Certainly! Let's proceed step-by-step to solve through the given parts:
### Part (a)
#### Vertex of [tex]\( R(x) \)[/tex]
First, consider the revenue function:
[tex]\[ R(x) = -x^2 + 6x \][/tex]
This is a quadratic function in standard form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 0 \)[/tex].
The vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] is at:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substitute [tex]\( a = -1 \)[/tex] and [tex]\( b = 6 \)[/tex]:
[tex]\[ x = -\frac{6}{2(-1)} = 3 \][/tex]
Now, find the corresponding [tex]\( y \)[/tex]-value (which is [tex]\( R(3) \)[/tex]):
[tex]\[ R(3) = -3^2 + 6(3) = -9 + 18 = 9 \][/tex]
So, the vertex of [tex]\( R(x) \)[/tex] is at:
[tex]\[ (3, 9) \][/tex]
#### y-intercept of [tex]\( C(x) \)[/tex]
Consider the cost function:
[tex]\[ C(x) = x + 4 \][/tex]
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ C(0) = 0 + 4 = 4 \][/tex]
So, the y-intercept of [tex]\( C(x) \)[/tex] is at [tex]\( y = 4 \)[/tex].
### Part (b)
#### Finding the Minimum Break-even Quantity
To find the break-even quantity, we need to determine the [tex]\( x \)[/tex]-values where the revenue [tex]\( R(x) \)[/tex] equals the cost [tex]\( C(x) \)[/tex]. Therefore, we set
[tex]\[ R(x) = C(x) \][/tex]
Using the given functions:
[tex]\[ -x^2 + 6x = x + 4 \][/tex]
This needs to be solved for [tex]\( x \)[/tex]. Simplify by moving all terms to one side of the equation:
[tex]\[ -x^2 + 6x - x - 4 = 0 \][/tex]
[tex]\[ -x^2 + 5x - 4 = 0 \][/tex]
Now, solve this quadratic equation:
[tex]\[ x^2 - 5x + 4 = 0 \][/tex]
Solve using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 4 \)[/tex]:
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{25 - 16}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm 3}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{5 + 3}{2} = 4 \][/tex]
[tex]\[ x = \frac{5 - 3}{2} = 1 \][/tex]
Thus, the break-even quantities where the revenue equals the cost are [tex]\( x = 4 \)[/tex] and [tex]\( x = 1 \)[/tex].
However, the minimum break-even quantity is the smallest [tex]\( x \)[/tex] value where this occurs, which is [tex]\( x = 1 \)[/tex].
Therefore, the minimum break-even quantity is:
[tex]\[ \boxed{1} \][/tex]
If any additional parts are required to graph the functions, use the identified critical points and general shapes of the parabola (concave down for [tex]\( R(x) \)[/tex]) and the linear function [tex]\( C(x) \)[/tex]. But primarily, the break-even analysis accurately identifies the minimum value where costs and revenues intersect.
