high school Irrational Functions

Irrational Function

An irrational function hides a variable under a root sign, like the side √S of a square with area S. The basic form y = √x is the inverse of y = x² (x ≥ 0), so the two graphs mirror across the line y = x, and the radicand must stay non-negative. The general form y = √(ax + b) + c starts at (−b/a, c) where the radicand is zero, and crossing a line means squaring, which can create extraneous roots to check. Toggle y = x² and move a, b, and c and the start point and direction shift.

Where do Irrational Functions Come From?

📐 Side Length of a Square

①What's the side length of a square with area S?

②side = √S — square root of the area

③Area 1 → 1, Area 4 → 2, Area 9 → 3

④But Area 2 → √2 ≈ 1.414… (irrational!)

⑤Functions involving square roots are 'irrational functions'

🪞 Inverse Relationship Insight

①Swap x and y in y = x² (x≥0)

②x = y² → y = √x (positive only)

③y = √x is the inverse of the 'positive half' of y = x²

④Both graphs are reflections across y = x

⑤That's the 'identity' of irrational functions — inverse of a quadratic!

Properties of y = √x

Basic Irrational Function

y = √x

Domain: x ≥ 0 / Range: y ≥ 0 / Starts at origin

Properties of y = √x

Features of the Basic Form

Domain

The radicand must be ≥ 0

x ≥ 0

Range

A square root is always ≥ 0

y ≥ 0

Monotonicity

y grows with x (but slower)

Always increasing

Growth Rate

1→4: y +1; 4→9: y +1 (gap widens)

Slows down

General Form y = √(ax + b) + c

a (sign sets direction)1

b (shifts start)0

c (vertical shift)0

General Irrational Function

y = √(ax + b) + c

Start: (−b/a, c) → a > 0 extends right, a < 0 extends left

🔑 Quick Way to Find the Start Point

①Solve radicand = 0: ax + b = 0 → x = −b/a

②Then y = √0 + c = c

③Start at (−b/a, c)

④a > 0 → x ≥ −b/a (grows right)

⑤a < 0 → x ≤ −b/a (grows left)

Intersection of Irrational Function & Line

Find Intersections

√(ax+b) + c = mx + n → square both sides

Squaring requires extraneous-root check!

⚠️ Trap of Squaring — Extraneous Roots

①Solve √x = x − 2 by squaring

②x = (x−2)² = x²−4x+4 → x²−5x+4=0

③x = 1 or x = 4

④Check: x=1 → √1 = 1, 1−2 = −1 → 1 ≠ −1 (extraneous!)

⑤x=4 → √4 = 2, 4−2 = 2 → ✓ (real)

⑥Squaring loses sign info — always plug back into the original equation!

Solution Steps

Steps for Irrational Equations

Step	Action	Caution
Step 1	Isolate the radical	√(…) = (expr)
Step 2	Square both sides	Check (expr) ≥ 0
Step 3	Solve the resulting equation	It becomes quadratic
Step 4	Verify in the original equation	Drop extraneous roots!

Work It Out

Example 1

Find the domain of the irrational function y = √(2x − 6) + 1.

The expression under the radical must be at least 0.

2x − 6 ≥ 0

Solve the inequality.

x ≥ 3

▸ Domain: {x | x ≥ 3}

For an irrational function, set the radicand ≥ 0 to fix the domain first.

Example 2

Describe how y = √(x − 2) + 3 is a translation of y = √x, and find its starting point.

x becomes x − 2 (shift +2 in x), and +3 outside (shift +3 in y).

y = √x → y = √(x − 2) + 3

The starting point is (0, 0) moved by the same amounts.

(0, 0) → (2, 3)

▸ Shift 2 in x and 3 in y; starting point (2, 3)

The starting point of y = √(x − p) + q is (p, q).

Wrap-up

Basic Form

y = √x

Dom x≥0, Ran y≥0

General Form

y = √(ax+b)+c

Start (−b/a, c)

Grade-10 school exam type

Find the x-coordinate where y = √(x + 1) meets the line y = x − 1.

①1

②2

③3

④4

⑤5

▸ ③ 3

Set the two equal and square both sides (with x − 1 ≥ 0).

√(x + 1) = x − 1 ⇒ x + 1 = (x − 1)²

This gives x² − 3x = 0, so x = 0 or 3; the condition x ≥ 1 leaves x = 3.

x² − 3x = 0 ⇒ x = 3 (x = 0 is extraneous)

🎯 Exam Points

①Radicand ≥ 0 → key domain condition

②Start (−b/a, c): a > 0 right, a < 0 left

③y = √x is the inverse of y = x² (x≥0) — reflected across y = x

④Irrational equations: square, then verify in the original (extraneous!)

⑤Shape: start point → slowly increasing curve

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Rational Functions

Exponents & Logarithms

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