AP PRECALCULUS — PREREQUISITE REVIEW
Linear and Quadratic Functions
Notes — Prerequisite Topic 1
💡 Learning Objectives By the end of this lesson you will be able to:
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1. Linear Functions
A linear function describes a relationship in which the output changes by a constant amount every time the input increases by 1. Its graph is a straight line, and its algebraic form captures two essential ideas: where the line crosses the y-axis and how steeply it rises or falls.
1.1 Forms of a linear equation
Form | Equation | What it tells you directly |
|---|---|---|
Slope–intercept | y = mx + b | slope m, y-intercept (0, b) |
Point–slope | y − y₁ = m(x − x₁) | slope m, a point (x₁, y₁) |
Standard | Ax + By = C | useful for intercepts & systems |
Horizontal | y = k | constant output; slope = 0 |
Vertical | x = k | undefined slope; not a function |
Three linear functions compared. A larger |m| means a steeper line; a negative m slopes downward; m = 0 gives a horizontal line.
💡 Slope between two points Given two points (x₁, y₁) and (x₂, y₂) on a line, the slope is: m = (y₂ − y₁) / (x₂ − x₁) In words: rise over run — the vertical change divided by the horizontal change. |
📘 Example — Writing a linear equation from two points Find an equation of the line through (−1, 4) and (3, −4). Step 1. Compute the slope: m = (−4 − 4) / (3 − (−1)) = −8 / 4 = −2. Step 2. Use point–slope form with the point (3, −4): y − (−4) = −2(x − 3) y + 4 = −2x + 6 y = −2x + 2. Check: at x = −1, y = −2(−1) + 2 = 4. ✓ |
1.2 Solving linear equations
To isolate the variable, undo operations in reverse order. Do the same thing to both sides to keep the equation balanced.
📘 Example — Multi-step linear equation Solve: 3(2x − 5) + 4 = 5x + 7. 6x − 15 + 4 = 5x + 7 (distribute) 6x − 11 = 5x + 7 (combine like terms) x = 18 (subtract 5x, add 11) |
⚠️ Watch out When an equation simplifies to a contradiction like 0 = 7, there is no solution. When it simplifies to an identity like 0 = 0, every real number is a solution. |
1.3 Solving linear inequalities
The rules for inequalities match those for equations, with one important twist.
⚠️ The flip rule When you multiply or divide both sides of an inequality by a negative number, reverse the inequality symbol. Example: −2x < 6 ⟹ x > −3 (symbol flipped). |
A number-line representation of 2x − 3 < 5. An open circle marks a strict inequality; a filled circle would mark ≤ or ≥.
📘 Example — Compound inequality Solve −1 ≤ 3 − 2x < 7. Subtract 3 from each part: −4 ≤ −2x < 4. Divide by −2 and flip both symbols: 2 ≥ x > −2. In standard form: −2 < x ≤ 2. Interval notation: (−2, 2]. |
2. Quadratic Functions
A quadratic function has the form f(x) = ax² + bx + c, where a ≠ 0. Its graph is a parabola — a U-shaped curve that opens up if a > 0 and opens down if a < 0.
Key features of the parabola y = x² − 4x + 1: the vertex, axis of symmetry, and two x-intercepts where the curve crosses the x-axis.
2.1 Three useful forms
Form | Equation | Reveals directly |
|---|---|---|
Standard | f(x) = ax² + bx + c | y-intercept = c |
Vertex | f(x) = a(x − h)² + k | vertex = (h, k) |
Factored | f(x) = a(x − r₁)(x − r₂) | x-intercepts r₁, r₂ |
💡 Vertex from standard form The x-coordinate of the vertex of f(x) = ax² + bx + c is: x = −b / (2a). Substitute this x-value back into f to find the y-coordinate. |
📘 Example — Finding the vertex For f(x) = 2x² − 8x + 5: x = −(−8) / (2 · 2) = 8/4 = 2. f(2) = 2(4) − 16 + 5 = −3. Vertex: (2, −3). Because a = 2 > 0, this is the minimum. |
2.2 Solving quadratic equations
A quadratic equation ax² + bx + c = 0 can have two, one, or zero real solutions. Three standard methods:
- Factoring: rewrite as a product equal to 0, then set each factor equal to 0.
- Completing the square: turn ax² + bx into a perfect square, then isolate x.
- Quadratic formula: always works for any quadratic.
💡 The Quadratic Formula For ax² + bx + c = 0 with a ≠ 0: x = ( −b ± √(b² − 4ac) ) / (2a) The quantity Δ = b² − 4ac is called the discriminant. |
How the discriminant Δ = b² − 4ac controls the number of real roots: two when Δ > 0, one when Δ = 0, and none when Δ < 0 (complex roots only).
📘 Example — Applying the formula Solve 3x² − 5x − 2 = 0. a = 3, b = −5, c = −2. Δ = (−5)² − 4(3)(−2) = 25 + 24 = 49. x = (5 ± 7) / 6 ⟹ x = 2 or x = −1/3. |
2.3 Quadratic inequalities
To solve a quadratic inequality, first find the zeros of the related equation. These zeros split the number line into regions. Test one value from each region to decide where the inequality holds.
📘 Example — A quadratic inequality Solve x² − 2x − 3 ≤ 0. Factor: (x − 3)(x + 1) ≤ 0. Zeros at x = −1 and x = 3. Test x = 0 (between the zeros): (0 − 3)(0 + 1) = −3 ≤ 0 ✓. Test x = −2 and x = 4: both give positive results. Solution: −1 ≤ x ≤ 3, or in interval notation [−1, 3]. |
3. Summary — What You Should Be Able to Do
- Move fluently between slope–intercept, point–slope, and standard forms of a line
- Solve any linear equation or inequality, remembering to flip the inequality when multiplying or dividing by a negative
- Read off the vertex, axis of symmetry, and intercepts of a parabola from each of its three forms
- Solve quadratics by factoring, completing the square, or the quadratic formula — and choose the fastest route for the given problem
- Use the discriminant to predict how many real solutions a quadratic has before solving