3.1 Quadratic Functions

A quadratic function is formally defined as any function of the form $f(x) = ax^2 + bx + c$ .
In this definition, $a$ , $b$ , and $c$ are real numbers, with the crucial condition that $a \ne 0$ (if $a$ were $0$ , the function would become linear, not quadratic).
The domain of any quadratic function is the set of all real numbers, denoted as $(-\infty, \infty)$ .

The graph of any quadratic function is a distinctive U-shaped curve called a parabola.
Shape (Concavity):
- The parabola can open either upward or downward.
- If a > 0, the parabola opens upward.
- If a < 0, the parabola opens downward.
Vertex:
- The vertex is the turning point of the parabola.
- It represents either the lowest point (if opening upward) or the highest point (if opening downward).
Axis of Symmetry:
- This is a vertical line that passes through the vertex.
- It divides the parabola into two mirror-image halves.
Transformations:
- Quadratic functions can be seen as transformations of the basic quadratic function $y = ax^2$ .
- These transformations lead to the standard form $y = a(x - h)^2 + k$ .

The quadratic function $f(x) = a(x - h)^2 + k$ , where $a \ne 0$ , is known as the standard form or vertex form.
Vertex: The graph of $f$ is a parabola whose vertex is the point $(h, k)$ .
Axis of Symmetry: The parabola is symmetric with respect to the vertical line $x = h$ .
Opening Direction:
- If a > 0, the parabola opens upward.
- If a < 0, the parabola opens downward.
Examples:
- Example 1: For $f(x) = -2(x - 3)^2 + 8$ :
  - Identify the vertex, intercepts, axis of symmetry ( $x = 3$ ), domain ( $(-\infty, \infty)$ ), and range.
- Example 2: For $f(x) = (x + 3)^2 + 1$ :
  - Identify the vertex, intercepts, axis of symmetry ( $x = -3$ ), domain ( $(-\infty, \infty)$ ), and range.

For a parabola defined by the quadratic function $f(x) = ax^2 + bx + c$ :
x-coordinate of the Vertex: The x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$ .
Vertex Coordinates: The full vertex is found at $(-\frac{b}{2a}, f(-\frac{b}{2a}))$ where $f(-\frac{b}{2a})$ is the y-coordinate obtained by substituting the x-coordinate back into the function.
Examples:
- Example 3: For $f(x) = -x^2 - 2x + 1$ :
  - Use the vertex and intercepts to sketch the graph, find the axis of symmetry, domain, and range.
- Example 4: For $f(x) = 3x^2 - 12x + 1$ :
  - Use the vertex and intercepts to sketch the graph, find the axis of symmetry, domain, and range.
- Example 5: For $y = 8x^2 - 40x - 130$ :
  - Find the vertex of the parabola. This information is then used to determine a reasonable viewing rectangle for a graphing utility.

For any quadratic function $f(x) = ax^2 + bx + c$ :
Condition for Minimum Value: If a > 0 (parabola opens upward), the function has a minimum value.
- This minimum occurs at $x = -\frac{b}{2a}$ .
- The minimum value itself is $f(-\frac{b}{2a})$ .
Condition for Maximum Value: If a < 0 (parabola opens downward), the function has a maximum value.
- This maximum occurs at $x = -\frac{b}{2a}$ .
- The maximum value itself is $f(-\frac{b}{2a})$ .
In both cases, $x = -\frac{b}{2a}$ provides the location (the x-coordinate) of the minimum or maximum value.
The value of $y$ , or $f(-\frac{b}{2a})$ , provides the actual minimum or maximum value (the y-coordinate).
Example 6: For the quadratic function $f(x) = -3x^2 + 6x - 13$ :
1. Determine Min/Max: Since a = -3 < 0, the parabola opens downward, and the function has a maximum value.
2. Find Value and Location: The maximum occurs at $x = -\frac{6}{2(-3)} = -\frac{6}{-6} = 1$ . The maximum value is $f(1) = -3(1)^2 + 6(1) - 13 = -3 + 6 - 13 = -10$ . So, the maximum value is $-10$ and it occurs at $x = 1$ .
3. Identify Domain and Range: The domain is $(-\infty, \infty)$ . Since the maximum value is $-10$ , the range is $(-\infty, -10]$ .

Example 7: An archer's arrow follows a parabolic path, with its height, $f(x)$ , in feet, modeled by $f(x) = -0.005x^2 + 2x + 5$ , where $x$ is the horizontal distance in feet.
1. Maximum Height of the Arrow:
  - To find the maximum height, we need the y-coordinate of the vertex.
  - The x-coordinate of the vertex is $x = -\frac{b}{2a} = -\frac{2}{2(-0.005)} = -\frac{2}{-0.01} = 200$ feet.
  - The maximum height (y-coordinate) is $f(200) = -0.005(200)^2 + 2(200) + 5 = -0.005(40000) + 400 + 5 = -200 + 400 + 5 = 205$ feet.
  - The maximum height of the arrow is $205$ feet.
2. Horizontal Distance to Hit the Ground:
  - The arrow hits the ground when its height $f(x) = 0$ .
  - We need to solve the quadratic equation $-0.005x^2 + 2x + 5 = 0$ .
  - Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . For this equation, $a = -0.005$ , $b = 2$ , $c = 5$ .
  - $x = \frac{-2 \pm \sqrt{2^2 - 4(-0.005)(5)}}{2(-0.005)}$
  - $x = \frac{-2 \pm \sqrt{4 + 0.1}}{-0.01}$
  - $x = \frac{-2 \pm \sqrt{4.1}}{-0.01}$
  - $x \approx \frac{-2 \pm 2.02485}{-0.01}$
  - Two possible solutions:
    - $x_1 \approx \frac{-2 + 2.02485}{-0.01} = \frac{0.02485}{-0.01} = -2.485$
    - $x_2 \approx \frac{-2 - 2.02485}{-0.01} = \frac{-4.02485}{-0.01} = 402.485$
  - Since horizontal distance cannot be negative, we take the positive value.
  - Rounding to the nearest foot, the arrow travels approximately $402$ feet before it hits the ground.