Intermediate Algebra and Functions

Intermediate Algebra and Functions accounts for about 15-20% of the ACT Math section. This area covers more advanced algebraic concepts including quadratics, complex numbers, logarithms, sequences, and function analysis.

Quadratic Equations

For ax² + bx + c = 0, the solutions are given by the Quadratic Formula:
x = (−b ± √(b² − 4ac)) / (2a)

The discriminant (b² − 4ac) tells you about the solutions:

• Positive → two distinct real solutions
• Zero → one repeated real solution
• Negative → no real solutions (two complex solutions)

Example 1: Solve 2x² − 5x − 3 = 0
Discriminant: (−5)² − 4(2)(−3) = 25 + 24 = 49 → two real solutions
Solutions: x = (5 ± 7)/4 → x = 3 or x = −1/2

Example 2: Solve x² + 6x + 5 = 0 by factoring.
Find two numbers that add to 6 and multiply to 5: 1 and 5.
(x + 1)(x + 5) = 0 → x = −1 or x = −5

Example 3: Solve x² − 4x + 1 = 0 using the quadratic formula.
a = 1, b = −4, c = 1. Discriminant = 16 − 4 = 12.
x = (4 ± √12)/2 = (4 ± 2√3)/2 = 2 ± √3

Completing the Square

This technique converts ax² + bx + c into vertex form a(x − h)² + k.
Example: Write x² + 8x + 3 in vertex form.
Step 1: Take half of b: 8/2 = 4. Square it: 16.
Step 2: Add and subtract 16: (x² + 8x + 16) − 16 + 3 = (x + 4)² − 13
Vertex form: (x + 4)² − 13, so the vertex is at (−4, −13).

Complex Numbers

i = √(−1), so i² = −1. To add/subtract complex numbers, combine real and imaginary parts separately: (3 + 2i) + (1 − 5i) = 4 − 3i. To multiply, use FOIL and replace i² with −1.

Example 1: (2 + 3i)(4 − i) = 8 − 2i + 12i − 3i² = 8 + 10i − 3(−1) = 11 + 10i

Powers of i cycle every 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1, i⁵ = i, ...
To find iⁿ, divide n by 4 and use the remainder. Example: i²³ → 23 ÷ 4 = 5 remainder 3, so i²³ = i³ = −i.

Functions and Their Properties

• Domain: all valid input values (watch for division by zero and negative square roots)
• Range: all possible output values
• Composition: f(g(x)) means "apply g first, then f"
• Inverse: f⁻¹(x) reverses the function — swap x and y, then solve for y

Example 1: If f(x) = 2x + 1 and g(x) = x², find f(g(3)).
g(3) = 9, then f(9) = 2(9) + 1 = 19.

Example 2: Find the inverse of f(x) = 3x − 7.
Write y = 3x − 7. Swap: x = 3y − 7. Solve: y = (x + 7)/3. So f⁻¹(x) = (x + 7)/3.

Example 3: What is the domain of f(x) = √(x − 4)?
The expression under the root must be ≥ 0: x − 4 ≥ 0 → x ≥ 4. Domain: [4, ∞).

Logarithms

log_b(x) = y means b^y = x. Key properties:

• log(ab) = log(a) + log(b)
• log(a/b) = log(a) − log(b)
• log(aⁿ) = n·log(a)
• log_b(b) = 1 and log_b(1) = 0

Example 1: Solve log₂(x) = 5. Convert to exponential form: 2⁵ = x, so x = 32.

Example 2: Simplify log₃(81).
81 = 3⁴, so log₃(81) = 4.

Example 3: If log(x) + log(x − 3) = log(10), find x.
Combine: log(x(x − 3)) = log(10) → x² − 3x = 10 → x² − 3x − 10 = 0 → (x − 5)(x + 2) = 0.
x = 5 or x = −2. Since log requires positive arguments, x = 5.

Sequences and Series

Example: In an arithmetic sequence, the 3rd term is 11 and the 7th term is 23. Find the 15th term.
d = (23 − 11)/(7 − 3) = 12/4 = 3. a₁ = 11 − 2(3) = 5.
a₁₅ = 5 + 14(3) = 47.

Absolute Value Functions and Rational Functions

The graph of y = |x − h| + k is a V-shape with vertex at (h, k). The function y = |f(x)| reflects any negative output above the x-axis.

For rational functions, vertical asymptotes occur where the denominator equals zero (after simplification). Horizontal asymptotes depend on the degree of the numerator vs. denominator: same degree → ratio of leading coefficients; numerator degree less → y = 0.

Example: Find the vertical asymptote(s) of f(x) = (x + 2)/(x² − 4).
Factor: f(x) = (x + 2)/((x + 2)(x − 2)). Cancel common factor: f(x) = 1/(x − 2) with a hole at x = −2.
Vertical asymptote: x = 2. (x = −2 is a hole, not an asymptote.)

Matrices (Basic): Matrix addition: add corresponding entries. Scalar multiplication: multiply each entry. The ACT rarely goes beyond 2×2 matrices.

Common Mistakes

Quick Reference — Formula Sheet

ACT Strategies for Intermediate Algebra

• Try factoring first: Before reaching for the quadratic formula, check if the expression factors easily — it is faster.
• Vertex shortcut: The vertex of y = ax² + bx + c is at x = −b/(2a). This alone answers many parabola questions.
• Log confusion? Convert to exponential form: log_b(x) = y becomes b^y = x.
• Sequence identification: Subtract consecutive terms. If the differences are constant → arithmetic. If the ratios are constant → geometric.
• Parentheses matter: When plugging values into functions, use parentheses carefully, especially with negatives: f(−2) = (−2)², not −2².
• Eliminate fast: Plug in simple numbers (2, 3, or 5) to test function expressions and quickly eliminate wrong answers.
• Powers of i: Divide the exponent by 4 and use the remainder. This instantly solves any iⁿ problem.
• Discriminant shortcut: If the ACT asks "how many solutions," compute b²−4ac only — you do not need to solve the equation.

Practice Walkthrough

ACT-Style Problem: What is the sum of the solutions of 2x² − 7x − 15 = 0?

A. −7/2    B. 7/2    C. −15/2    D. 5

Strategy — Vieta's formulas shortcut: For ax² + bx + c = 0, the sum of the roots = −b/a.
Sum = −(−7)/2 = 7/2. Answer: B

Verification by solving: Factor or use the formula: 2x² − 7x − 15 = (2x + 3)(x − 5) = 0 → x = −3/2 or x = 5.
Sum = −3/2 + 5 = −3/2 + 10/2 = 7/2. Confirmed.

Eliminating wrong answers: Choice A (−7/2) forgets the negative sign in −b/a. Choice C (−15/2) confuses the sum with the product formula (product = c/a). Choice D (5) is just one of the roots. Only B (7/2) is correct.