Quadratic Functions and Equations

TL;DR

Quadratic functions create a U-shaped curve called a parabola, and quadratic equations help us find where this curve crosses the x-axis (its roots). You can solve these equations by factoring, using the quadratic formula, or completing the square. Understanding them helps you model real-world situations like projectile motion and optimizing areas.

1. The Mental Model

Imagine throwing a ball; its path through the air is a curve. That curve is a parabola, and a quadratic function is the math behind it. When you want to know exactly when the ball hits the ground, you're solving a quadratic equation.

2. The Core Material

Quadratic functions are polynomial functions of degree 2. Their standard form is $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants and $a \neq 0$. The graph of a quadratic function is always a parabola.

When you set a quadratic function equal to zero ($ax^2 + bx + c = 0$), you get a quadratic equation. The solutions to this equation are called the roots, zeros, or x-intercepts of the function, which are the points where the parabola crosses the x-axis. A quadratic equation can have two, one, or no real solutions.

Understanding the Parts of a Quadratic Function ($f(x) = ax^2 + bx + c$)

• $a$: Controls the parabola's direction and width.
  • If $a > 0$, the parabola opens upwards (like a "U").
  • If $a < 0$, the parabola opens downwards (like an "n").
  • A larger absolute value of $a$ makes the parabola narrower; a smaller absolute value makes it wider.
• $c$: This is the y-intercept. It's where the parabola crosses the y-axis (when $x=0$, $f(0) = c$).
• Vertex: The highest or lowest point of the parabola. Its x-coordinate is given by $x = -b / (2a)$. You find the y-coordinate by plugging this x-value back into the function.
• Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two mirror images. Its equation is $x = -b / (2a)$.

Methods for Solving Quadratic Equations ($ax^2 + bx + c = 0$)

There are three main ways to find the roots:

1. Factoring

This method works well when the quadratic expression can be easily factored into two linear expressions.
1. Set the equation to 0.
2. Factor the quadratic expression.
3. Set each factor equal to 0 and solve for $x$.

Example: Solve $x^2 - 5x + 6 = 0$
1. Already set to 0.
2. Factor: $(x - 2)(x - 3) = 0$
3. Set factors to zero:
* $x - 2 = 0 \implies x = 2$
* $x - 3 = 0 \implies x = 3$
The roots are $x=2$ and $x=3$.

2. The Quadratic Formula

This is your reliable go-to method; it always works, even when factoring is difficult or impossible.
The formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The part under the square root, $b^2 - 4ac$, is called the discriminant ($\Delta$). It tells you about the nature of the roots:
* If $\Delta > 0$, there are two distinct real roots. (The parabola crosses the x-axis twice.)
* If $\Delta = 0$, there is exactly one real root (a repeated root). (The parabola touches the x-axis at its vertex.)
* If $\Delta < 0$, there are no real roots (two complex roots). (The parabola never crosses the x-axis.)

Example: Solve $2x^2 + 3x - 5 = 0$ using the quadratic formula.
Here, $a=2$, $b=3$, $c=-5$.
$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)}$
$x = \frac{-3 \pm \sqrt{9 + 40}}{4}$
$x = \frac{-3 \pm \sqrt{49}}{4}$
$x = \frac{-3 \pm 7}{4}$

Two solutions:
* $x_1 = \frac{-3 + 7}{4} = \frac{4}{4} = 1$
* $x_2 = \frac{-3 - 7}{4} = \frac{-10}{4} = -\frac{5}{2}$
The roots are $x=1$ and $x = -5/2$.

3. Completing the Square

This method transforms the quadratic equation into a "perfect square trinomial" plus a constant. It's often used to convert the standard form of a quadratic function into vertex form ($f(x) = a(x-h)^2 + k$), where $(h, k)$ is the vertex.

1. Move the constant term to the right side of the equation.
2. Divide all terms by $a$ (if $a \neq 1$).
3. Take half of the coefficient of the $x$ term, square it, and add it to both sides of the equation.
4. Factor the left side as a perfect square trinomial.
5. Take the square root of both sides and solve for $x$.

Example: Solve $x^2 + 6x + 5 = 0$ by completing the square.
1. $x^2 + 6x = -5$
2. (a is already 1)
3. Half of $6$ is $3$; $3^2 = 9$. Add $9$ to both sides:
$x^2 + 6x + 9 = -5 + 9$
$x^2 + 6x + 9 = 4$
4. Factor the left side:
$(x + 3)^2 = 4$
5. Take the square root of both sides:
$x + 3 = \pm\sqrt{4}$
$x + 3 = \pm 2$

Two solutions:
* $x + 3 = 2 \implies x = 2 - 3 \implies x = -1$
* $x + 3 = -2 \implies x = -2 - 3 \implies x = -5$
The roots are $x=-1$ and $x=-5$.

graph TD
    A["Start: Quadratic Equation (ax² + bx + c = 0)"] --> B{Can it be easily factored?};
    B -- Yes --> C["Method 1: Factoring"];
    C --> D["Find factors (e.g., (x-p)(x-q)=0)"];
    D --> E["Set each factor to zero"];
    E --> F["Solve for x (roots)"];

    B -- No --> G{Is completing the square preferred? (e.g., for vertex form)};
    G -- Yes --> H["Method 3: Completing the Square"];
    H --> I["Isolate x² and x terms"];
    I --> J["Add (b/2a)² to both sides"];
    J --> K["Factor as a perfect square"];
    K --> L["Take square root, solve for x"];
    L --> F;

    G -- No --> M["Method 2: Quadratic Formula"];
    M --> N["Identify a, b, c values"];
    N --> O["Substitute into: x = [-b ± sqrt(b²-4ac)] / 2a"];
    O --> P{Calculate Discriminant (b²-4ac)};
    P -- >0 --> Q["Two real roots"];
    Q --> F;
    P -- =0 --> R["One real root"];
    R --> F;
    P -- <0 --> S["No real roots (complex roots)"];
    S --> T["End"];

    F --> T;

3. Worked Example

Let's say you're designing a rectangular garden plot. You want the length to be 5 meters longer than the width, and the total area of the garden should be 84 square meters. What are the dimensions of the garden?

1. Define variables:
   • Let the width be $w$ meters.
   • The length is $w + 5$ meters.
2. Set up the equation:
   • Area = length $\times$ width
   • $84 = (w + 5) \times w$
   • $84 = w^2 + 5w$
3. Rearrange into standard quadratic form ($ax^2 + bx + c = 0$):
   • $w^2 + 5w - 84 = 0$
4. Solve the quadratic equation. Let's use factoring for this one:
   • We need two numbers that multiply to -84 and add to 5.
   • Consider pairs: (1, -84), (-1, 84), (2, -42), (-2, 42), (3, -28), (-3, 28), (4, -21), (-4, 21), (6, -14), (-6, 14), (7, -12), (-7, 12).
   • The pair 12 and -7 works: $12 \times (-7) = -84$ and $12 + (-7) = 5$.
   • So, the equation factors to: $(w + 12)(w - 7) = 0$
5. Find the possible values for $w$:
   • $w + 12 = 0 \implies w = -12$
   • $w - 7 = 0 \implies w = 7$
6. Interpret the solution:
   • Since width cannot be negative, $w = -12$ is not a valid physical dimension.
   • Therefore, the width $w = 7$ meters.
   • The length is $w + 5 = 7 + 5 = 12$ meters.

Answer: The garden should be 7 meters wide and 12 meters long. (Check: $7 \times 12 = 84$, which is correct.)

4. Key Takeaways

• Quadratic functions ($f(x) = ax^2 + bx + c$) graph as parabolas.
• Quadratic equations ($ax^2 + bx + c = 0$) help you find the x-intercepts (roots) of these parabolas.
• The sign of '$a$' tells you if the parabola opens up ($a>0$) or down ($a<0$).
• The constant '$c$' is the y-intercept of the parabola.
• You have three main tools to solve quadratic equations: factoring, the quadratic formula, and completing the square.
• The quadratic formula is universally applicable, and the discriminant ($b^2 - 4ac$) tells you how many real roots to expect.
• Real-world problems often involve quadratic equations, like calculating areas or projectile paths.

Common Mistakes to Avoid:
- Forgetting the $\pm$ in the quadratic formula or when taking a square root. This halves your solutions!
- Incorrectly identifying $a$, $b$, and $c$ in the quadratic formula, especially when terms are missing (e.g., if $3x^2 - 12 = 0$, then $b=0$).
- Not setting the quadratic equation to zero before attempting to factor or use the quadratic formula.
- Dividing by a variable in the equation, which can cause you to lose a valid solution.
- Mistakes with negative signs, especially in the $b^2 - 4ac$ part of the formula.

5. Now Try It

You're designing a picture frame. The picture itself is 8 inches by 10 inches. You want to add a uniform border around it, and the total area of the framed picture should be 120 square inches. What should the width of the border be?

What to do:
1. Draw a diagram to visualize the problem.
2. Define a variable for the border width.
3. Write expressions for the total width and total length of the framed picture.
4. Set up a quadratic equation using the total area.
5. Solve the equation to find the border width.

What success looks like:
You should find a positive value for the border width, expressed in inches. You can check your answer by plugging it back into your area equation to see if it equals 120.