Breakeven Analysis & Quadratic Functions

Breakeven Analysis: Review from Last Time

• 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 $(x, ext{revenue}) = (2000, 20000)$.

Section 2.6: Quadratic Functions

Definition and General Form

• 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 $a <br>eq 0$ for it to be a second-degree polynomial.

  • $b$ or $c$ (or both) can be equal to zero.

Graph of a Quadratic Function: The Parabola

• 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.

Sketching Graphs of Parabolas

• 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).

Case 1: Simplest Form ($f(x) = ax^2$)

• In this case, $b=0$ and $c=0$.

• The vertex is always at the origin, $(0,0)$.

• Example 1: Sketch $f(x) = rac{1}{2}x^2$

  • Here, $a = rac{1}{2}$, which is positive, so the parabola opens upward.

  • Vertex: $(0,0)$.

  • Choose points equidistant from $x=0$ (e.g., $x=2$ and $x=-2$):

  • $f(2) = rac{1}{2}(2)^2 = rac{1}{2}(4) = 2$

  • $f(-2) = rac{1}{2}(-2)^2 = rac{1}{2}(4) = 2$

  • Additional points: $(2,2)$ and $(-2,2)$.

  • Plot these three points and connect them to sketch the parabola.

• Example 2: Sketch $f(x) = -4x^2$

  • Here, $a = -4$, which is negative, so the parabola opens downward.

  • Vertex: $(0,0)$.

  • Choose points equidistant from $x=0$ (e.g., $x=1$ and $x=-1$):

  • $f(1) = -4(1)^2 = -4$

  • $f(-1) = -4(-1)^2 = -4$

  • Additional points: $(1,-4)$ and $(-1,-4)$.

  • Plot these three points and connect them.

Case 2: Vertex Form ($f(x) = a(x-h)^2 + k$)

• 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.

Case 3: General Form ($f(x) = ax^2 + bx + c$)

• 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.

Application Problems: Optimization Using Quadratic Functions

• Quadratic functions are used in optimization problems, where the goal is to:

  • Maximize a