To solve this problem, we need to derive two equations: one for the quadratic boundary of the closed airspace and another for the linear path of the commuter jet. Follow these steps:
### Step 1: Find the Quadratic Equation
The quadratic boundary of the closed airspace has its vertex at [tex]\((10, 6)\)[/tex] and passes through the point [tex]\((12, 7)\)[/tex]. The general form of a quadratic equation with a vertex [tex]\((h, k)\)[/tex] is:
[tex]\[ y = a(x - h)^2 + k \][/tex]
Here, [tex]\( (h, k) = (10, 6) \)[/tex].
Substitute the vertex into the equation:
[tex]\[ y = a(x - 10)^2 + 6 \][/tex]
To find the value of [tex]\(a\)[/tex], use the point [tex]\((12, 7)\)[/tex]:
[tex]\[ 7 = a(12 - 10)^2 + 6 \][/tex]
Simplify:
[tex]\[ 7 = a(2)^2 + 6 \][/tex]
[tex]\[ 7 = 4a + 6 \][/tex]
[tex]\[ 1 = 4a \][/tex]
[tex]\[ a = \frac{1}{4} \][/tex]
Thus, the equation of the quadratic boundary is:
[tex]\[ y = \frac{1}{4}(x - 10)^2 + 6 \][/tex]
### Step 2: Find the Linear Equation
The linear path of the commuter jet goes from [tex]\((-18, 14)\)[/tex] to [tex]\((16, -13)\)[/tex]. Use the slope formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Here, [tex]\((x_1, y_1) = (-18, 14)\)[/tex] and [tex]\((x_2, y_2) = (16, -13)\)[/tex].
Calculate the slope [tex]\(m\)[/tex]:
[tex]\[ m = \frac{-13 - 14}{16 - (-18)} \][/tex]
[tex]\[ m = \frac{-27}{16 + 18} \][/tex]
[tex]\[ m = \frac{-27}{34} \][/tex]
Using the point-slope form [tex]\(y - y_1 = m(x - x_1)\)[/tex] with the point [tex]\((-18, 14)\)[/tex], the equation is:
[tex]\[ y - 14 = \frac{-27}{34}(x - (-18)) \][/tex]
[tex]\[ y - 14 = \frac{-27}{34}(x + 18) \][/tex]
Simplify:
[tex]\[ y = \frac{-27}{34}x + \frac{-27}{34} \cdot 18 + 14 \][/tex]
[tex]\[ y = \frac{-27}{34}x - \frac{486}{34} + 14 \][/tex]
[tex]\[ y = \frac{-27}{34}x - \frac{243}{17} + 14 \][/tex]
[tex]\[ y = \frac{-27}{34}x - 14.2941176470588235 \][/tex]
### Step 3: Form the System of Equations
The system of equations that represents the boundary of the closed airspace and the path of the commuter jet is:
[tex]\[
\left\{
\begin{array}{l}
y = \frac{1}{4}(x - 10)^2 + 6 \\
y = \frac{-27}{34}x - \frac{5}{17}
\end{array}
\right.
\][/tex]
None of the given options perfectly match the derived equations. Thus, the correct system of equations based on the given points should be:
[tex]\[
\left\{
\begin{array}{l}
y = \frac{1}{4}(x - 10)^2 + 6 \\
y = \frac{-27}{34} x - \frac{5}{17}
\end{array}
\right.
\][/tex]
It's important to match the form exactly to the correct problem parameters.
