficient is $2$. (This table comes from $y = 2x^3 - 3x^2 - x + 1$.)

Example 2: Reading features from the equation

For $f(x) = -3x^5 + 2x^2 - x + 4$, describe the end behaviour, and give the possible numbers of $x$-intercepts and turning points.

Solution. The degree is $5$ (odd) and the leading coefficient is $-3$ (negative).

• End behaviour: as $x \to -\infty$, $y \to \infty$, and as $x \to \infty$, $y \to -\infty$. The graph goes from Q2 to Q4.
• $x$-intercepts: at least $1$, at most $5$.
• Turning points: an even number up to $4$, so $0$, $2$, or $4$.
• Range: $\{y \in \mathbb{R}\}$, since the degree is odd.

Example 3: Working backwards from a description

A polynomial’s graph goes from Q3 to Q1, has $4$ turning points, and crosses the $x$-axis $3$ times. What is the least possible degree, and what is the sign of its leading coefficient?

Solution.

• Q3 to Q1 means the ends go in opposite directions, so the degree is odd, and the right end rises, so the leading coefficient is positive.
• $4$ turning points means $n - 1 \ge 4$, so $n \ge 5$.

The least odd degree that’s at least $5$ is $5$. So the least possible degree is $5$, with a positive leading coefficient. (Three $x$-intercepts is fine: a quintic can have from $1$ to $5$.)

Example 4: Range of an even-degree polynomial

Find the range of $f(x) = x^4 - 2x^2$.

Solution. The degree is even and $a = 1 \gt 0$, so there is a minimum value. Treat $x^2$ like a single variable and complete the square:

$$x^4 - 2x^2 = (x^2)^2 - 2(x^2) + 1 - 1 = (x^2 - 1)^2 - 1$$

A square is never negative, so $(x^2 - 1)^2 \ge 0$ and $f(x) \ge -1$. The value $-1$ happens when $x^2 = 1$, at $x = \pm 1$.

Range: $\{y \in \mathbb{R} \mid y \ge -1\}$.

Common mistakes

Using unequally spaced $x$-values for finite differences. Differences only work if the $x$-values go up by the same step every time. Check the $x$-row first.

Forgetting the factorial. Constant third differences of $12$ mean $6a = 12$, so $a = 2$, not $12$. The constant is $a \cdot n!$.

Reading end behaviour from the first term written. For $f(x) = 4 + x - 2x^3$, the leading term is $-2x^3$, so the graph goes from Q2 to Q4. Always find the highest power first.

Saying a polynomial of degree $n$ has exactly $n$ $x$-intercepts or exactly $n - 1$ turning points. These are maximums. The parabola $y = x^2 + 1$ has degree $2$ but no $x$-intercepts.

Giving an odd-degree polynomial a limited range. If the degree is odd, the graph goes off to $\infty$ at one end and $-\infty$ at the other, so the range is all real numbers. Only even-degree polynomials have a maximum or minimum value.

Practice

1. (Warm-up) Describe the end behaviour of $y = 4x^3 - x$, using quadrants.

Solution

Degree $3$ (odd), leading coefficient $4$ (positive): as $x \to -\infty$, $y \to -\infty$, and as $x \to \infty$, $y \to \infty$. The graph goes from Q3 to Q1.

2. (Warm-up) For $y = -x^6 + 2x$, describe the end behaviour, and give the maximum number of $x$-intercepts and of turning points.

Solution

Degree $6$ (even), leading coefficient $-1$ (negative): both ends go down, so the graph goes from Q3 to Q4.

At most $6$ $x$-intercepts and at most $5$ turning points.

3. (Warm-up) A cubic function’s table of values (with $x$ going up by $1$) has constant third differences of $-24$. Find the leading coefficient.

Solution

$$6a = -24 \quad\Rightarrow\quad a = -4$$

4. (Core) Find the degree and the leading coefficient of the polynomial function in this table.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$18$	$-2$	$0$	$0$	$22$	$138$

Solution

Differences
first	$-20$	$2$	$0$	$22$	$116$
second	$22$	$-2$	$22$	$94$
third	$-24$	$24$	$72$
fourth	$48$	$48$

The fourth differences are constant, so the degree is $4$:

$$24a = 48 \quad\Rightarrow\quad a = 2$$

(The table comes from $y = 2x^4 - 3x^2 + x$.)

5. (Core) A quartic function has a negative leading coefficient. Give its end behaviour, the possible numbers of $x$-intercepts and turning points, and say whether it has a maximum or a minimum value.

Solution

Even degree, negative leading coefficient: Q3 to Q4 (both ends down).

$x$-intercepts: $0$, $1$, $2$, $3$, or $4$. Turning points: $1$ or $3$ (an odd number, at most $3$).

Both ends go down, so it has a maximum value; the range is $\{y \in \mathbb{R} \mid y \le M\}$, where $M$ is that maximum value.

6. (Core) An open-top box is made from a $20$ cm by $20$ cm sheet of cardboard by cutting a square of side $x$ cm from each corner and folding up the sides. Its volume is $V(x) = x(20 - 2x)^2$.

• (a) Find the degree and the leading coefficient of $V$, and describe the end behaviour of $y = V(x)$.
• (b) Which values of $x$ make sense for the box?

Solution

(a) Expand:

$$V(x) = x(400 - 80x + 4x^2) = 4x^3 - 80x^2 + 400x$$

Degree $3$, leading coefficient $4$. As a function, $y = V(x)$ goes from Q3 to Q1.

(b) The cut must be positive, and you can’t cut more than half of each $20$ cm side, so $0 \lt x \lt 10$. Outside that interval the formula still gives numbers, but they don’t describe a real box.

7. (Core) A polynomial’s graph starts in Q2 and ends in Q1. It has $3$ turning points and $2$ $x$-intercepts. Find the least possible degree and the sign of the leading coefficient.

Solution

Q2 to Q1: both ends go up, so the degree is even and the leading coefficient is positive.

$3$ turning points means $n - 1 \ge 3$, so $n \ge 4$. The least possible degree is $4$, with a positive leading coefficient.

8. (Challenge) Find the range of $f(x) = -x^4 + 8x^2 - 7$.

Solution

Even degree with a negative leading coefficient, so there is a maximum. Complete the square in $x^2$:

$$\begin{aligned} f(x) &= -(x^4 - 8x^2) - 7 \\ &= -(x^4 - 8x^2 + 16 - 16) - 7 \\ &= -(x^2 - 4)^2 + 16 - 7 \\ &= -(x^2 - 4)^2 + 9 \end{aligned}$$

Since $-(x^2 - 4)^2 \le 0$, $f(x) \le 9$, with $f(x) = 9$ at $x = \pm 2$.

Range: $\{y \in \mathbb{R} \mid y \le 9\}$.

9. (Challenge) Make a table of values for $y = x^3$ using $x = 0, 2, 4, 6, 8$. Show that the third differences are constant. Why is the constant not $6$?

Solution

$y$-values: $0, 8, 64, 216, 512$.

• First differences: $8, 56, 152, 296$
• Second differences: $48, 96, 144$
• Third differences: $48, 48$

The third differences are constant, so the degree is $3$. But the step is $h = 2$, not $1$, so the constant is

$$a \cdot 3! \cdot h^3 = 1 \cdot 6 \cdot 8 = 48$$

The value $6a$ only applies when the $x$-values go up by $1$.