Question

an air show is scheduled for an airport located on a coordinate system measured in miles. the air traffic controllers have closed the airspace, modeled by a quadratic equation, to non - air show traffic. the boundary of the closed airspace starts at the vertex at (10, 6) and passes through the point (12, 7). a commuter jet has filed a flight plan that takes it along a linear path from (-18, 14) to (16, -13). which system of equations can be used to determine whether the commuter jet’s flight path intersects the closed airspace?\
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$$\begin{cases}y = \\frac{1}{4}(x - 5)^2+10\\\\y=-\\frac{1}{2}x + 5\\end{cases}$$

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$$\begin{cases}y = \\frac{1}{4}(x - 5)^2+10\\\\y=-2x - 22\\end{cases}$$

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$$\begin{cases}y = \\frac{1}{4}(x - 10)^2+6\\\\y=-\\frac{27}{34}x-\\frac{5}{17}\\end{cases}$$

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Explanation:

Step1: Find the quadratic equation

The vertex form of a quadratic equation is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. The vertex here is \((10, 6)\), so \( h = 10 \) and \( k = 6 \). So the equation starts as \( y = a(x - 10)^2 + 6 \). Now we use the point \((12, 7)\) to find \( a \). Plug in \( x = 12 \) and \( y = 7 \):

$$ 7 = a(12 - 10)^2 + 6 $$

$$ 7 = a(2)^2 + 6 $$

$$ 7 = 4a + 6 $$

Subtract 6 from both sides: \( 1 = 4a \), so \( a = \frac{1}{4} \). Thus, the quadratic equation is \( y = \frac{1}{4}(x - 10)^2 + 6 \).

Step2: Find the linear equation

The linear path is from \((-18, 14)\) to \((16, -13)\). First, find the slope \( m \) using \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Let \((x_1, y_1) = (-18, 14)\) and \((x_2, y_2) = (16, -13)\):

$$ m = \frac{-13 - 14}{16 - (-18)} = \frac{-27}{34} $$

Now use the point - slope form \( y - y_1 = m(x - x_1) \). Using \((-18, 14)\):

$$ y - 14 = \frac{-27}{34}(x + 18) $$

$$ y = \frac{-27}{34}x - \frac{27\times18}{34} + 14 $$

Calculate \( \frac{27\times18}{34}=\frac{486}{34}=\frac{243}{17} \), and \( 14=\frac{238}{17} \). So:

$$ y = \frac{-27}{34}x - \frac{243}{17}+\frac{238}{17} $$

$$ y = \frac{-27}{34}x - \frac{5}{17} $$

Answer:

The system of equations is \(

$$\begin{cases}y=\frac{1}{4}(x - 10)^2+6\\y =-\frac{27}{34}x-\frac{5}{17}\end{cases}$$

\) (the third option in the given choices)