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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\\(\

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

\\)

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 becomes \( y = a(x - 10)^2 + 6 \).
Now, we know the parabola passes through \((12, 7)\). Substitute \( x = 12 \) and \( y = 7 \) into the equation:
\( 7 = a(12 - 10)^2 + 6 \)
\( 7 = a(2)^2 + 6 \)
\( 7 = 4a + 6 \)
Subtract 6 from both sides: \( 1 = 4a \)
So \( a=\frac{1}{4} \). Thus, the quadratic equation is \( y=\frac{1}{4}(x - 10)^2 + 6 \)? Wait, no, wait the options have \( y=\frac{1}{4}(x - 10)^2 + 6 \)? Wait, no, looking at the third option: \( y=\frac{1}{4}(x - 10)^2 + 6 \)? Wait, the third system has \( y=\frac{1}{4}(x - 10)^2 + 6 \)? Wait, let's check the linear equation.

Step2: Find the linear equation

The linear path is from \((-18, 14)\) to \((16, -13)\). The slope \( m \) is \( \frac{-13 - 14}{16 - (-18)}=\frac{-27}{34} \).
Using point - slope form \( y - y_1=m(x - x_1) \), using the point \((-18, 14)\):
\( y - 14=\frac{-27}{34}(x + 18) \)
\( y=\frac{-27}{34}x-\frac{27\times18}{34}+14 \)
\( \frac{27\times18}{34}=\frac{486}{34}=\frac{243}{17} \), and \( 14=\frac{238}{17} \)
\( y=\frac{-27}{34}x-\frac{243}{17}+\frac{238}{17} \)
\( y=\frac{-27}{34}x-\frac{5}{17} \)

And the quadratic equation with vertex \((10,6)\) and \( a = \frac{1}{4} \) is \( y=\frac{1}{4}(x - 10)^2+6 \) (wait, in the third system, the quadratic is \( y=\frac{1}{4}(x - 10)^2 + 6 \)? Wait, the third system is:
\(

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

\)

Let's re - check the quadratic:
Vertex \((h,k)=(10,6)\), so \( y = a(x - 10)^2+6 \). We found \( a=\frac{1}{4} \) by plugging in \((12,7)\):
\( 7=a(12 - 10)^2+6\Rightarrow7 = 4a+6\Rightarrow a=\frac{1}{4} \). So the quadratic is \( y=\frac{1}{4}(x - 10)^2+6 \).

The linear equation: slope \( m=\frac{-13 - 14}{16+18}=\frac{-27}{34} \), and using point - slope form, we derived \( y =-\frac{27}{34}x-\frac{5}{17} \).

So the system of equations is the third one:
\(

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

\) (the third system in the given options)

Answer:

\(

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

\) (the third system of equations, i.e., the system \(

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

\))