Quadratic Functions and Their Properties

Quadratic Functions

Definition and Standard Form

Quadratic functions are a fundamental class of polynomial functions with degree two. The general form of a quadratic function is:

• Standard Form: $f(x) = ax^2 + bx + c$, where $a \neq 0$

• The graph of a quadratic function is called a parabola.

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

Vertex and Axis of Symmetry

The vertex of a parabola is its highest or lowest point, depending on the direction it opens. The axis of symmetry is a vertical line passing through the vertex, dividing the parabola into two mirror images.

• Axis of Symmetry: $x = h$

• Vertex: $(h, k)$

• For $f(x) = a(x-h)^2 + k$, the vertex is at $(h, k)$.

• For $f(x) = ax^2 + bx + c$, the vertex is at $\left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right)$.

Transformations of Quadratic Functions

Quadratic functions can be transformed by shifting, stretching, or reflecting their graphs.

• Horizontal Shift: $h > 0$ shifts left, $h < 0$ shifts right.

• Vertical Shift: $k > 0$ shifts up, $k < 0$ shifts down.

• Vertical Stretch/Compression: $|a| > 1$ stretches, $0 < |a| < 1$ compresses.

• Reflection: If $a < 0$, the graph reflects over the x-axis.

Intercepts

Intercepts are points where the graph crosses the axes.

• Y-intercept: Set $x = 0$, compute $f(0)$.

• X-intercepts: Set $f(x) = 0$ and solve for $x$ (may use factoring, completing the square, or the quadratic formula).

Example: Finding Intercepts and Vertex

• Given $f(x) = -2(x-3)^2 + 8$:

  • $a = -2$ (opens downward, stretches vertically)

  • $h = 3$ (shifts right 3 units)

  • $k = 8$ (shifts up 8 units)

  • Vertex: $(3, 8)$

  • Y-intercept: $f(0) = -2(0-3)^2 + 8 = -2(9) + 8 = -18 + 8 = -10$; point $(0, -10)$

  • X-intercepts: Set $f(x) = 0$ and solve:

    • $0 = -2(x-3)^2 + 8$

    • $-2(x-3)^2 = -8$

    • $(x-3)^2 = 4$

    • $x-3 = \pm 2$

    • $x = 5$ and $x = 1$; points $(5, 0)$ and $(1, 0)$

Quadratic Formula

The quadratic formula is used to find the roots (x-intercepts) of any quadratic equation $ax^2 + bx + c = 0$:

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

Example: Complex Roots

• Given $f(x) = (x+3)^2 + 1$:

  • Y-intercept: $f(0) = (0+3)^2 + 1 = 9 + 1 = 10$; point $(0, 10)$

  • X-intercepts: $0 = (x+3)^2 + 1 \implies (x+3)^2 = -1 \implies x+3 = \pm i$

  • Thus, $x = -3 + i$ and $x = -3 - i$ (no real x-intercepts)

Vertex from Standard Form

• Given $f(x) = -x^2 - 2x + 1$:

  • $a = -1$, $b = -2$, $c = 1$

  • Vertex: $x = -\frac{-2}{2(-1)} = -1$

  • $f(-1) = -(-1)^2 - 2(-1) + 1 = -1 + 2 + 1 = 2$

  • Vertex: $(-1, 2)$

  • Y-intercept: $f(0) = 1$; point $(0, 1)$

  • X-intercepts: Use quadratic formula:

    • $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-1)(1)}}{2(-1)} = \frac{2 \pm \sqrt{4 + 4}}{-2} = \frac{2 \pm 2\sqrt{2}}{-2}$

    • Approximate roots: $x \approx -2.4$ and $x \approx 0.4$

Domain and Range

• Domain: $(-\infty, \infty)$ for all quadratic functions

• Range: If $a > 0$, $[k, \infty)$; if $a < 0$, $(-\infty, k]$ where $k$ is the y-coordinate of the vertex

Minimum and Maximum Values

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

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

• Example: $f(x) = -3x^2 + 6x - 13$

  • $a = -3$ (maximum)

  • Vertex at $x = -\frac{6}{2 \times -3} = 1$

  • $f(1) = -3(1)^2 + 6(1) - 13 = -3 + 6 - 13 = -10$

  • Maximum value: $-10$ at $(1, -10)$

  • Domain: $(-\infty, \infty)$; Range: $(-\infty, -10]$

Applications of Quadratic Functions

Quadratic functions are widely used in modeling real-world phenomena, such as projectile motion and optimization problems.

• Projectile Example: The path of a punted football: $f(x) = -0.1x^2 + 1.18x + 2$

  • Y-intercept (initial height): $f(0) = 2$

  • Maximum height at $x = -\frac{1.18}{2 \times -0.1} = 5.9$ (should be $x = 5.9$, but the notes say $x = 59$; likely a calculation error)

  • Maximum height: $f(5.9) = -0.1(5.9)^2 + 1.18(5.9) + 2 \approx 5.47$

  • To block the punt at $x = 6$: $f(6) = -0.1(6)^2 + 1.18(6) + 2 = -3.6 + 7.08 + 2 = 5.48$

  • Distance traveled: Solve $0 = -0.1x^2 + 1.18x + 2$ for $x$ (use quadratic formula)

Optimization Example: Maximizing Area

• Given 100 yards of fencing for a rectangular enclosure:

  • Perimeter: $2x + 2y = 100 \implies y = 50 - x$

  • Area: $A = x \cdot y = x(50 - x) = 50x - x^2$

  • Maximum area occurs at vertex: $x = -\frac{50}{2 \times -1} = 25$

  • $y = 50 - 25 = 25$

  • Maximum area: $A = 25 \times 25 = 625$ square yards

Summary Table: Key Properties of Quadratic Functions

Property	Formula/Description
Standard Form	$f(x) = ax^2 + bx + c$
Vertex	$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$
Axis of Symmetry	$x = -\frac{b}{2a}$
Y-intercept	$(0, c)$
X-intercepts	Solve $ax^2 + bx + c = 0$
Opens Up/Down	$a > 0$ up, $a < 0$ down
Domain	$(-\infty, \infty)$
Range	$[k, \infty)$ if $a > 0$; $(-\infty, k]$ if $a < 0$