Graphing Polynomial Functions

Learning goals

Read the blueprint off factored form
Change sign only at a real zero, and not always there
Cross at odd multiplicity and touch at even
Settle a whole interval with one test point
Note that nonreal roots give no intercepts
Leave the turning points to calculus rather than faking them

The blueprint hidden in the factored form

Write a polynomial in factored form and read what it says:

P(x) = 2(x+3)(x-1)^2(x-4)^3.

The degree is $1 + 2 + 3 = 6$ , because multiplying out takes one $x$ from the first factor, two from the second, and three from the third. The leading coefficient is $2$ , the number in front. The real zeros are $-3$ , $1$ , and $4$ , with multiplicities $1$ , $2$ , and $3$ . Every fact you need is sitting right there in the notation, and a sketch is just those facts drawn.

A sketch has to answer four questions, and the factored form answers all four:

Where do the two arms go? The degree and the leading coefficient settle the end behavior, which you already know how to read.
Where does the curve meet the x-axis? At the real zeros, and nowhere else.
How does it meet the axis at each zero? That is what the multiplicity is for, and it is the one genuinely new idea in this lesson.
Which side of the axis is the curve on between consecutive zeros? One test point per interval settles it.

There is also a free fifth anchor: the y-intercept, $P(0)$ , which is just the constant term of the expanded polynomial. In the example above, $P(0) = 2(3)(-1)^2(-4)^3 = 2 \cdot 3 \cdot 1 \cdot (-64) = -384$ , so the curve passes through $(0, -384)$ .

Questions 2, 3, and 4 all rest on a single fact from the start of the last chapter: the graph of a polynomial is one unbroken curve. No jumps, no holes, no gaps. Everything below is squeezed out of that one sentence.

Where a polynomial is allowed to change sign

A polynomial changes sign only at a real zero#

Suppose $a < b$ and that $P(a)$ and $P(b)$ have opposite signs, say $P(a) < 0$ and $P(b) > 0$ . The point $(a, P(a))$ lies below the x-axis and the point $(b, P(b))$ lies above it. Since the graph of a polynomial is a single unbroken curve, the piece of the curve joining those two points cannot leap across the axis. Instead that connecting piece of the polynomial’s curve has to meet the axis somewhere between those two points. A point where the curve meets the axis is a point where the height is zero, so there is some $c$ with $a < c < b$ and $P(c) = 0$ . That $c$ is a real zero.

Now read the statement backwards. If an interval contains no real zero of $P$ , then no two points of that interval can give values of opposite sign. The reason is that a sign change would force a real zero of $P$ to sit between those two points. So on an interval free of real zeros, $P$ keeps one and the same sign at every single point.

This is the engine of the whole lesson, so it is worth being precise about what it does and does not say. It says a sign change requires a real zero. It does not say that a real zero forces a sign change. The converse is false, and the next section is about exactly the zeros where the sign refuses to flip.

The immediate payoff is practical. The real zeros cut the number line into intervals. Inside one of those intervals there is no zero, so $P$ holds one sign the whole way across. Test one point and you have settled the entire interval. Pick whatever number is easiest to evaluate, and never pick a zero itself, since that returns $0$ and tells you nothing.

How the curve meets the axis at a zero

Recall what multiplicity means. Saying that $r$ is a zero of multiplicity $m$ means

P(x) = (x - r)^m \, G(x), \qquad G(r) \neq 0,

where $G$ is a polynomial: the factor $(x-r)$ divides $P$ exactly $m$ times and no more. The whole question of how the curve behaves at $r$ comes down to the sign of that product just to the left and just to the right of $r$ .

Odd multiplicity crosses, even multiplicity touches#

Let $r$ be a real zero of $P$ with multiplicity $m$ , and write $P(x) = (x-r)^m G(x)$ with $G(r) \neq 0$ .

First, pin down $G$ near $r$ . A polynomial of degree $k$ has at most $k$ real zeros, so $G$ has only finitely many, and $r$ is not one of them. Choose a distance $d$ smaller than the gap from $r$ to the nearest real zero of $G$ . If $G$ has no real zeros at all, then the distance $d$ can be anything you like. On the interval $r - d < x < r + d$ the polynomial $G$ has no zero, so by the result above it holds a single sign there. And since $r$ sits inside that interval, the single sign the polynomial $G$ holds there is the sign of $G(r)$ .

Next, the sign of the other factor. For $x > r$ the quantity $x - r$ is positive, so $(x-r)^m$ is positive no matter what $m$ is. For $x < r$ the quantity $x - r$ is negative, and a negative number raised to the power $m$ is positive when $m$ is even and negative when $m$ is odd.

Multiply the two signs together on our small interval, where $G$ never changes sign.

If $m$ is odd, then just to the right of $r$ the product $P(x)$ carries the sign of $G(r)$ , and just to the left it carries the opposite sign. The curve is on one side of the axis before $r$ and on the other side after it, and it is zero at $r$ itself. It crosses.

If $m$ is even, then $(x-r)^m$ is positive on both sides, so $P(x)$ carries the sign of $G(r)$ on both sides of $r$ . The curve approaches the axis, reaches it at $r$ , and leaves on the side it came from. It touches and turns back. In fact $P(r) = 0$ while $P$ has one constant sign nearby, so $r$ is a genuine turning point of the graph. That turning point is a local minimum if the constant sign is positive, and a local maximum if that sign is negative.

At a real zero there are only two possibilities, crossing or touching, and there are only two possibilities for $m$ , odd or even. The two implications above therefore run both ways: the graph crosses the x-axis at a real zero $r$ if and only if the multiplicity of $r$ is odd. And the graph touches without crossing at that same zero $r$ if and only if the multiplicity of $r$ is even.

Read the scope of that theorem carefully. It is a statement about a real zero of $P$ , and it describes only what happens at the axis. It says nothing whatever about turning points elsewhere: a graph can perfectly well turn around at a point high above the axis. At such a point no zero and no multiplicity is involved at all.

The proof also hands you a mental picture. Near $r$ , the factor $G(x)$ barely moves, sitting close to the fixed number $G(r)$ , while $(x-r)^m$ does all the work. So near a zero of multiplicity $m$ , the curve looks like the power curve $y = G(r)(x-r)^m$ , shifted to sit at $r$ :

Near a zero of multiplicity m, the curve looks like the power curve y = c(x - r) to the m-th power. Odd powers pass through the axis, even powers rebound off it, and the higher the power the more the curve flattens against the axis first. These pictures assume c is positive; a negative c flips each panel vertically, which swaps the sides the curve travels on but never changes whether it crosses.

Why a higher multiplicity flattens the curve

Look again at $|P(x)| = |x-r|^m \cdot |G(x)|$ . Close to $r$ the second factor is near the fixed number $|G(r)|$ , so the size of $P$ near $r$ is governed entirely by $|x-r|^m$ . Now watch what the exponent does to a small distance. At $d = 0.2$ away from the zero,

d^1 = 0.2, \qquad d^2 = 0.04, \qquad d^3 = 0.008, \qquad d^5 = 0.000\,32 .

Each extra power multiplies the height by another factor of $0.2$ . Raising the multiplicity from $1$ to $5$ therefore pulls the curve $0.2^4 = 1/625$ of the way toward the axis at that same distance. The same $G$ does the rest of the work, so the curve lies down almost flat against the axis before it finally pulls away. That is the visual signature of a high multiplicity: not a different kind of behavior, just a much slower start. The parity still decides whether the curve crosses or rebounds, and the size of $m$ only decides how flattened it looks on the way.

Check your understanding

For $P(x) = (x+1)^3(x-4)^2(x-7)$ , what does the graph do at $x = 4$ ?

A It crosses the x-axis at

x = 4

. B It touches the x-axis at

x = 4

and turns back. C It crosses the x-axis at

x = 4

, flattened against it. D It has no zero at

x = 4

The sign chart

Put the two ideas together. The real zeros chop the number line into intervals; on each interval $P$ holds one sign. So one test point per interval reveals the entire picture of which side of the axis the curve travels on. Take

Q(x) = (x+1)(x-2)^2 .

The real zeros are $-1$ and $2$ , which leave three intervals to test. Choose easy numbers inside each:

Q(-2) = (-1)(-4)^2 = -16, \qquad Q(0) = (1)(-2)^2 = 4, \qquad Q(3) = (4)(1)^2 = 4 .

So $Q$ is negative to the left of $-1$ , positive between $-1$ and $2$ , and positive again to the right of $2$ .

The sign chart for Q(x) = (x + 1)(x - 2)(x - 2). One test point settles each interval. The sign flips across the simple zero at x = -1, and it does not flip across the double zero at x = 2, which is precisely the difference between crossing and touching.

The sign chart and the multiplicity rule are two views of the same fact, and each checks the other. The zero at $-1$ has multiplicity $1$ , which is odd, so the curve crosses, and sure enough the sign flips from negative to positive across it. The zero at $2$ has multiplicity $2$ , which is even, so the curve touches, and sure enough the sign stays positive on both sides of it. If your test points ever contradict your multiplicities, you have made an arithmetic slip, and you should find it before you draw anything.

Check your understanding

For $P(x) = (x-1)(x-3)(x-5)$ , what is true on the interval $3 < x < 5$ ?

P(x) > 0

for every

x

in the interval. B

P(x) < 0

for every

x

in the interval. C

P

changes sign somewhere inside the interval. D You cannot decide without testing many points in the interval.

Non-real roots leave no footprint

The Fundamental Theorem of Algebra promises a degree $n$ polynomial exactly $n$ roots, counted with multiplicity. It does not promise you $n$ x-intercepts, and here is the reason. The graph lives in the real plane, and an x-intercept is a real number $x$ where the height $P(x)$ is zero. A non-real root such as $2 + 3i$ is not a location on the x-axis at all. It never shows up in the picture. Only real zeros produce x-intercepts.

So the counting works like this. Suppose the real zeros of $P$ have multiplicities adding up to $R$ . The remaining $n - R$ roots are non-real, and in a real polynomial they always come in conjugate pairs. If $z = a + bi$ is a root of a real polynomial then so is its conjugate $\overline{z}$ , and multiplying those two matching factors gives

(x - z)(x - \overline{z}) = x^2 - (z + \overline{z})x + z\overline{z} = x^2 - 2ax + (a^2 + b^2),

which has real coefficients. Dividing $P$ by that real quadratic leaves another real polynomial, two degrees smaller, so repeating the step pairs off every non-real root. The non-real roots therefore total an even number, which forces $R$ and $n$ to have the same parity.

That single observation pays a dividend. If $n$ is odd, $R$ cannot be zero (zero is even), so an odd-degree polynomial always has at least one real zero. More than that: the real multiplicities cannot all be even, or they would sum to an even $R$ , so at least one real zero has odd multiplicity. By the theorem above, the graph must cross the axis somewhere. That agrees exactly with the end behavior you already know. An odd-degree graph sends one arm up and the other down, so it has to travel from below the axis to above it. Any journey from one side of the axis to the other has to cross. Two completely different arguments, the same conclusion.

Check your understanding

A degree $6$ polynomial with real coefficients has real zeros only at $x = -1$ (multiplicity $1$ ) and $x = 2$ (multiplicity $3$ ). How many non-real roots does it have, counted with multiplicity?

0

1

2

4

Getting the zeros when the polynomial is not factored

A polynomial rarely arrives already factored. The route to its real zeros is the one this chapter and the last one built:

The Rational Root Theorem lists the candidate rational roots $p/q$ , where $p$ divides the constant term and $q$ divides the leading coefficient.
Synthetic division tests a candidate. A remainder of $0$ confirms a root and hands you the depressed polynomial, one degree lower, at the same time.
Divide by the same root again. If the remainder is $0$ a second time, the root has multiplicity at least $2$ , and you keep going until the remainder is nonzero. Repeated synthetic division is how you measure multiplicity in practice.
Once you are down to a quadratic, factor it or use the quadratic formula. A positive discriminant gives two real zeros, which may be irrational, and irrational zeros are still perfectly good x-intercepts. A negative discriminant gives a conjugate pair of non-real roots, which contribute no intercept at all.

Keep one honest limitation in view. The Rational Root Theorem only ever proposes rational candidates. A polynomial such as $(x^2 - 2)(x^2 + 1)$ has the real zeros $\sqrt{2}$ and $-\sqrt{2}$ , which are genuine x-intercepts, and no rational candidate will ever find them. They surface only when the quadratic factor is solved.

Putting it together

Here is the routine, in the order that makes each step cheap.

Ends. Read the degree and the leading coefficient, and fix the two arms.
Real zeros. Factor as far as you can, and mark each real zero on the axis.
Multiplicity. At each real zero, odd means cross, even means touch and turn back. A large multiplicity flattens the curve there.
Sign. Test one point in each interval between consecutive zeros, and one in each unbounded interval beyond the outermost zeros. Check the outer signs against the end behavior from step 1.
Anchor. Plot $P(0)$ , and any other point you would like, for extra accuracy.

Worked example 1 Sketch $P(x) = (x+2)(x-1)^2(x-3)$

Ends. Multiplying the leading terms gives $x \cdot x^2 \cdot x = x^4$ , so the degree is $4$ and the leading coefficient is $1$ . An even degree with a positive leading coefficient sends both arms upward.

Real zeros and multiplicity. The zeros are $-2$ with multiplicity $1$ , $1$ with multiplicity $2$ , and $3$ with multiplicity $1$ . So the curve crosses at $-2$ , touches and turns back at $1$ , and crosses at $3$ .

Sign. The three zeros carve out four intervals. Test one point in each:

P(-3) = (-1)(16)(-6) = 96, \qquad P(0) = (2)(1)(-3) = -6,

P(2) = (4)(1)(-1) = -4, \qquad P(4) = (6)(9)(1) = 54 .

So the curve runs above the axis to the left of $-2$ . It runs below the axis from $-2$ all the way to $3$ (dipping, touching the axis at $1$ , and dipping again), and above the axis to the right of $3$ .

Check. The outer signs are both positive, which is what “both arms up” demanded. The sign flips at $-2$ and at $3$ , the two odd zeros, and does not flip at $1$ , the even one. Everything agrees.

Anchor. $P(0) = -6$ , so the curve passes through $(0, -6)$ .

Since the curve is below the axis on both sides of $x = 1$ and equal to zero there, the point $(1, 0)$ is a local maximum. It is the one turning point in this whole picture whose location we can name exactly.

The finished sketch: the curve crosses at x = -2, touches the axis and turns back at the double zero x = 1, crosses again at x = 3, passes through (0, -6), and sends both arms upward. The vertical scale is compressed to fit. The heights of the two dips are drawn from a few extra plotted points, not computed, because algebra cannot find them.

Worked example 2 Sketch $P(x) = x^3 - 4x^2 + x + 6$

Factor first. The Rational Root Theorem offers the divisors of $6$ over the divisors of $1$ , so the candidates are $\pm 1, \pm 2, \pm 3, \pm 6$ . Test $x = -1$ by synthetic division on the coefficients $1, -4, 1, 6$ : bring down $1$ , multiply by $-1$ and add to get $-5$ . Multiply by $-1$ and add to get $6$ , then multiply by $-1$ and add to get $0$ . The remainder is $0$ , so $-1$ is a root and the depressed polynomial is $x^2 - 5x + 6$ :

P(x) = (x+1)(x^2 - 5x + 6) = (x+1)(x-2)(x-3).

Ends. Degree $3$ with leading coefficient $1$ : the left arm goes down, the right arm goes up.

Zeros and multiplicity. The zeros are $-1$ , $2$ , and $3$ , each of multiplicity $1$ . All three are odd, so the curve crosses at every one of them. There is no touching anywhere.

Sign. Four intervals, four test points:

P(-2) = (-1)(-4)(-5) = -20, \qquad P(0) = (1)(-2)(-3) = 6,

P(2.5) = (3.5)(0.5)(-0.5) = -0.875, \qquad P(4) = (5)(2)(1) = 10 .

The signs run negative, positive, negative, positive, alternating exactly as three simple zeros in a row demand. The far-left sign is negative and the far-right sign is positive, matching the end behavior.

Anchor. $P(0) = 6$ .

The sketch: the curve rises from the bottom left, crosses at $-1$ , arches above the axis through $(0,6)$ , and comes back down to cross at $2$ . From there the curve dips shallowly below the axis, crosses again at $3$ , and climbs. The dip between $2$ and $3$ is a shallow one, and the test value $P(2.5) = -0.875$ is the evidence for that.

Worked example 3 Sketch $R(x) = -(x-2)^2(x^2+1)$

Ends. The leading term is $-(x^2)(x^2) = -x^4$ . The degree is $4$ , even, and the leading coefficient is $-1$ , negative, so both arms point downward.

Zeros. The factor $(x-2)^2$ gives the real zero $2$ with multiplicity $2$ . The factor $x^2 + 1$ has discriminant $0^2 - 4(1)(1) = -4 < 0$ , so its roots are the non-real pair $i$ and $-i$ . They contribute no x-intercept. The graph therefore has exactly one x-intercept, at $x = 2$ , even though the polynomial has four roots counted with multiplicity.

Multiplicity. The multiplicity of $2$ is even, so the curve touches the axis there and turns back.

Sign. Here the factors settle the sign without any test points at all. For every real $x$ we have $x^2 + 1 > 0$ and $(x-2)^2 \geq 0$ , so

R(x) = -\underbrace{(x-2)^2}_{\geq \, 0}\ \underbrace{(x^2+1)}_{> \, 0} \leq 0 ,

with equality only at $x = 2$ . The graph never rises above the x-axis. It climbs from below, kisses the axis at $(2,0)$ , and falls away again.

Anchor. $R(0) = -(4)(1) = -4$ .

Note that $(2,0)$ is a local maximum, since $R$ is negative on both sides and zero there. This is the even-multiplicity turning point of the theorem, and it is the highest the curve ever gets.

The graph of the negative quartic with a double zero at x = 2 and a non-real pair of roots. The two non-real roots are nowhere in the picture, which is the whole point: only real zeros make intercepts. The curve rises to touch the axis once and falls away, and it passes through (0, -4).

What algebra cannot tell you

Look at the sketch in Worked Example 1 again and ask a natural question: how deep is the dip between $-2$ and $1$ , and where exactly is the bottom of it? Nothing in this lesson can answer that. The tools of this chapter locate the curve’s intercepts and its sign, and that is all. They say nothing about how high a hump climbs or where a valley bottoms out.

It is tempting to guess that the bottom sits halfway between the two zeros, at $x = -0.5$ . That guess is false, and you can refute it with arithmetic alone. Compare the value there with the value at $x = -1$ :

P(-0.5) = (1.5)(2.25)(-3.5) = -11.8125, \qquad P(-1) = (1)(4)(-4) = -16 .

The curve is already lower at $x = -1$ than at the midpoint, so the midpoint is certainly not the lowest point. The symmetry that puts a parabola’s vertex halfway between its roots is special to degree $2$ , and it does not survive to higher degrees.

What we can prove is weaker and still useful. Between two consecutive real zeros the curve leaves the axis and comes back to it, so somewhere in between it has to turn around. There is at least one turning point in each gap between consecutive real zeros. And you already know the ceiling on the count, at most $n - 1$ turning points for a degree $n$ polynomial.

Exactly two kinds of turning point are within algebra’s reach, and both are worth naming. A parabola’s vertex sits halfway between its two roots, by the symmetry argument from the quadratics chapter. And a real zero $r$ of even multiplicity is itself a turning point, at the exactly known coordinates $(r, 0)$ , which is how Worked Example 1 could name $(1, 0)$ . Every other turning point is out of reach. Where it sits, and how high or low it goes, is a question about the slope of the curve. And finding a point where the slope is zero is the job of the derivative, a calculus tool you have not met.

So be honest about what your sketch claims. It claims the correct ends, the correct intercepts, the correct behavior at each intercept, and the correct side of the axis on every interval. It does not claim the exact height of any hump. If you want a more accurate picture, there is one legitimate way to get it: plot more points by evaluating $P$ . Anything else is decoration.

Reading a graph backwards

Run the machine in reverse and you can recover a great deal of a polynomial from its picture, as long as you also say clearly what you cannot recover. Suppose you are shown a graph and told that every x-intercept is visible in the window.

What the picture gives you. Each x-intercept is a real zero. A crossing means odd multiplicity, a touch means even multiplicity. The two arms give you the sign of the leading coefficient and the parity of the degree. Adding one to the count for each crossing and two for each touch gives the least possible degree. The true degree can exceed that minimum only by an even number, since any extra multiplicity comes in steps of two and any non-real roots come in pairs.

What the picture hides. It cannot tell you a crossing’s multiplicity is $1$ rather than $3$ or $5$ (a flattened crossing is a hint, never a proof). And it cannot show you a single non-real root. It also cannot give you the leading coefficient itself, only its sign, until you are handed one more point.

Worked example 4 Build a cubic from its graph

A cubic crosses the x-axis at $x = -2$ , touches the axis at $x = 1$ without crossing, and passes through $(0, 6)$ . Find it.

Turn each feature into a factor. Crossing at $-2$ means a zero of odd multiplicity there; touching at $1$ means a zero of even multiplicity. The smallest choices are $1$ and $2$ , and they use up $1 + 2 = 3$ degrees, which is the whole cubic. Nothing is left over, so those are the only options:

P(x) = a(x+2)(x-1)^2 .

Use the extra point to pin $a$ . Substituting $x = 0$ and $P(0) = 6$ ,

a(2)(-1)^2 = 6 \quad \Longrightarrow \quad 2a = 6 \quad \Longrightarrow \quad a = 3 .

So $P(x) = 3(x+2)(x-1)^2$ , which expands to $3x^3 - 9x + 6$ .

Check the picture. The degree is odd and $a = 3 > 0$ , so the left arm goes down and the right arm goes up. Testing $P(-3) = 3(-1)(16) = -48$ and $P(0) = 6$ and $P(2) = 3(4)(1) = 12$ gives the signs negative, positive, positive: the curve crosses at $-2$ and only touches at $1$ , exactly as described. Since the curve is positive on both sides of $x = 1$ and zero there, the point $(1,0)$ is a local minimum.

One caution on the word “find”. This is the only cubic with those features, because a cubic has just three degrees to spend. If the problem had said “a polynomial” instead of “a cubic”, there would be infinitely many answers. That is because we could raise the multiplicity at $-2$ to $3$ , or multiply by any quadratic with no real roots, without changing a single visible feature of the graph.

Common mistakes

Treating every zero as a crossing. A zero of even multiplicity does not cross. Reading $P(x) = (x-3)^2(x+1)$ as “crosses at $3$ and at $-1$ ” throws away the multiplicity, which is the one piece of information this chapter was built to give you.
Counting non-real roots as x-intercepts. A degree $5$ polynomial with one real zero has exactly one x-intercept, not five. The Fundamental Theorem of Algebra counts roots in the complex numbers; the x-axis only ever shows you the real ones.
Testing a point that is itself a zero. A test point must lie strictly inside the interval. Feeding a zero into $P$ returns $0$ , which reports no sign at all.
Believing a hump peaks halfway between two zeros. True for a parabola, false in general. For $P(x) = (x+2)(x-1)^2(x-3)$ the value at the midpoint $x = -0.5$ is $-11.8125$ , while $P(-1) = -16$ is lower still, so the bottom is not at the midpoint.
Inventing a maximum. Drawing a hump is fine; labelling its peak with coordinates you never computed is not. Without calculus, the only heights you may claim are the ones you actually evaluated.
Reading an exact degree off a picture. A graph with two crossings and one touch has degree at least $4$ , and could be degree $6$ , $8$ , or more. The visible features fix a minimum and a parity, never the exact value.
Forgetting to reconcile the signs with the ends. If your sign chart says $P$ is negative far to the right while your leading term says the right arm rises, one of the two is wrong. Find the error before you draw.

Key takeaways

The factored form is a blueprint: it shows the degree, the leading coefficient, every real zero, and every multiplicity, which is everything a sketch needs.
A polynomial’s graph is unbroken, so the sign of $P$ can change only at a real zero. The converse fails, and multiplicity is what decides.
At a real zero $r$ of multiplicity $m$ , write $P(x) = (x-r)^m G(x)$ with $G(r) \neq 0$ . The graph crosses at $r$ if and only if $m$ is odd, and touches and turns back if and only if $m$ is even. A larger $m$ flattens the curve against the axis either way.
The real zeros cut the line into intervals, and $P$ holds one sign on each. One test point settles an entire interval.
Non-real roots produce no x-intercepts. They come in conjugate pairs, so the multiplicities of the real zeros always add up to a number with the same parity as the degree. That parity is why every odd-degree polynomial must cross the axis at least once.
An honest sketch gets the ends, the intercepts, the behavior at each intercept, and the sign on every interval right. A parabola’s vertex and the turning points that sit on the axis at even zeros are within algebra’s reach. Apart from those, where the curve turns and how high the humps climb is a question for calculus, and it should not be faked.

Graphing Polynomial Functions

Learning goals

The blueprint hidden in the factored form

Where a polynomial is allowed to change sign

How the curve meets the axis at a zero

Why a higher multiplicity flattens the curve

The sign chart

Non-real roots leave no footprint

Getting the zeros when the polynomial is not factored

Putting it together

What algebra cannot tell you

Reading a graph backwards

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)