Question

a radio tower is located on a coordinate system measured in miles. the range of a signal in a particular direction is modeled by a quadratic function where the boundary of the signal starts at the vertex at (4, 2). it passes through the point (5, 4). a linear road connects points (-3, 7) and (8, 2). which system of equations can be used to determine whether the road intersects the boundary of the tower’s signal?\
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$$\begin{cases} y - 2(x - 4)^2 = 2 \\\\ 5x + 11y = 82 \\end{cases}$$

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$$\begin{cases} y - (x - 4)^2 = 2 \\\\ 5x + 11y = 82 \\end{cases}$$

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$$\begin{cases} y - (x - 4)^2 = 2 \\\\ 11x + 5y = 2 \\end{cases}$$

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$$\begin{cases} y - 2(x - 4)^2 = 2 \\\\ 11x + 5y = 2 \\end{cases}$$

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Explanation:

Step1: Find the quadratic function

The vertex form of a quadratic function is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Here, the vertex is \((4, 2)\), so \( h = 4 \) and \( k = 2 \). The function passes through \((5, 4)\). Substitute \( x = 5 \), \( y = 4 \), \( h = 4 \), and \( k = 2 \) into the vertex form:

$$ 4 = a(5 - 4)^2 + 2 $$

$$ 4 = a(1)^2 + 2 $$

$$ 4 = a + 2 $$

Subtract 2 from both sides: \( a = 2 \). So the quadratic function is \( y = 2(x - 4)^2 + 2 \), which can be rewritten as \( y - 2(x - 4)^2 = 2 \).

Step2: Find the linear equation

The linear road connects \((-3, 7)\) and \((8, 2)\). First, find the slope \( m \):

$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 7}{8 - (-3)} = \frac{-5}{11} $$

Using the point - slope form \( y - y_1 = m(x - x_1) \) with the point \((8, 2)\):

$$ y - 2 = -\frac{5}{11}(x - 8) $$

Multiply both sides by 11 to eliminate the fraction:

$$ 11(y - 2)=- 5(x - 8) $$

$$ 11y-22=-5x + 40 $$

Rearrange to standard form \( Ax+By = C \):

$$ 5x+11y=40 + 22 $$

$$ 5x + 11y=62 $$

Answer:

\(

$$\begin{cases}y - 2(x - 4)^2=2\\5x + 11y = 62\end{cases}$$

\) (the first system of equations)