Polynomial and rational functions · AP Precalculus

Chapter 1 · Polynomial and rational functions

how outputs change with inputs

A function pairs every input with one output. Precalculus asks how the outputs change as the inputs change: where they rise and fall, how fast, and what happens far away.

The graph is the cubic $p(x) = \tfrac{1}{2}(x + 2)(x - 1)(x - 3)$ . Its factors show its zeros, −2, 1, 3, where it meets the x-axis. The red secant slides along it; its slope is the average rate of change over its interval. Over [0, 2]:

\frac{p(2) - p(0)}{2 - 0} = \frac{-2 - 3}{2} = -2.5

This chapter is AP Precalculus Unit 1, the largest unit on the exam (about 30–40%). It builds the tools for polynomial functions (sums of powers of x) and rational functions (one polynomial over another): rates of change, zeros, end behavior, asymptotes and holes, transformations, and finally choosing and building a function to model real data.

1.1 · Change in tandem

inputs and outputs, varying together

A function f sends each input x to one output f(x). As the input changes, the output changes with it, in tandem. A graph shows both at once: the input along the x-axis, the output as the height.

Follow the point on $f(x) = x^3 - 3x$ as x moves. Its gray guides carry the input down to the x-axis and the output across to the y-axis.

x	−2	−1	0	1	2
f(x)	−2	2	0	−2	2

f is increasing on an interval if, as the inputs increase, the outputs always increase: for all a < b in it, f(a) < f(b). It is decreasing if the outputs always decrease. Here f increases for x < -1 (red), decreases between -1 and 1 (blue), and increases again after 1.

The output is positive where the graph is above the x-axis and negative where it is below. It changes sign at the zeros, the inputs with f(x) = 0: here 0 and ±√3 ≈ ±1.73.

In context the same words apply: as water is poured into a vase at a steady rate, the depth increases, quickly where the vase is narrow and slowly where it is wide. A graph of depth against volume is a picture of how the two quantities change together.

1.2 · Rates of change

the slope of a secant, and the slope at a point

The average rate of change of f over [a, b] is the change in output over the change in input: the slope of the secant line from (a, f(a)) to (b, f(b)).

\frac{f(b) - f(a)}{b - a}

The rate of change at a point is approximated by average rates of change over smaller and smaller intervals around it. For $f(x) = \tfrac{1}{2}x^2$ at x = 2, take secants over [2 − h, 2] (blue) and [2, 2 + h] (red):

h	1	0.5	0.1	0.01
left	1.5	1.75	1.95	1.995
right	2.5	2.25	2.05	2.005

The two columns close in from either side on 2: the rate of change at x = 2 is 2, the slope of the dashed tangent line. The same work at x = 1 gives 1.

Comparing the two: the rate at 2 is greater than the rate at 1, so the graph gets steeper as x grows. Two links to remember:

a positive rate of change means f is increasing there; a negative one, decreasing;
where the rate of change is increasing the graph is concave up (it bends upward, like this one); where it is decreasing, concave down.

1.3 · Rates of change in linear and quadratic functions

constant change, and change that changes steadily

Step the input by equal amounts and watch the output. The pattern of the changes tells a linear function from a quadratic one.

x	0	1	2	3	4
2x + 1	1	3	5	7	9
change	+2	+2	+2	+2
x² − 4x + 5	5	2	1	2	5
change	−3	−1	+1	+3
change of change	+2	+2	+2

Linear: the average rate of change is the same over every interval, the slope, 2. Equal input steps give equal output changes.

Quadratic: over consecutive intervals of equal length, the average rates of change are −3, −1, 1, 3. They are not constant, but they change at a constant rate, 2 per step. For $ax^2 + bx + c$ with steps of 1 that constant is 2a.

The rates also locate the vertex. They are negative up to x = 2 (the function is decreasing) and positive after (increasing), so the minimum is at x = 2. Because the rates increase, the parabola is concave up; with a < 0 they would decrease, and it would be concave down.

1.4 · Polynomial functions and rates of change

turning points, inflection points, global extremes

A polynomial function is $p(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$ , with real coefficients and a_n ≠ 0. Its degree is n, and a_n is the leading coefficient. A nonzero constant is a polynomial of degree 0.

The graph is $p(x) = \tfrac{1}{4}x^4 - 2x^2 + 1$ . The red line rides along with the slope of the curve, its rate of change:

a local maximum is where the rate of change goes from positive to negative: (0, 1);
a local minimum, from negative to positive: (−2, −3) and (2, −3);
a point of inflection is where the rate of change goes from increasing to decreasing or back, so the concavity changes: at x = ±2/√3 ≈ ±1.15.

The lowest value the function ever takes, −3, is its global minimum. An even-degree polynomial always has a global maximum or a global minimum (a minimum when a_n > 0, as here); an odd-degree one has neither, since its outputs run off to both ∞ and −∞.

Its zeros are ±(√3 − 1) ≈ ±0.73 and ±(√3 + 1) ≈ ±2.73. Between every two distinct real zeros of a nonconstant polynomial there is a local maximum or minimum: here the minimum at −2, the maximum at 0, the minimum at 2.

1.5 · Polynomial functions and complex zeros

n zeros for degree n, counted with multiplicity

A polynomial of degree n has exactly n complex zeros, counted with multiplicity. Some may be real, some not.

Take $p(x) = \tfrac{1}{10}(x + 2)^2(x - 1)(x^2 + 1)$ , of degree 2 + 1 + 2 = 5. Its zeros:

zero	multiplicity	on the graph
−2	2	touches the x-axis, doesn’t cross
1	1	crosses the x-axis
i, −i	1 each	no x-intercept: not real

a is a zero of p exactly when (x − a) is a factor of p(x); when a is real, (a, 0) is an x-intercept.
If (x − a) appears k times, a has multiplicity k. At a real zero of even multiplicity the graph touches the axis and turns back; at odd multiplicity it crosses.
With real coefficients, non-real zeros come in conjugate pairs: if a + bi is a zero, so is a − bi.

The figure swaps the last factor for $x^2 + q$ and slides q. For q < 0 its zeros ±√(−q) are real, two red x-intercepts; they meet at q = 0 and then leave the real axis together, up and down the imaginary axis as ±i√q. The degree, and the count of zeros, never changes.

Even and odd functions. f is even if f(−x) = f(x) for every x (symmetric about the y-axis), like 1.4’s $\tfrac{1}{4}x^4 - 2x^2 + 1$ ; odd if f(−x) = −f(x) (symmetric through the origin), like 1.1’s $x^3 - 3x$ . A polynomial with only even powers is even; only odd powers, odd.

1.6 · Polynomial functions and end behavior

far out, the leading term is all that matters

The end behavior of a function is what its outputs do as x → ∞ and x → −∞. For a polynomial, it is decided by the leading term alone.

The red curve is $p(x) = -x^3 + 3x^2 + x - 3 = -(x + 1)(x - 1)(x - 3)$ . The window it sits in grows to 12 times its size and back: watch the numbers on the axes. Zoomed out, the zeros crowd into the origin, the wiggles vanish, and the curve lies on its leading term $-x^3$ (blue). At x = 48, p(x) is 0.937 times −x³.

\begin{aligned} \lim_{x\to\infty} p(x) &= -\infty \\ \lim_{x\to -\infty} p(x) &= \infty \end{aligned}

	a_n > 0	a_n < 0
n even	up on both ends	down on both ends
n odd	down on the left, up on the right	up on the left, down on the right

The reason: $p(x) = -x^3\left(1 - \tfrac{3}{x} - \tfrac{1}{x^2} + \tfrac{3}{x^3}\right)$ , and every term in the bracket after the 1 shrinks to 0 as |x| grows. Read limit notation as “as x increases without bound, p(x) decreases without bound.”

1.7 · Rational functions and end behavior

the ratio of the leading terms

A rational function is a ratio of polynomials, $r(x) = \dfrac{p(x)}{q(x)}$ with q not the zero polynomial. Far out, each polynomial behaves like its leading term, so r behaves like the ratio of the two.

degrees	horizontal asymptote	far out
top < bottom	y = 0	3x/(x² + 1) ≈ 3/x → 0
top = bottom	y = ratio of leading coefficients	(2x² − 1)/(x² + 2) ≈ 2x²/x² = 2
top > bottom	none; behaves like a polynomial	(x² + x + 1)/(x − 1) ≈ x²/x = x

In each sheet, red and blue points run out toward −∞ and ∞, and settle along the red dashed line. With equal degrees, $r(100) = 1.9995$ : close to 2, never quite there from below.

\lim_{x\to\pm\infty} \frac{2x^2 - 1}{x^2 + 2} = 2

When the top’s degree is exactly one more than the bottom’s, the graph approaches a slant asymptote, a line y = mx + b found by dividing (1.11). Here it is y = x + 2. A graph may cross a horizontal or slant asymptote; the asymptote only describes the ends.

1.8 · Rational functions and zeros

where the top is zero and the bottom isn’t

The real zeros of $r(x) = \dfrac{p(x)}{q(x)}$ are the real zeros of the numerator p that are in the domain of r: inputs where p is 0 and q is not.

For $r(x) = \dfrac{(x + 2)(x - 1)}{(x + 1)(x - 3)}$ , the zeros are −2 and 1. The inputs −1 and 3 are not in the domain: there the bottom is 0, and the graph has vertical asymptotes (1.9).

The output can change sign only at a zero or where r is undefined. So those four inputs cut the number line into intervals, and one test input in each gives the sign on the whole interval:

interval	x < −2	−2 < x < −1	−1 < x < 1	1 < x < 3	x > 3
test x	−3	−1.5	0	2	4
r(x)	0.33	−0.56	0.67	−1.33	3.6
sign	+	−	+	−	+

That solves inequalities. $r(x) \ge 0$ where the graph is on or above the axis, red: x ≤ −2, or −1 < x ≤ 1, or x > 3. The zeros are included (there r = 0); the asymptotes never are.

A zero of the numerator that is also a zero of the denominator is not a zero of r. That is a hole or an asymptote (1.10).

1.9 · Rational functions and vertical asymptotes

where the bottom is zero, and stays zero

If a is a real zero of the denominator and not of the numerator, or is a zero of the denominator with higher multiplicity, then the line x = a is a vertical asymptote of r. Near it, the outputs grow without bound.

For $r(x) = \dfrac{4}{(x + 1)(x - 2)^2}$ the bottom is 0 at −1 and 2, and the top never is:

x	−1.01	−0.99	1.99	2.01
r(x)	−44	45	13378	13289

\begin{aligned} \lim_{x\to -1^-} r(x) &= -\infty \\ \lim_{x\to -1^+} r(x) &= \infty \\ \lim_{x\to 2^\pm} r(x) &= \infty \end{aligned}

The sign on each side comes from the factors. Near x = −1 the factor x + 1 changes sign, so the two sides go opposite ways. Near x = 2 the factor is squared, positive on both sides, so both sides go up. Odd multiplicity: opposite directions; even: the same direction.

$\lim_{x\to a^+}$ means x approaches a from the right, through values greater than a; $\lim_{x\to a^-}$ , from the left.

1.10 · Rational functions and holes

a factor that cancels

If a is a zero of both the numerator and the denominator, with multiplicity in the numerator at least as large as in the denominator, the graph has a hole at x = a: a single missing point.

\begin{aligned} r(x) &= \frac{x^2 + x - 6}{x^2 - x - 2} \\ &= \frac{(x - 2)(x + 3)}{(x - 2)(x + 1)} \\ &= \frac{x + 3}{x + 1}, \quad x \ne 2 \end{aligned}

The factor x − 2 cancels, once on top and once below, so x = 2 is a hole. Its height is the value the simplified form gives there, which is also the limit both sides approach:

\lim_{x\to 2} r(x) = \frac{2 + 3}{2 + 1} = \frac{5}{3}

The factor x + 1 does not cancel, so x = -1 is a vertical asymptote; x + 3 is left on top, so x = -3 is the only zero. Note that 2 is not a zero, though it makes the numerator 0.

If the multiplicity is higher in the denominator, the canceling leaves a factor below, and the result is an asymptote, not a hole: $\dfrac{x - 2}{(x - 2)^2} = \dfrac{1}{x - 2}$ has a vertical asymptote at x = 2.

1.11 · Equivalent representations of polynomial and rational expressions

factored, standard, divided, expanded

The same function can be written several ways. Each form shows something the others hide.

Standard form $x^3 - 2x^2 - 5x + 6$ shows the leading term (end behavior) and the y-intercept, 6.
Factored form $(x + 2)(x - 1)(x - 3)$ shows the zeros, and in a rational function the asymptotes and holes.

Polynomial long division rewrites a rational function as a polynomial plus a remainder fraction. Divide $x^2 + x + 1$ by $x - 1$ : x(x − 1) takes away x² − x, leaving 2x + 1; 2(x − 1) takes away 2x − 2, leaving 3.

\begin{gathered} x^2 + x + 1 = (x - 1)(x + 2) + 3 \\ \frac{x^2 + x + 1}{x - 1} = x + 2 + \frac{3}{x - 1} \end{gathered}

As |x| grows the remainder term (the blue gap) shrinks to 0, so the quotient y = x + 2 is the slant asymptote. In general, if p = q·s + t, then p/q = s + t/q. A remainder of 0 means the divisor is a factor.

The binomial theorem expands $(a + b)^n$ with the numbers in row n of Pascal’s triangle, each the sum of the two above it:

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Row 3 is 1, 3, 3, 1, so $(x + 2)^3 = x^3 + 3x^2(2) + 3x(2)^2 + 2^3 = x^3 + 6x^2 + 12x + 8$ .

1.12 · Transformations of functions

shifts, stretches and reflections

A new function can be built from a parent f by shifting and stretching its graph. The red graph is $g(x) = a\,f\big(b(x - h)\big) + k$ , built from the gray $f(x) = x^3 - 3x$ on [−2, 2]; the figure changes h, k, a and b one at a time.

form	graph of f is
f(x) + k	translated vertically by k (up if k > 0)
f(x − h)	translated horizontally by h (right if h > 0); f(x + h) moves it left
a·f(x)	dilated vertically by \|a\|; reflected over the x-axis if a < 0
f(bx)	dilated horizontally by 1/\|b\|; reflected over the y-axis if b < 0

Additive changes translate; multiplicative ones dilate. Changes outside f act on outputs and do what they say; changes inside act on inputs and do the opposite: x − 1.5 moves right, and b = 2 squeezes the graph to half its width.

The domain and range move with the graph. At the end of the figure’s run, h = 1.5, k = 1, a = 1.5, b = 2:

g(x) = 1.5\,f\big(2(x - 1.5)\big) + 1

The domain [−2, 2] is squeezed by 1/2 and shifted 1.5: [0.5, 2.5]. The range [−2, 2] is stretched by 1.5 and shifted up 1: [−2, 4]. The blue bars on the axes show both.

1.13 · Function model selection and assumption articulation

which kind of function fits, and on what domain

To model data, choose the kind of function whose pattern of change matches the data’s, then state the assumptions and the domain on which the model makes sense.

A ball is thrown upward; its height is measured each second (hypothetical data):

t (s)	0	1	2	3	4
h (m)	1	16	21	16	1
change	15	5	−5	−15
change of change	−10	−10	−10

The first differences are not constant, so a linear model is wrong: the dashed line through the first two points predicts 61 m at t = 4 against the measured 1 m. The second differences are constant, −10, so a quadratic fits exactly:

h(t) = -5t^2 + 20t + 1

linear: roughly constant rate of change;
quadratic: rates changing at a roughly constant rate, or data rising then falling symmetrically about one extreme value;
cubic or higher: more turns, or rates whose own rate changes;
piecewise: different patterns over different stretches of input;
rational: quantities that are inversely proportional, or level off toward a value.

Assumptions. This model ignores air resistance and holds only while the ball is in the air: from the throw to when h(t) = 0, 0 ≤ t ≤ 4.05. Outside that domain its outputs (negative heights) mean nothing.

1.14 · Function model construction and application

from a situation to a function, and back

Building a model means naming the input, writing the output in terms of it from the situation’s geometry or rules, and finding the domain the situation allows. Then the model answers questions.

Cut an x × x square from each corner of a 12 × 8 inch sheet and fold up the sides. The box is x deep, 12 − 2x long and 8 − 2x wide:

\begin{gathered} V(x) = x(12 - 2x)(8 - 2x) \\ 0 < x < 4 \end{gathered}

The domain comes from the situation: a cut must have positive size, and the 8-inch side runs out at x = 4. The cubic’s other behavior (negative volumes beyond) is outside the model.

Largest box: from the graph (or a calculator), the maximum is at x ≈ 1.57 in, V ≈ 67.6 in³.
A 60 in³ box: solve V(x) = 60. One solution is exact, x = 1: 1 · 10 · 6 = 60. The other is x ≈ 2.21.

Other models are built the same way. A polynomial of degree n can be fitted through n + 1 points with distinct inputs; a rational model suits a ratio, like an average cost $C(n) = \dfrac{500 + 3n}{n}$ dollars per item, whose horizontal asymptote, 3, is the cost per item once the fixed $500 is spread thin. Always report answers in the situation’s units, and check them against its limits.