Breakeven Analysis & Quadratic Functions

To find the breakeven quantity (x), set the cost function equal to the revenue function (C(x) = R(x)) and solve for x.
To find the breakeven revenue, plug the breakeven quantity (x) into either the revenue function (R(x)) or the cost function (C(x)).
Example (continued from previous lecture):
- Breakeven quantity was previously found to be 2000.
- Given a revenue function, assumed to be R(x) = 10x based on the calculation.
- Breakeven revenue: R(2000) = 10 imes 2000 = 20000.
- The breakeven point is (x, ext{revenue}) = (2000, 20000).

A quadratic function is a function whose formula is given by a second-degree polynomial.
General formula: f(x) = ax^2 + bx + c
- The requirement is that a
  eq 0 for it to be a second-degree polynomial.
- b or c (or both) can be equal to zero.

The graph of a quadratic function is a parabola.
Opening Direction:
- If the leading coefficient a is positive (a > 0), the parabola opens upward.
- If the leading coefficient a is negative (a < 0), the parabola opens downward.
Vertex:
- The lowest point on an upward-opening parabola.
- The highest point on a downward-opening parabola.
- The vertex is crucial in optimization problems.

To sketch the graph of a parabola, you typically need three points:
1. The vertex.
2. One point to the left of the vertex.
3. One point to the right of the vertex (preferably equidistant from the vertex due to symmetry).
Parabolas are symmetric about a vertical line that passes through their vertex (the axis of symmetry).

This form is called the vertex form because it directly reveals the vertex.
The vertex is at the point (h,k).
Once the vertex is found, select points to the left and right, evaluate f(x), and plot.
Example 1: Sketch f(x) = -2(x-3)^2 + 2
- This is in vertex form. By comparing it to a(x-h)^2 + k, we get:
- h = 3
- k = 2
- So, the vertex is (3,2).
- Here, a = -2, which is negative, so the parabola opens downward.
- Choose points equidistant from x=3 (e.g., x=2 and x=4):
- f(4) = -2(4-3)^2 + 2 = -2(1)^2 + 2 = -2 + 2 = 0
- By symmetry, f(2) = 0
- Additional points: (2,0) and (4,0). These are the x-intercepts.
- Plot (3,2), (2,0), and (4,0) and connect them.
Example 2: Sketch f(x) = (x+3)^2 + 1
- (Note: Based on the instructor's calculations and plotted points, the function was implicitly treated as f(x) = (x+3)^2 + 1 despite implied negative a elsewhere in the lecture's general discussion).
- This is in vertex form: f(x) = 1(x-(-3))^2 + 1
- h = -3
- k = 1
- So, the vertex is (-3,1).
- Here, a = 1, which is positive, so the parabola opens upward.
- Choose points equidistant from x=-3 (e.g., x=-2 and x=-4):
- f(-2) = (-2+3)^2 + 1 = (1)^2 + 1 = 1 + 1 = 2
- By symmetry, f(-4) = 2
- Additional points: (-2,2) and (-4,2).
- Plot (-3,1), (-2,2), and (-4,2) and connect them.

When the formula is not in vertex form, you can find the coordinates of the vertex using specific formulas derived from the general coefficients a, b, c.
Vertex Coordinates ((h,k)):
- h = rac{-b}{2a}
- k = rac{4ac - b^2}{4a} (Alternatively, substitute h into the function: k = f(h)).
Once the vertex is found, the process for sketching is the same as above: find a point on the left and a point on the right, and connect.
Example: Sketch f(x) = x^2 - 6x + 8
- Identify coefficients: a=1, b=-6, c=8.
- Calculate h (x-coordinate of vertex):
- h = rac{-(-6)}{2(1)} = rac{6}{2} = 3
- Calculate k (y-coordinate of vertex) using k = f(h):
- k = f(3) = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1
- The vertex is (3,-1).
- Here, a=1, which is positive, so the parabola opens upward.
- Choose points equidistant from x=3 (e.g., x=2 and x=4):
- f(2) = (2)^2 - 6(2) + 8 = 4 - 12 + 8 = 0
- By symmetry, f(4) = 0
- Additional points: (2,0) and (4,0). These are the x-intercepts.
- Plot (3,-1), (2,0), and (4,0) and connect them.

Quadratic functions are used in optimization problems, where the goal is to:
- Maximize a