Math 30-1 · Polynomial, radical, rational
Three families of functions
Polynomials, square roots and fractions with x on the bottom. Each family has one or two signature features, and exam questions almost always ask about exactly those.
Polynomial graphs
The idea: the degree and leading coefficient decide how the ends behave; the factors decide where the graph meets the x-axis and whether it crosses or bounces.
| Degree | Leading coefficient | Ends |
|---|---|---|
| odd | positive | down left, up right |
| odd | negative | up left, down right |
| even | positive | up on both ends |
| even | negative | down on both ends |
Going the other way: a zero of −2 with multiplicity 2 and a zero of 3 gives y = (x + 2)²(x − 3). The sign inside each factor is the opposite of the zero.
The remainder and factor theorems
The idea: dividing P(x) by x − a leaves a remainder of P(a). If that remainder is 0, then x − a is a factor.
- Remainder theorem: x² + 3x − 1 divided by x − 2 leaves P(2) = 4 + 6 − 1 = 9. No long division.
- Factor theorem: for x³ − 7x + 6, P(1) = 1 − 7 + 6 = 0, so x − 1 is a factor.
- Integral zero theorem: any integer zero must divide the constant term. For x³ − 2x² − 5x + 6, test ±1, ±2, ±3, ±6.
Finding an unknown coefficient
If x³ + kx² − 4 divided by x − 2 leaves 8, then P(2) = 8: 8 + 4k − 4 = 8, so k = 1.
Factoring a cubic
- Test the possible integer zeros until one gives 0 — say x = 1 for x³ − 2x² − 5x + 6.
- Divide by x − 1 (synthetic division is fastest) to get x² − x − 6.
- Factor what is left: (x − 3)(x + 2). So P(x) = (x − 1)(x − 3)(x + 2).
Watch out: dividing by x + 2 means a = −2, so the remainder is P(−2), not P(2).
Radical functions
The idea: y = √x starts at (0, 0) and only goes right and up, because you cannot square-root a negative. Every radical question is about that starting point and that restriction.
Transformations work exactly as in the transformations unit. For y = √(x − 2) + 3, the starting point moves from (0, 0) to (2, 3), so the domain is x ≥ 2 and the range is y ≥ 3.
From y = f(x) to y = √f(x)
- Wherever f(x) is negative, √f(x) does not exist.
- Points where y = 0 or y = 1 stay put — those are the invariant points, since √0 = 0 and √1 = 1.
- Between 0 and 1 the graph rises (√0.25 = 0.5); above 1 it drops (√9 = 3).
Solving radical equations
Isolate the root, square both sides, solve — then check every answer. For √(x + 5) = x − 1, squaring gives x² − 3x − 4 = 0, so x = 4 or −1. But x = −1 gives √4 = −2, which is false. Only x = 4 works.
Watch out: squaring creates extraneous roots. A square root is never negative, so any answer that makes the right side negative has to go.
Rational functions
The idea: where the denominator is zero, the graph breaks. If that factor cancels, it is a hole; if it does not, it is a vertical asymptote.
Asymptote or hole?
For y = (x − 2) / ((x − 2)(x + 1)):
- x − 2 cancels → a hole (point of discontinuity) at x = 2.
- x + 1 does not cancel → a vertical asymptote at x = −1.
Horizontal asymptotes
- Degree on top is lower than the bottom → y = 0. Example: y = 1/x.
- Same degree on top and bottom → the ratio of leading coefficients. For y = (2x + 1)/(x − 3), it is y = 2.
Watch out: calling every zero of the denominator an asymptote. Always factor and cancel first — a cancelled factor is a hole, not an asymptote.