Quadratic Functions: Graphs, Properties, and Applications

Quadratic Functions

Introduction to Quadratic Functions

A quadratic function is a polynomial function of degree two, generally written as f(x) = ax^2 + bx + c, where a, b, and c are real numbers and a ≠ 0. The graph of a quadratic function is a parabola that opens upward if a > 0 and downward if a < 0.

Characteristics of Parabolas

• Vertex: The highest or lowest point on the parabola, depending on whether it opens downward or upward.

• Axis of Symmetry: A vertical line passing through the vertex, given by x = h in vertex form.

• Direction: Determined by the sign of a in the quadratic function.

• Intercepts: Points where the parabola crosses the x-axis (x-intercepts) and y-axis (y-intercept).

Quadratic Function in Vertex Form

Definition and Properties

The vertex form of a quadratic function is:

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

• The vertex is at the point (h, k).

• The axis of symmetry is the line x = h.

• If a > 0, the parabola opens upward; if a < 0, it opens downward.

Graphing Steps

1. Identify the direction of opening from the sign of a.

2. Locate the vertex at (h, k).

3. Find x-intercepts by solving $a(x - h)^2 + k = 0$.

4. Find the y-intercept by evaluating $f(0)$.

5. Plot the vertex, intercepts, and additional points as needed. Draw a smooth curve through these points.

Example: Graphing a Quadratic Function in Vertex Form

• Given:

• Step 1: Since a = -1, the parabola opens downward.

• Step 2: Vertex is at (1, 4).

• Step 3: Find x-intercepts by solving $- (x - 1)^2 + 4 = 0$.

• Step 4: Find y-intercept: $f(0) = - (0 - 1)^2 + 4 = -1 + 4 = 3$; so the y-intercept is (0, 3).

• Summary: The parabola opens downward, vertex at (1, 4), passes through (3, 0), (0, 3), and is symmetric about x = 1.

Quadratic Function in General Form

Definition and Properties

The general form of a quadratic function is:

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

• The vertex's x-coordinate is $x = -\frac{b}{2a}$.

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

• The y-intercept is at (0, c).

Graphing Steps

1. Determine the direction of opening from the sign of a.

2. Find the vertex using $x = -\frac{b}{2a}$ and $y = f\left(-\frac{b}{2a}\right)$.

3. Find x-intercepts by solving $ax^2 + bx + c = 0$.

4. Find the y-intercept at (0, c).

5. Plot the vertex, intercepts, and additional points as needed. Draw a smooth curve through these points.

Example: Graphing a Quadratic Function in General Form

• Given: $f(x) = -x^2 + 4x - 3$

• Step 1: Since a = -1, the parabola opens downward.

• Step 2: Vertex x-coordinate: $x = -\frac{4}{2 \times (-1)} = 2$

• Vertex y-coordinate: $f(2) = - (2)^2 + 4 \times 2 - 3 = -4 + 8 - 3 = 1$

• Vertex: (2, 1)

• Step 3: Find x-intercepts by solving $-x^2 + 4x - 3 = 0$.

• Step 4: y-intercept: $f(0) = -3$

• Summary: The parabola opens downward, vertex at (2, 1), y-intercept at (0, -3), and x-intercepts found by solving the quadratic equation.

Minimum and Maximum Values of Quadratic Functions

Finding Extrema

• If a > 0, the parabola opens upward and the function has a minimum at the vertex.

• If a < 0, the parabola opens downward and the function has a maximum at the vertex.

• The x-coordinate of the vertex is $x = -\frac{b}{2a}$.

• The minimum or maximum value is $f\left(-\frac{b}{2a}\right)$.

Example: Determining Minimum or Maximum

• Given: $f(x) = 4x^2 - 8x + 5$

• Since a = 4 > 0, the function has a minimum value.

• Minimum occurs at $x = -\frac{-8}{2 \times 4} = 1$

• Minimum value: $f(1) = 4(1)^2 - 8(1) + 5 = 4 - 8 + 5 = 1$

Domain and Range of Quadratic Functions

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

• Range: If a > 0, $[k, \infty)$ where k is the minimum value. If a < 0, $(-\infty, k]$ where k is the maximum value.

Applications: Maximizing or Minimizing Quadratic Functions

Strategy for Optimization Problems

1. Identify the quantity to be maximized or minimized.

2. Express this quantity as a function of one variable, using any given constraints.

3. Rewrite the function in standard quadratic form.

4. Find the vertex to determine the maximum or minimum value.

5. Interpret the result in the context of the problem.

Example: Maximizing Area

• Problem: You have 120 feet of fencing to enclose a rectangular region. Find the dimensions that maximize the area.

• Step 1: Let x = length, y = width. Perimeter: $2x + 2y = 120$

• Step 2: Solve for y: $y = 60 - x$

• Step 3: Area function: $A(x) = x(60 - x) = 60x - x^2$

• Step 4: Maximum area at $x = -\frac{60}{2 \times (-1)} = 30$

• Step 5: Dimensions: 30 ft by 30 ft (a square). Maximum area: $30 \times 30 = 900$ square feet.

Additional info: In optimization problems involving quadratic functions, the vertex always gives the maximum or minimum value, depending on the sign of a. For area and revenue problems, always express the quantity to be optimized as a function of a single variable before applying vertex analysis.