Quadratic Functions: Standard Form, Vertex, and Optimization

Quadratic Functions

The Standard Form of a Quadratic Function

A quadratic function is a polynomial function of degree two, typically written in the standard form:

• $f(x) = ax^2 + bx + c$

• Where a, b, and c are real numbers, and a \neq 0.

• The graph of a quadratic function is a parabola.

Example 1: Graph the quadratic function $f(x) = -4(x-1)^2 + 4$.

This function is in vertex form: $f(x) = a(x-h)^2 + k$, where the vertex is $(h, k)$.

The Vertex of a Parabola

The vertex of a parabola given by $f(x) = ax^2 + bx + c$ is the point where the function reaches its maximum or minimum value.

• The vertex can be found using the formula:

  • $x = -\frac{b}{2a}$

  • The y-coordinate is $f(-\frac{b}{2a})$

• If $a > 0$, the parabola opens upward and the vertex is a minimum.

• If $a < 0$, the parabola opens downward and the vertex is a maximum.

Example 2: Graph the quadratic function $f(x) = -4x^2 + x + 4$.

Minimum and Maximum Values of Quadratic Functions

Quadratic functions have either a minimum or maximum value, depending on the sign of the leading coefficient a.

• If $a > 0$, the function has a minimum at the vertex.

• If $a < 0$, the function has a maximum at the vertex.

• The minimum or maximum value occurs at $x = -\frac{b}{2a}$.

Example 3: Consider the quadratic function $f(x) = -4x^2 + 16x + 1000$. Determine, without graphing, whether the function has a minimum or maximum value and where it occurs. State the domain and range.

• Since $a = -4 < 0$, the function has a maximum.

• The maximum occurs at $x = -\frac{16}{2 \times (-4)} = 2$.

• Domain: All real numbers ($-\infty < x < \infty$).

• Range: All real numbers less than or equal to the maximum value.

Strategy for Solving Maximization or Minimization Problems

Quadratic functions are often used to solve optimization problems, such as maximizing area or minimizing cost.

1. Read the problem carefully and decide which quantity is to be maximized or minimized.

2. Use the conditions of the problem to express the quantity as a function in one variable.

3. Rewrite the function in the form $f(x) = ax^2 + bx + c$.

4. Calculate $x = -\frac{b}{2a}$. If $a > 0$, $f$ has a minimum. If $a < 0$, $f$ has a maximum. It occurs at $f(-\frac{b}{2a})$.

5. Answer the question posed in the problem.

Example 4: You have 120 feet of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

• Let $x$ be the width, then the length is $60 - x$ (since $2x + 2y = 120$).

• Area $A = x(60 - x) = 60x - x^2$.

• Maximum area occurs at $x = -\frac{60}{2 \times (-1)} = 30$.

• Dimensions: $30$ft by $30$ ft. Maximum area: $900$ sq ft.