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AP Precalculus: Quadratic Functions

Master parabolas, vertex form, graphing techniques, and all solving methods for quadratic equations

📐 3 Forms 📊 Graphing 🧮 4 Solving Methods 🎯 Discriminant

📚 Understanding Quadratic Functions

Quadratic functions are polynomial functions of degree 2, producing parabola-shaped graphs. They are fundamental to AP Precalculus and appear in countless real-world applications including projectile motion, optimization problems, and economics. This guide covers all essential formulas and solving techniques.

🎯 The Three Forms of Quadratic Functions

Standard Form

\(f(x) = ax^2 + bx + c\)

Best for: Finding y-intercept (\(c\)), using quadratic formula

Vertex Form

\(f(x) = a(x - h)^2 + k\)

Best for: Identifying vertex \((h, k)\), graphing quickly

Factored Form

\(f(x) = a(x - r_1)(x - r_2)\)

Best for: Finding x-intercepts (roots) \(r_1\) and \(r_2\)

1 Standard Form & Key Features

The standard form \(f(x) = ax^2 + bx + c\) where \(a \neq 0\) is the most common way to write a quadratic function. The coefficient \(a\) determines the parabola's direction and width.

Key Properties from Standard Form

  • Direction: Opens upward if \(a > 0\), downward if \(a < 0\)
  • Y-intercept: The point \((0, c)\) — always equal to the constant term \(c\)
  • Axis of symmetry: \(x = -\frac{b}{2a}\) — the vertical line through the vertex
  • Width: Larger \(|a|\) = narrower parabola; smaller \(|a|\) = wider parabola
📌 Example

Given: \(f(x) = 2x^2 - 8x + 6\)

\(a = 2 > 0\) → Opens upward

Y-intercept: \((0, 6)\)

Axis of symmetry: \(x = -\frac{-8}{2(2)} = \frac{8}{4} = 2\)

2 Vertex & Maximum/Minimum Value

The vertex is the highest or lowest point on a parabola. It represents the maximum or minimum value of the quadratic function.

Vertex Formula (from Standard Form) \[x_{\text{vertex}} = -\frac{b}{2a}\] \[y_{\text{vertex}} = f\left(-\frac{b}{2a}\right)\]

Maximum vs. Minimum

  • If \(a > 0\) (opens up): vertex is a minimum point
  • If \(a < 0\) (opens down): vertex is a maximum point
  • The y-coordinate of the vertex is the max/min value of the function
  • The x-coordinate tells you where the max/min occurs
📌 Example

Find the vertex of: \(f(x) = -3x^2 + 12x - 7\)

\(x_{\text{vertex}} = -\frac{12}{2(-3)} = -\frac{12}{-6} = 2\)

\(y_{\text{vertex}} = f(2) = -3(2)^2 + 12(2) - 7 = -12 + 24 - 7 = 5\)

Vertex: \((2, 5)\)

Since \(a = -3 < 0\), this is a maximum value of 5 at \(x = 2\)

💡 Optimization Tip

In word problems asking for maximum profit, minimum cost, or optimal dimensions, find the vertex! The x-coordinate gives the optimal input, and the y-coordinate gives the optimal output value.

3 Graphing Quadratic Functions

Every quadratic function graphs as a parabola — a U-shaped curve that is symmetric about its axis of symmetry.

Steps to Graph a Quadratic

  1. Determine direction: Check the sign of \(a\) (up if \(a > 0\), down if \(a < 0\))
  2. Find the vertex: Use \(x = -\frac{b}{2a}\) and calculate \(y\)
  3. Draw the axis of symmetry: Vertical line through the vertex
  4. Find the y-intercept: Evaluate \(f(0) = c\)
  5. Find x-intercepts (if any): Solve \(f(x) = 0\)
  6. Plot additional points: Use symmetry to reflect points across the axis

Matching Functions to Graphs

  • Sign of \(a\): Determines if parabola opens up or down
  • Magnitude of \(|a|\): Larger = narrower, smaller = wider
  • Value of \(c\): Y-intercept location
  • Vertex location: Use to match specific graphs

4 Solving by Square Roots

When a quadratic equation has no linear term (\(b = 0\)), you can isolate \(x^2\) and take the square root of both sides.

Square Root Method If \(ax^2 = k\), then \(x = \pm\sqrt{\frac{k}{a}}\)

When to Use This Method

  • Equation is in the form \(ax^2 = k\) or \(ax^2 + c = 0\)
  • There is no \(x\) term (no \(bx\))
  • Can also be used with \((x - h)^2 = k\) (vertex form)
📌 Examples

Example 1: Solve \(3x^2 = 27\)

\(x^2 = 9\) → \(x = \pm\sqrt{9} = \pm 3\)

Example 2: Solve \(2x^2 - 50 = 0\)

\(2x^2 = 50\) → \(x^2 = 25\) → \(x = \pm 5\)

Example 3: Solve \((x - 4)^2 = 16\)

\(x - 4 = \pm 4\) → \(x = 4 + 4 = 8\) or \(x = 4 - 4 = 0\)

⚠️ Don't Forget the ±

When taking the square root of both sides, always include both the positive and negative roots: \(\pm\sqrt{k}\)

5 Solving by Factoring

Factoring converts a quadratic into a product of two linear factors. Apply the Zero Product Property: if \(AB = 0\), then \(A = 0\) or \(B = 0\).

Zero Product Property If \((dx + p)(ex + q) = 0\), then \(dx + p = 0\) or \(ex + q = 0\)

Factoring Techniques

  • GCF Factoring: Factor out the greatest common factor first
  • Simple Trinomials: \(x^2 + bx + c = (x + m)(x + n)\) where \(mn = c\) and \(m + n = b\)
  • AC Method: For \(ax^2 + bx + c\), find factors of \(ac\) that sum to \(b\)
  • Difference of Squares: \(a^2 - b^2 = (a+b)(a-b)\)
📌 Examples

Example 1: Solve \(x^2 + 5x + 6 = 0\)

Factor: \((x + 2)(x + 3) = 0\)

\(x + 2 = 0\) → \(x = -2\) or \(x + 3 = 0\) → \(x = -3\)

Example 2: Solve \(2x^2 - 7x + 3 = 0\)

Factor: \((2x - 1)(x - 3) = 0\)

\(2x - 1 = 0\) → \(x = \frac{1}{2}\) or \(x - 3 = 0\) → \(x = 3\)

💡 When to Factor

Factoring is fastest when the quadratic has integer roots. If you don't find integer factors after a minute, switch to the quadratic formula.

6 Completing the Square

Completing the square transforms a quadratic from standard form to vertex form. This method always works and reveals the vertex directly.

Steps to Complete the Square

  1. Start with \(ax^2 + bx + c = 0\). If \(a \neq 1\), divide everything by \(a\)
  2. Move the constant to the right side: \(x^2 + \frac{b}{a}x = -\frac{c}{a}\)
  3. Take half of the coefficient of \(x\), square it: \(\left(\frac{b}{2a}\right)^2\)
  4. Add this value to both sides of the equation
  5. Factor the left side as a perfect square: \(\left(x + \frac{b}{2a}\right)^2\)
  6. Take the square root of both sides and solve for \(x\)
📌 Example

Solve: \(x^2 + 6x + 5 = 0\)

Step 1-2: \(x^2 + 6x = -5\)

Step 3: Half of 6 is 3; \(3^2 = 9\)

Step 4: \(x^2 + 6x + 9 = -5 + 9 = 4\)

Step 5: \((x + 3)^2 = 4\)

Step 6: \(x + 3 = \pm 2\) → \(x = -3 + 2 = -1\) or \(x = -3 - 2 = -5\)

7 The Quadratic Formula

The quadratic formula is the universal method for solving any quadratic equation \(ax^2 + bx + c = 0\). It always works, regardless of whether the quadratic factors nicely.

The Quadratic Formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Using the Formula

  1. Write the equation in standard form: \(ax^2 + bx + c = 0\)
  2. Identify \(a\), \(b\), and \(c\) (include their signs!)
  3. Substitute into the formula
  4. Simplify under the radical first, then continue
  5. The \(\pm\) gives two solutions: one with +, one with −
📌 Example

Solve: \(2x^2 + 5x - 3 = 0\)

\(a = 2\), \(b = 5\), \(c = -3\)

\(x = \frac{-5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}\)

\(x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}\) or \(x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\)

⚠️ Common Mistakes

• Don't forget the negative sign in front of \(b\)
• The entire numerator (including both terms) is divided by \(2a\)
• Double-check the sign of \(c\) when substituting

8 The Discriminant

The discriminant is the expression under the square root in the quadratic formula: \(D = b^2 - 4ac\). It tells you the nature and number of solutions without solving the equation.

Discriminant Formula \[D = b^2 - 4ac\]

What the Discriminant Tells You

\(D > 0\)
Two distinct real solutions
Parabola crosses x-axis twice
\(D = 0\)
One repeated real solution
Parabola touches x-axis once (at vertex)
\(D < 0\)
Two complex solutions
Parabola never crosses x-axis
📌 Examples

\(x^2 - 5x + 6 = 0\): \(D = 25 - 24 = 1 > 0\) → Two real solutions

\(x^2 - 4x + 4 = 0\): \(D = 16 - 16 = 0\) → One real solution (double root)

\(x^2 + x + 1 = 0\): \(D = 1 - 4 = -3 < 0\) → Two complex solutions

💡 Quick Check

Before solving a quadratic, calculate the discriminant first. It tells you what type of answer to expect and can help you catch calculation errors.

9 Quadratic Word Problems

Quadratic functions model many real-world situations, especially those involving projectile motion, area optimization, and business applications.

Common Word Problem Types

Projectile Motion

\(h(t) = -16t^2 + v_0t + h_0\)
Find max height (vertex) or when it hits ground (\(h = 0\))

Area Problems

Maximize area given perimeter constraint; solve for dimensions

Revenue/Profit

\(R(x) = (\text{price})(\text{quantity})\)
Find price that maximizes revenue

Number Problems

Find two numbers with given sum/product relationship

📌 Example: Projectile Motion

Problem: A ball is thrown upward with initial velocity 64 ft/s from a height of 80 ft. The height is \(h(t) = -16t^2 + 64t + 80\). Find the maximum height.

Solution: Use the vertex formula.

\(t = -\frac{64}{2(-16)} = -\frac{64}{-32} = 2\) seconds

\(h(2) = -16(4) + 64(2) + 80 = -64 + 128 + 80 = 144\) feet

Answer: Maximum height is 144 feet at \(t = 2\) seconds

📋 Which Solving Method Should You Use?

Square Roots

No \(x\) term (only \(x^2\)) or equation is in vertex form

Factoring

Coefficients are small integers and factors are easy to find

Completing the Square

Need to convert to vertex form or derive the quadratic formula

Quadratic Formula

Always works! Use when factoring is difficult or impossible

💡 AP Exam Strategy

On multiple choice, the quadratic formula is often fastest. On free response, show your work with factoring or completing the square when possible to earn method points.

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