AP Precal Study Guide 1.8-1.10AP Precal Study Guide 1.8-1.10

STUDY GUIDE

Rational Functions

Precalculus 1.8 - 1.10

End Behavior, Asymptotes, Holes & Equivalent Representations

Overview

Lessons 1.8-1.10 build on each other into one skill: given a rational function

y = f(x) / g(x) (f and g are both polynomials, g(x) ≠ 0)

you should be able to find every key feature of its graph - end behavior, vertical asymptotes, holes, x-intercepts, and slant asymptotes - and go the other direction too, writing an equation for a function when you're given its features or its graph. 1.10 is really just 1.8 + 1.9 combined into one repeatable procedure, so master those two first.

Throughout this guide, look for boxes like this one - stop and answer them yourself before reading the answer that follows: Quick check: What is 7 + 8?

→ Answer: 15. (See - that's all these are: a one-step pause to make sure the idea that just got explained

actually landed, before moving on to the next one.)

1.8 - Rational Functions and End Behavior

Rational function: the quotient of two polynomials, y = f(x)/g(x), where g(x) ≠ 0. Because polynomial functions have domain "all real numbers," but rational functions divide by another polynomial, the domain of a rational function excludes any x-value that makes the denominator zero.

Quick check: Is y = (x + 1)/5 a rational function in the sense this lesson cares about (the kind with domain restrictions)? →

Answer: No - the "denominator" here is just the constant 5, which is never zero, so this behaves like an ordinary polynomial with no restricted domain. The interesting cases are when the denominator is a polynomial with a variable in it.

End behavior: compare the degrees of the leading terms

Write the function using just its leading terms, f(x) = a·xⁿ / (b·xᵈ), where n is the degree of the numerator and d is the degree of the denominator. There are exactly three cases:

Case Degree comparison End behavior

I n = d (same degree) Horizontal asy mptote: y = a/b (ratio of leading coefficients)

II n < d (denominator's degree wins) Horizontal asy mptote: y = 0

III n > d (numerator's degree wins) No horizontal asy mptote. f(x) behaves like the poly nomial y = (a/b)xⁿ⁻ᵈ as x → ±∞

Quick check: If n = 4 and d = 4, which case are you in, and if the leading coefficients are 9 and 3, what's the horizontal asymptote?→ Answer: Case I (same degree) → y = 9/3 = y = 3.

Quick check: If n = 2 and d = 5, which case, and what's the horizontal asymptote - no need to even know the actual coefficients this time.

→ Answer: Case II (denominator degree wins)

→ y = 0, regardless of what the leading coefficients are.

Special case of III - the slant (oblique) asymptote: if the numerator's degree is exactly one more than the denominator's degree (n = d + 1), the function's end behavior is a straight line, not a curve. That line is called a slant asymptote.

Quick check: n = 5, d = 4. Does this function have a horizontal asymptote, a slant asymptote, or neither?

→ Answer: n is exactly

one more than d (5 = 4 + 1), so slant asymptote - not "no asymptote at all." That phrase only applies once the gap is 2 or more. Describing end behavior with limit notation

Once you know the horizontal asymptote (or the polynomial the function behaves like), write it as two limit statements:

lim f(x) = [value] lim f(x) = [value]

x→-∞ x→∞

If there's a horizontal asymptote y = k, both limits equal k. If the function grows or shrinks without bound instead, the limit is ∞ or -∞. Rational Functions Study Guide 2 of 11

Worked Example 1

Determine whether each function has a horizontal asymptote, a slant asymptote, or neither. If it has a horizontal asymptote, state it. a) f(x) = (4x² - x + 6) / (2x² + 5)

Both leading terms are degree 2 (Case I): H.A. → y = 4/2 = y = 2

b) g(x) = (3x + 1) / (x² - 4)

Numerator degree 1 < denominator degree 2 (Case II): y = 0

c) h(x) = (5x³ + 2x - 1) / (x² + 9)

Numerator degree 3 > denominator degree 2, and 3 is exactly one more than 2, so this is a slant asymptote case, not Case III's general "no asymptote." (See Example 2 for finding its equation.)

Worked Example 2 - finding a slant asymptote

Find the slant asymptote of f(x) = (x² + x - 8) / (x - 2).

Use polynomial (or synthetic) long division and ignore the remainder:

x² + x - 8 ÷ (x - 2) → quotient: x + 3, remainder: -2

Slant asymptote: y = x + 3

Quick check: After dividing, you get a quotient of 2x - 1 with a remainder of 7. What's the slant asymptote? → Answer: y = 2x - 1.

The remainder never appears in the asymptote - only the quotient does.

Worked Example 3 - limit statements from a table

A function g(x) has these values:

x -1000 -100 100 1000

g(x) 4.98 4.8 5.2 5.02

As x → ±∞, g(x) is settling toward 5 from both directions, so:

lim g(x) = 5 lim g(x) = 5

x→-∞ x→∞

Rational Functions Study Guide 3 of 11

1.9 - Vertical Asymptotes and Removable Discontinuities

Vertical asymptotes and holes both happen at x-values that make the denominator zero - the difference is entirely about whether that factor also appears in the numerator.

A hole occurs when... A vertical asymptote occurs when...

Rule the factor in the denominator cancels with a matching factor in the

numerator

the factor in the denominator cannot cancel with

any thing in the numerator

What the

graph does

a single open circle at that x-value; the function is still approaching a

real y -value from both sides

the graph shoots off toward +∞ or -∞ on at least one side

Limit

notation

lim f(x) = [a real number] lim f(x) = ±∞

Quick check: f(x) = [(x - 4)(x + 2)] / [(x + 2)(x - 6)]. The factor (x + 2) appears in both the top and bottom. Hole or vertical asymptote - and at what x-value?

→ Answer: It cancels, so it's a hole, at x = -2. (The leftover (x - 6) in the denominator is what gives the vertical asymptote instead, at x = 6.)

Finding the coordinates of a hole: cancel the common factor first, then plug the x-value into what's left over (the simplified function) - not the original. Plugging into the original just gives 0/0.