Core Ideas in Plain Terms

Study Sheet

Core Ideas in Plain Terms

The next layer, in everyday language

Functions and Their Graphs

Concept
A function is an input-output machine

f(x)f(x) means "the output when the input is xx." Small changes to the formula move the graph in predictable ways: f(x)+kf(x) + k shifts it up, f(xh)f(x - h) shifts it right, and a minus sign flips it. Learn the handful of "parent" shapes and everything else is a shifted, stretched, or flipped version.

Example
Shift a parabola
xy(2, 3)

y=(x2)2+3y = (x-2)^2 + 3 is y=x2y = x^2 moved right 22 and up 33, so its lowest point is (2,3)(2, 3).

Complex Numbers

Concept
i is the square root of −1

Some equations have no real solution because you'd need the square root of a negative. Mathematicians define i=1i = \sqrt{-1}, so i2=1i^2 = -1. A complex number a+bia + bi just has a real part and an "imaginary" part; you add and multiply them like binomials, using i2=1i^2 = -1.

Example
Multiply complex numbers

(2+i)(3i)=62i+3ii2=6+i+1=7+i(2+i)(3-i) = 6 - 2i + 3i - i^2 = 6 + i + 1 = 7 + i.

Exponentials and Logarithms

Concept
Logs undo exponents

logbx=y\log_b x = y is just another way to write by=xb^y = x — it asks "what power turns bb into xx?" That is why logs are the tool for solving equations where the unknown is stuck in an exponent, and why they turn multiplication into addition.

Example
Solve 2^x = 8

log28=3\log_2 8 = 3 because 23=82^3 = 8.

Going Deeper: Polynomials and Rational Functions

Concept
The remainder theorem in one line

Plugging aa into a polynomial gives the remainder when dividing by (xa)(x - a): P(a)P(a) is the remainder. If P(a)=0P(a) = 0, the division is exact and (xa)(x - a) is a factor — that is how you hunt for roots.

Reminder — Remainder and Factor theorems:P(x)=(xc)Q(x)+P(c)(xc)P(x)    P(c)=0P(x)=(x-c)Q(x)+P(c) \quad\Longrightarrow\quad (x-c)\mid P(x) \iff P(c)=0
Example
Check it

P(x)=x32x2+x2P(x) = x^3 - 2x^2 + x - 2 at x=2x = 2: 88+22=08 - 8 + 2 - 2 = 0. So (x2)(x - 2) is a factor, and dividing gives P(x)=(x2)(x2+1)P(x) = (x-2)(x^2+1) — the other two roots are the complex pair ±i\pm i.

Concept
Asymptotes, plainly

A rational function blows up where its denominator hits 00 (vertical asymptotes), and for huge xx it flattens toward the ratio of the leading terms (horizontal asymptote). y=2x+1x3y = \dfrac{2x+1}{x-3}: vertical at x=3x = 3, horizontal at y=2y = 2.

Example
A line meets a parabola (multi-step)
xy(4, 5)

Find the intersections of y=x22x3y = x^2 - 2x - 3 and y=x+1y = x + 1.

Step 1 — set the formulas equal: x22x3=x+1x^2 - 2x - 3 = x + 1.

Step 2 — bring everything to one side: x23x4=0x^2 - 3x - 4 = 0, which factors as (x4)(x+1)=0(x - 4)(x + 1) = 0.

Step 3 — zero-product property: x=4x = 4 or x=1x = -1, so the points are (4,5)(4, 5) and (1,0)(-1, 0).

Problem-Solving Playbook

Concept
Substitute to simplify

A scary equation is often a familiar one in disguise. x413x2+36=0x^4 - 13x^2 + 36 = 0 is a quadratic in u=x2u = x^2. Spot the pattern, substitute, solve the easy version, then translate back.

Example
Worked: quadratic in disguise

x413x2+36=0x^4 - 13x^2 + 36 = 0. Let u=x2u = x^2: u213u+36=(u4)(u9)=0u^2 - 13u + 36 = (u - 4)(u - 9) = 0, so u=4u = 4 or u=9u = 9. Back-substitute: x2=4x^2 = 4 or x2=9x^2 = 9, giving x=±2,±3x = \pm 2, \pm 3.