3.2 Quadratic Functions

Quadratic functions are the building blocks of curved relationships in math. They produce U-shaped graphs called parabolas, each with a vertex, axis of symmetry, and predictable behavior determined by just a few coefficients. Mastering these functions is essential for modeling real-world situations and solving optimization problems, both of which show up heavily in pre-calc and beyond.

Quadratic Functions and Parabolas

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Features of parabolas

A parabola is a U-shaped curve that's symmetric about a vertical line called the axis of symmetry. Every parabola has a few key features you need to know:

• Vertex: The turning point of the parabola, where it changes direction. This is either the minimum or maximum point of the function.
  • If $a > 0$ in $f(x) = ax^2 + bx + c$, the parabola opens upward (cup-shaped), and the vertex is a minimum.
  • If $a < 0$, the parabola opens downward (cap-shaped), and the vertex is a maximum.
• Axis of symmetry: A vertical line through the vertex that divides the parabola into two mirror-image halves. Its equation is $x = -\frac{b}{2a}$.
• Y-intercept: The point where the parabola crosses the y-axis. You find it by evaluating $f(0) = c$.
• X-intercepts (roots/zeros): The points where the parabola crosses the x-axis. A quadratic can have 0, 1, or 2 real x-intercepts depending on the discriminant.

The domain of any quadratic function is all real numbers. The range depends on the vertex and direction of opening:

• If $a > 0$: range is $[k, \infty)$
• If $a < 0$: range is $(-\infty, k]$

where $k$ is the y-coordinate of the vertex.

Graphing quadratic functions

There are two main forms you'll work with, and each one reveals different information at a glance.

Standard form: $f(x) = ax^2 + bx + c$

• $a$ controls the direction of opening and the width of the parabola.
  • $|a| > 1$: narrower than the parent function $f(x) = x^2$
  • $0 < |a| < 1$: wider than the parent function
• $b$ works together with $a$ to determine the horizontal position of the vertex. The axis of symmetry is at $x = -\frac{b}{2a}$.
• $c$ is the y-intercept (the value of $f(0)$).

Vertex form: $f(x) = a(x - h)^2 + k$

This form directly gives you the vertex at $(h, k)$. The value of $a$ still controls width and direction, just like in standard form. Vertex form is the most convenient for graphing because you can read the vertex right off the equation.

Steps to graph a quadratic function:

1. Identify the vertex. In vertex form, read $(h, k)$ directly. In standard form, compute $x = -\frac{b}{2a}$, then plug that x-value back in to get the y-coordinate.
2. Draw the axis of symmetry as a dashed vertical line through the vertex.
3. Plot the y-intercept $(0, c)$.
4. Use the axis of symmetry to reflect the y-intercept to a symmetric point on the other side of the vertex.
5. Plot one or two additional points on one side of the vertex and reflect them as well.
6. Connect the points in a smooth U-shape.

Extrema of quadratic functions

The minimum or maximum value of a quadratic function always occurs at the vertex. To find it:

1. Calculate the x-coordinate of the vertex: $x = -\frac{b}{2a}$
2. Substitute that x-value back into the function to get the y-coordinate: $f\left(-\frac{b}{2a}\right)$

That y-coordinate is the extreme value of the function.

• $a > 0$: the vertex gives a minimum (the graph goes up in both directions from there).
• $a < 0$: the vertex gives a maximum (the graph goes down in both directions from there).

These extrema are what make quadratics so useful for optimization: any quantity that can be modeled as a quadratic has a guaranteed max or min that you can calculate exactly.

Applications of quadratic optimization

Optimization problems ask you to find the maximum or minimum value of some quantity. When that quantity can be expressed as a quadratic function, the vertex gives you the answer directly.

General approach:

1. Identify what you're trying to maximize or minimize (area, profit, height, etc.).
2. Define your variables and write a constraint equation from the given information.
3. Use the constraint to express the quantity as a quadratic function of a single variable.
4. Find the vertex to determine the optimal value.
5. Interpret the result in context, including units.

Example: A farmer has 100 m of fencing to enclose a rectangular garden. What dimensions maximize the area?

1. Let $x$ = width and $y$ = length.

2. The perimeter constraint gives $2x + 2y = 100$, so $y = 50 - x$.

3. Area as a function of $x$: $A(x) = x(50 - x) = -x^2 + 50x$.

4. Find the vertex: $x = -\frac{50}{2(-1)} = 25$. Then $y = 50 - 25 = 25$.

5. Maximum area: $A(25) = -625 + 1250 = 625 \text{ m}^2$.

The optimal garden is a 25 m × 25 m square. Notice that the coefficient $a = -1$ is negative, confirming this vertex is indeed a maximum.

Solving Quadratic Equations

There are several methods for solving $ax^2 + bx + c = 0$, and each has its strengths.

Factoring works when the quadratic breaks neatly into linear factors. For example, $x^2 - 5x + 6 = 0$ factors as $(x - 2)(x - 3) = 0$, giving roots $x = 2$ and $x = 3$. This is the fastest method when it works, but not every quadratic factors cleanly over the integers.

Completing the square rewrites the equation in vertex form. This method always works and is especially useful when you need the vertex form of the function. The process: take $ax^2 + bx + c = 0$, isolate the $x$-terms, then add $\left(\frac{b}{2a}\right)^2$ to both sides to create a perfect square trinomial.

Quadratic formula is the universal method that works for any quadratic equation:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The discriminant $\Delta = b^2 - 4ac$ tells you the nature of the roots before you even solve:

• $\Delta > 0$: two distinct real roots (the parabola crosses the x-axis twice)
• $\Delta = 0$: one repeated real root (the parabola touches the x-axis at the vertex)
• $\Delta < 0$: no real roots, two complex conjugate roots (the parabola doesn't cross the x-axis)

For an Honors Pre-Calc course, you should be comfortable choosing the most efficient method for a given problem. If the equation looks factorable, try that first. If you need exact answers and the numbers are messy, go straight to the quadratic formula.