Graphs of Functions

Intuition

In an earlier course you graphed a function by making a table, plotting the points, and connecting them. That skill still works, and this lesson does not repeat it. It asks a harder question the table method quietly skips: how do you know the curve does what your smooth line says it does between the points you plotted? The honest answer is that a handful of points cannot know. Infinitely many different functions pass through any finite set of plotted points, so joining the dots is a guess. This lesson replaces the guess with proof. From the factored form of a rule you can prove where it is zero, and where it is positive or negative. The same form also settles whether the rule is symmetric, how it behaves far out, and where its domain punches a hole in the picture. Those features are earned, and the smooth curve is drawn to fit them, not the other way around.

Learning goals

Draw the graph as the relation $(x, f(x))$
Show why a finite table cannot pin a function down
Find zeros from a factored form and read each sign interval
Tell a crossing from a touch by the factor's parity
Test even and odd as reflection and rotation symmetry
Read end behavior by factoring out the highest power

The graph is the relation, drawn

In Relations and Functions you saw that a function’s rule and domain together are a set of ordered pairs, one pair for each input. The graph is nothing more than that set of pairs plotted as points: the graph of $f$ is the collection of all points $(x, f(x))$ as $x$ runs over the domain. There is no new object here, only the same set of pairs shown on paper.

Seeing the graph this way settles what the vertical line test really is. A vertical line $x = a$ collects every plotted point sharing the input $a$ . So the test “no vertical line meets the graph more than once” is just the picture of “no input has two outputs.” The vertical line test is therefore a fact about the picture of a relation, not the definition of a function. Relations and Functions proved that equivalence in full, so we take it as known and spend this lesson on the reverse direction. Given a rule, what can we prove about the picture before we ever plot a point?

A finite table cannot pin down a function

Start with the flaw in connect-the-dots, because everything else is a repair for it. Suppose you know a function $f$ only through a table: its values at finitely many inputs $x_1, x_2, \dots, x_n$ . The table looks like solid evidence, but it leaves the function almost completely open.

A finite table is matched by infinitely many different functions#

Let $f$ be any function and let $x_1, x_2, \dots, x_n$ be the sampled inputs. Pick any constant $c$ and build a new function

g(x) = f(x) + c\,(x - x_1)(x - x_2)\cdots(x - x_n).

Evaluate $g$ at any sampled input $x_k$ . Among the factors of the product there sits $(x_k - x_k) = 0$ , so the entire product is $0$ , and therefore

g(x_k) = f(x_k) + c\cdot 0 = f(x_k).

This holds at every sampled input and for every choice of $c$ , so $g$ reproduces the table exactly, no matter which $c$ you chose. Yet away from the samples the product $(x - x_1)\cdots(x - x_n)$ is not zero, so for any input $x$ that was not sampled and any $c \ne 0$ ,

g(x) - f(x) = c\,(x - x_1)\cdots(x - x_n) \ne 0,

which means $g$ and $f$ disagree there. Different values of $c$ give different functions, all sharing the one table, so infinitely many functions fit it. A finite table simply does not determine the function it came from.

Make it concrete with the friendliest function there is, the identity $f(x) = x$ , sampled at $x = 1, 2, 3$ . Its table is $(1,1)$ , $(2,2)$ , $(3,3)$ , three points marching up a straight line. Nothing about that table forces a line, though. Take $c = 1$ and the three sampled inputs and build

g(x) = x + (x - 1)(x - 2)(x - 3).

At each of $x = 1, 2, 3$ one factor vanishes, so $g$ agrees with $f$ there, yet $g$ is a cubic that leaves the line everywhere else. The next figure plots both functions through the identical three points.

Two different functions through the same three table points. The straight line is f(x) = x; the curve is g(x) = x + (x - 1)(x - 2)(x - 3). Both pass exactly through (1, 1), (2, 2), and (3, 3), yet the cubic plunges to g(0) = -6 while f(0) = 0. The table cannot tell them apart.

Worked example 1 Two functions, one table

Show that $f(x) = x$ and $g(x) = x + (x - 1)(x - 2)(x - 3)$ share the table at $x = 1, 2, 3$ but part ways off it.

Check the three sampled inputs. At each one, a single factor of the product is zero, so the whole product drops out:

\begin{aligned} g(1) &= 1 + (0)(-1)(-2) = 1, \\ g(2) &= 2 + (1)(0)(-1) = 2, \\ g(3) &= 3 + (2)(1)(0) = 3. \end{aligned}

So $g$ matches $f$ at every row of the table. Now expand the product to see $g$ as a single rule:

g(x) = x + (x^3 - 6x^2 + 11x - 6) = x^3 - 6x^2 + 12x - 6.

Evaluate both functions at inputs the table never sampled:

g(0) = -6 \ \text{ while } \ f(0) = 0, \qquad g(4) = 10 \ \text{ while } \ f(4) = 4.

The two functions sit $6$ units apart at $x = 0$ and again at $x = 4$ , yet their tables at $x = 1, 2, 3$ are identical. A table is evidence about finitely many inputs; a graph is a claim about infinitely many, and only proof closes the gap between them.

Graph the structure you can prove

If plotting points cannot justify a curve, what can? Four features of a rule can be derived exactly, with algebra you already own. Each one pins the curve down at infinitely many inputs at once, which is what a table could never do.

Zeros: where the graph meets the horizontal axis

The zeros of $f$ are the inputs with $f(x) = 0$ . A point sits on the horizontal axis exactly when its height is $0$ . So the zeros are precisely the $x$ -coordinates where the graph crosses or touches that axis, the x-intercepts. Finding them is solving $f(x) = 0$ , and the fastest route is a factored form, because a product is zero exactly when one of its factors is zero. Our running example for the rest of the lesson is

f(x) = x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2),

whose zeros are $x = 0$ , $x = 2$ , and $x = -2$ . Those three inputs are the only places the graph can meet the horizontal axis.

Sign: which side of the axis, from the factors

A function written as a product of linear factors keeps to one side of the axis on each interval between consecutive roots. The factors tell you which side, without any appeal to the shape of the curve. This is special to factored forms, not a fact about every function. A function can switch sign without passing through a zero, at a point where it is undefined, the way $\frac{1}{x}$ jumps from negative to positive across $0$ . Two facts settle the sign of a factored form, and both are pure algebra.

The sign of a factored function is the product of its factor signs#

A single linear factor $(x - r)$ is negative exactly when $x < r$ , zero exactly when $x = r$ , and positive exactly when $x > r$ . This is just the statement that $x - r < 0$ means the same as $x < r$ , read three ways, so each factor changes sign only at its own root.

Now suppose $f(x) = a\,(x - r_1)(x - r_2)\cdots$ is a product of a nonzero constant $a$ and linear factors. From pre-algebra, the sign of a product of nonzero numbers is the product of their signs: two negatives make a positive, and each extra negative flips the result. Pick any interval lying strictly between two consecutive roots. No factor has a root inside that interval, so no factor changes sign across it, so every factor holds one fixed sign there. The sign of $f$ on the interval is then the product of those fixed signs, a single computation with no reference to continuity or to the graph’s shape.

Multiplicity follows from the same rule. Group a repeated root as $(x - r)^k$ . If $k$ is even, then $(x - r)^k = \big((x - r)^2\big)^{k/2}$ is a square raised to a power, so it is positive for every $x \ne r$ and never negative. It therefore contributes the same sign, $+$ , on both sides of $r$ , so $f$ keeps one sign across $r$ . Geometrically, the graph meets the axis at $r$ but stays on the same side of it, touching and turning back rather than crossing. If $k$ is odd, then $(x - r)^k = (x - r)\big((x - r)^2\big)^{(k-1)/2}$ has the same sign as the single factor $(x - r)$ . That factor is negative to the left of $r$ and positive to the right, so $f$ flips sign across $r$ and the graph crosses.

A factor that is never zero and never negative is easy to handle: it just contributes a $+$ everywhere. The quadratic $x^2 + 1$ is one, since $x^2 + 1 \ge 1 > 0$ for every real $x$ . It has no real root, so it never splits an interval and never changes the sign of the product. You can therefore drop that factor from the sign chart entirely, keeping only the linear factors that do cross zero.

Worked example 2 Zeros and sign of $f(x) = x^3 - 4x$

Use the factored form $f(x) = x(x - 2)(x + 2)$ . The roots $-2, 0, 2$ split the number line into four intervals. On each one, read the sign of each factor and multiply.

For $x < -2$ , all three of $x$ , $x - 2$ , and $x + 2$ are negative:

(-)(-)(-) = -, \qquad \text{so } f < 0 \text{ on } x < -2.

For $-2 < x < 0$ , only $x + 2$ has turned positive:

(-)(-)(+) = +, \qquad \text{so } f > 0 \text{ on } -2 < x < 0.

For $0 < x < 2$ , now $x$ is positive too but $x - 2$ is still negative:

(+)(-)(+) = -, \qquad \text{so } f < 0 \text{ on } 0 < x < 2.

For $x > 2$ , every factor is positive, so $f > 0$ . Each root has power $1$ , an odd power, so the sign flips at all three and the graph crosses the axis at each. The sign pattern below the axis, above, below, above reads off directly.

A sign chart for f(x) = x(x - 2)(x + 2). Each factor's sign is marked on the three intervals it defines, and the bottom row multiplies them to give the sign of f. The product is negative, positive, negative, positive across the four intervals, and since every root has odd power the graph crosses at each.

Worked example 3 An even power that touches: $f(x) = (x - 1)^2(x + 2)$

The zeros are $x = 1$ from the factor $(x - 1)^2$ and $x = -2$ from $(x + 2)$ . Read the sign on the three intervals, remembering that $(x - 1)^2$ is a square, so it is positive everywhere except at $x = 1$ and never contributes a negative.

For $x < -2$ , the factor $(x + 2)$ is negative while $(x - 1)^2$ is positive:

(+)(-) = -, \qquad \text{so } f < 0 \text{ on } x < -2.

For $-2 < x < 1$ and again for $x > 1$ , the factor $(x + 2)$ is positive and $(x - 1)^2$ is positive:

(+)(+) = +, \qquad \text{so } f > 0 \text{ on both } -2 < x < 1 \text{ and } x > 1.

The sign is positive on both sides of $x = 1$ . At the simple root $x = -2$ (power $1$ , odd) the sign flips, so the graph crosses. At the double root $x = 1$ (power $2$ , even) the sign is the same on both sides. So the graph touches the axis at $(1, 0)$ and stays above it, turning back instead of crossing. The multiplicity of a root, not the fact that it is a root, decides crossing versus touching.

Check your understanding

For $f(x) = (x + 3)(x - 4)^2$ , what does the graph do at $x = 4$ ?

A It crosses the axis, because

4

is a zero B It touches the axis at

(4, 0)

and stays on one side C It has a vertical asymptote at

x = 4

D It jumps from one side of the axis to the other

Symmetry: proving half the graph determines the other half

Some graphs repeat themselves under a reflection or a rotation, and when they do you can prove one half of the curve from the other, halving the work. Two kinds of symmetry have names. A function is even when $f(-x) = f(x)$ for every $x$ in its domain, and odd when $f(-x) = -f(x)$ for every $x$ .

Both definitions ask about $f(-x)$ , so both are meaningless unless $-x$ is in the domain whenever $x$ is. Say the domain is symmetric about $0$ when $x$ is in it exactly when $-x$ is. If it is not, for instance the domain $x \ge 0$ of $\sqrt{x}$ , then $f(-x)$ can be undefined and neither question can even be asked. Assume a symmetric domain from here on, and the two symmetries are exact.

Even means $y$ -axis symmetry; odd means $180^\circ$ symmetry#

Assume the domain $D$ is symmetric about $0$ . Reflection across the $y$ -axis sends a point $(x, y)$ to $(-x, y)$ . Apply it to the graph $\{(x, f(x)) : x \in D\}$ to get the reflected set $\{(-x, f(x)) : x \in D\}$ . Reindex with $u = -x$ ; because $D$ is symmetric, $u$ ranges over all of $D$ , and the reflected set becomes $\{(u, f(-u)) : u \in D\}$ , which is the graph of the rule $u \mapsto f(-u)$ . Two functions on the same domain are equal exactly when they agree at every input, the equality test from Function Notation. So this reflected graph equals the original graph $\{(u, f(u))\}$ exactly when $f(-u) = f(u)$ for all $u$ in $D$ . That condition is the definition of even, and the “exactly when” runs in both directions. So $f$ is even if and only if its graph is unchanged by reflection across the $y$ -axis.

The odd case is the same argument with a $180^\circ$ rotation about the origin, which sends $(x, y)$ to $(-x, -y)$ . Rotating the graph gives $\{(-x, -f(x)) : x \in D\}$ , and the same reindex turns it into $\{(u, -f(-u)) : u \in D\}$ , the graph of $u \mapsto -f(-u)$ . It equals the original graph exactly when $-f(-u) = f(u)$ for all $u$ , that is $f(-u) = -f(u)$ , the definition of odd. So $f$ is odd if and only if its graph is unchanged by a $180^\circ$ rotation about the origin.

Finally, suppose $f$ is both even and odd. Then $f(-x) = f(x)$ and $f(-x) = -f(x)$ for every $x$ in $D$ , so $f(x) = -f(x)$ , which gives $2f(x) = 0$ and $f(x) = 0$ . Every value is $0$ , so $f$ is the zero function, and the zero function plainly satisfies both definitions. The only function that is both even and odd on a symmetric domain is the zero function.

One quick consequence is worth recording. If an odd function has $0$ in its domain, then setting $x = 0$ in $f(-x) = -f(x)$ gives $f(0) = -f(0)$ , so $2f(0) = 0$ and $f(0) = 0$ . So whenever $0$ is in the domain, an odd function’s graph passes through the origin. An odd function whose domain omits $0$ , like $\frac{1}{x}$ , need not, and an even function carries no such forced value at all, since $f(0) = f(0)$ says nothing.

A warning about a tempting shortcut. It is often said that even powers give even functions and odd powers give odd functions. The claim is true for a single power $x^n$ , and it stays true for a polynomial whose terms all share the same exponent parity. It fails the moment the parities mix. The function $f(x) = x^2 + x$ has one even-degree term and one odd-degree term, and it is neither even nor odd:

f(-x) = (-x)^2 + (-x) = x^2 - x,

which equals $f(x) = x^2 + x$ only at $x = 0$ , and equals $-f(x) = -x^2 - x$ only at $x = 0$ . A typical function is neither even nor odd; symmetry is the special case, and you confirm it by testing $f(-x)$ , not by glancing at exponents.

Worked example 4 Testing even, odd, or neither

Classify each function on its (symmetric) domain by computing $f(-x)$ .

First, the running cubic $f(x) = x^3 - 4x$ , with domain all real numbers:

f(-x) = (-x)^3 - 4(-x) = -x^3 + 4x = -(x^3 - 4x) = -f(x).

Since $f(-x) = -f(x)$ , the cubic is odd, so its graph has $180^\circ$ symmetry about the origin.

Next, the rational function $f(x) = \dfrac{x}{x^2 + 1}$ . The denominator $x^2 + 1$ is never $0$ , so the domain is all real numbers, which is symmetric. Substitute $-x$ , squaring inside the denominator:

f(-x) = \frac{-x}{(-x)^2 + 1} = \frac{-x}{x^2 + 1} = -f(x),

so this one is odd as well. Now the even partner $g(x) = \dfrac{1}{x^2 + 1}$ :

g(-x) = \frac{1}{(-x)^2 + 1} = \frac{1}{x^2 + 1} = g(x),

so $g$ is even, with a graph symmetric across the $y$ -axis. The parity is decided by what happens to the sign under $x \mapsto -x$ , and a rational rule is tested exactly the same way a polynomial is.

Check your understanding

On the domain of all real numbers, which describes $f(x) = x^4 - 3x^2$ ?

A Odd, because it has a rotation symmetry B Even, because

f(-x) = f(x)

C Neither even nor odd D Both even and odd

End behavior: what the far ends do, without limits

The last provable feature is what the graph does far from the origin, and you can settle it with a factoring trick and an explicit inequality, no limits required. Factor out the highest power. For the running cubic,

f(x) = x^3 - 4x = x^3\left(1 - \frac{4}{x^2}\right).

Look at the bracket. Whenever $\lvert x \rvert > 2$ , we have $x^2 > 4$ , so $\dfrac{4}{x^2} < 1$ and the bracket $1 - \dfrac{4}{x^2}$ is a positive number less than $1$ . A positive bracket cannot change the sign of $x^3$ . So for $\lvert x \rvert > 2$ the value $f(x)$ has the same sign as $x^3$ : positive and large when $x > 2$ , negative and large when $x < -2$ . For the magnitude, take $\lvert x \rvert \ge 3$ ; then $\dfrac{4}{x^2} \le \dfrac{4}{9}$ , so the bracket is at least $\dfrac{5}{9}$ , giving

\lvert f(x) \rvert = \lvert x \rvert^3 \left(1 - \frac{4}{x^2}\right) \ge \frac{5}{9}\,\lvert x \rvert^3,

which grows past any bound as $\lvert x \rvert$ increases. So far to the right the graph climbs like $x^3$ , and far to the left it falls like $x^3$ . The middle wiggle near the origin is irrelevant to the ends. This is a complete argument with a stated threshold; a later course rephrases it with limits, but the algebra already decides it. The sign of the leading coefficient rides along untouched. Had the cubic been $-x^3 + 4x = x^3\left(-1 + \dfrac{4}{x^2}\right)$ , the bracket would be negative for $\lvert x \rvert > 2$ , and both ends would flip, falling on the right and rising on the left.

Worked example 5 End behavior of the quartic $f(x) = x^4 - 5x^2 + 4$

Factor out the highest power, $x^4$ :

f(x) = x^4\left(1 - \frac{5}{x^2} + \frac{4}{x^4}\right).

The leading factor $x^4$ is never negative, so the sign far out depends only on the bracket. The term $\dfrac{4}{x^4}$ is positive, so dropping it only lowers the bracket, giving the clean bound

1 - \frac{5}{x^2} + \frac{4}{x^4} \ge 1 - \frac{5}{x^2}.

For $\lvert x \rvert \ge 3$ we have $\dfrac{5}{x^2} \le \dfrac{5}{9}$ , so the bracket is at least $1 - \dfrac{5}{9} = \dfrac{4}{9} > 0$ . Therefore for $\lvert x \rvert \ge 3$ ,

f(x) = x^4\left(1 - \frac{5}{x^2} + \frac{4}{x^4}\right) \ge \frac{4}{9}\,x^4,

which is positive and grows without bound. Both ends of the quartic rise to large positive values, matching its leading term $x^4$ . A degree-four graph with a positive leading coefficient goes up on the left and up on the right, and this inequality is why.

The method survives an interior term of odd power, where the correction changes sign with $x$ . Take $2x^4 - 3x = x^4\left(2 - \dfrac{3}{x^3}\right)$ : the term $\dfrac{3}{x^3}$ is negative for $x > 0$ and positive for $x < 0$ , but only its size matters. Once $\lvert x \rvert \ge 2$ we have $\left\lvert \dfrac{3}{x^3} \right\rvert \le \dfrac{3}{8}$ , so the bracket stays between $2 - \dfrac{3}{8}$ and $2 + \dfrac{3}{8}$ , positive at both ends. The sign of the correction does not change the conclusion; the graph rises to large positive values on the far left as well as the far right.

Check your understanding

Far to the right, where $x$ is large and positive, how does $f(x) = x^3 - 4x$ behave?

A It rises to large positive values, like

x^3

B It falls to large negative values C It levels off toward the height

y = 0

D It settles onto the line

y = -4x

The domain is visible in the graph

A rule can look like a familiar curve yet be missing a single input, and that missing input shows up on the graph as a gap. This is the sharpest payoff of taking the domain seriously, a theme from Relations and Functions and Function Notation.

Consider $f(x) = \dfrac{x^2 - 4}{x - 2}$ . The denominator is $0$ at $x = 2$ , so $2$ is not in the domain. For every other input the numerator factors and the common factor cancels:

\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2, \qquad x \ne 2.

So the graph agrees with the line $y = x + 2$ at every allowed input. But $x = 2$ is not allowed, and the point the line would occupy there, $(2, 4)$ , is removed, leaving a hole. Function Notation proved that $\dfrac{x^2 - 4}{x - 2}$ and $x + 2$ are different functions precisely because their domains differ at $x = 2$ , and the hole is where that difference becomes visible.

The graph of f(x) = (x squared minus 4) divided by (x minus 2). Everywhere it is defined the rule equals x + 2, so the graph is the line y = x + 2, but x = 2 is excluded from the domain, so the point (2, 4) is punched out as an open hole.

Worked example 6 Locating a hole

Find the hole in the graph of $f(x) = \dfrac{x^2 - 4}{x - 2}$ , and give its exact coordinates.

First find the excluded input. The denominator $x - 2$ is $0$ at $x = 2$ , so $2$ is not in the domain, and that is the only excluded input. Next simplify the rule for the inputs that remain, factoring the numerator as a difference of squares:

f(x) = \frac{(x - 2)(x + 2)}{x - 2} = x + 2, \qquad x \ne 2.

The graph is therefore the line $y = x + 2$ with one point missing. To find the missing point’s height, evaluate the simplified rule at the excluded input:

x + 2 \Big|_{x = 2} = 2 + 2 = 4.

The hole sits at $(2, 4)$ . It is a genuine feature of the picture, not bookkeeping: the excluded input is invisible in the simplified formula $x + 2$ but plainly visible as a gap in the graph.

Not every excluded input makes a hole. The rule $y = \dfrac{1}{x - 2}$ also excludes $x = 2$ . But there the factor causing trouble does not cancel, so instead of a single missing point the graph shoots off along a vertical line at $x = 2$ , a vertical asymptote. Two rules can exclude the same input and look completely different near it. Which one you get depends on whether the offending factor cancels, and the full study of these graphs waits for a later chapter. The point for now is only that the domain is never invisible; it leaves a mark.

Putting it together

Collect what we proved about the running cubic $f(x) = x^3 - 4x = x(x - 2)(x + 2)$ : its zeros are $-2, 0, 2$ , and its sign runs below, above, below, above the axis across those roots. The cubic is odd, so its graph has $180^\circ$ symmetry about the origin. Far out it climbs like $x^3$ on the right and falls like $x^3$ on the left. Every one of those is a proven statement about infinitely many inputs. The figure marks them, then draws a smooth curve to fit. That final smooth stroke is the one thing on the picture we did not prove. It is an interpolation between features we did prove, and its only job is to respect them.

The assembled graph of f(x) = x cubed minus 4x. The three x-intercepts, the sign on each interval, the 180-degree rotational symmetry about the origin, and the ends rising and falling like x cubed are all proved. The smooth curve joining them is drawn to fit those proven features, not to establish them.

Common mistakes

Connecting the dots as if a table proved the curve. A finite table is matched by infinitely many functions, so a smooth line through plotted points is a guess until you prove the features between them.
Justifying a sign change by “the graph must cross zero to switch sides.” That is the Intermediate Value Theorem, a later idea about continuity; here the sign comes only from multiplying the signs of the factors, which needs no such theorem.
Treating every zero as a crossing. A factor to an even power keeps the same sign on both sides of its root, so the graph touches the axis and turns back there. Only an odd power flips the sign and crosses.
Testing even or odd on a domain that is not symmetric about $0$ . If $x$ is allowed but $-x$ is not, then $f(-x)$ is undefined and the question has no answer, so check the domain before you test.
Reading even and odd off the exponents of a mixed expression. The shortcut holds only for a single power or a polynomial whose terms all share exponent parity. For instance $x^2 + x$ is neither even nor odd, despite having one even and one odd term.
Calling $\dfrac{x^2 - 4}{x - 2}$ the same function as $x + 2$ . They agree wherever both are defined, but $x = 2$ is missing from the first domain, so its graph carries a hole at $(2, 4)$ that the plain line does not.
Claiming a function is “both even and odd” casually. On a symmetric domain the only function that is both is the zero function, since $f(-x) = f(x)$ and $f(-x) = -f(x)$ together force $f(x) = 0$ .

Key takeaways

The graph of $f$ is the relation drawn: the set of points $(x, f(x))$ . The vertical line test is a fact about that picture, not the definition of a function.
A finite table cannot pin down a function. For samples $x_1, \dots, x_n$ , the function $g(x) = f(x) + c(x - x_1)\cdots(x - x_n)$ matches every sample for every $c$ yet differs everywhere else, so infinitely many functions share the table.
Zeros are the solutions of $f(x) = 0$ and are the $x$ -intercepts; a factored form finds them because a product is zero exactly when a factor is.
For a factored form, the sign on each interval between consecutive roots is the product of the factor signs. An odd-power factor flips the sign so the graph crosses; an even-power factor keeps it so the graph touches and turns back.
On a domain symmetric about $0$ , $f$ is even exactly when its graph is unchanged by reflection across the $y$ -axis. On that same domain, $f$ is odd exactly when its graph is unchanged by a $180^\circ$ rotation about the origin. The only function that is both is the zero function.
End behavior comes from factoring out the highest power: since $x^3 - 4x = x^3(1 - 4/x^2)$ and the bracket is positive for $\lvert x \rvert > 2$ , the ends track $x^3$ , with no limit needed.
The domain is visible: where a rule simplifies but drops an input, like $\dfrac{x^2 - 4}{x - 2} = x + 2$ missing $x = 2$ , the graph shows a hole. The smooth curve between proven features is an interpolation, not a proof.

Graphs of Functions

Learning goals

The graph is the relation, drawn

A finite table cannot pin down a function

Graph the structure you can prove

Zeros: where the graph meets the horizontal axis

Sign: which side of the axis, from the factors

Symmetry: proving half the graph determines the other half

End behavior: what the far ends do, without limits

The domain is visible in the graph

Putting it together

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)