Let's carefully derive the system of equations to determine whether the linear road intersects the boundary of the tower's signal.
1. Quadratic Function for Boundary of the Signal:
The quadratic function is given with the vertex at [tex]\((4, 2)\)[/tex] and passing through the point [tex]\((5, 4)\)[/tex].
The general form of a quadratic function with vertex [tex]\((h, k)\)[/tex] is:
[tex]\[
y = a(x - h)^2 + k
\][/tex]
Plugging in the vertex [tex]\((4, 2)\)[/tex], this becomes:
[tex]\[
y = a(x - 4)^2 + 2
\][/tex]
We know the function passes through the point [tex]\((5, 4)\)[/tex]. Plugging in [tex]\(x = 5\)[/tex] and [tex]\(y = 4\)[/tex] into the equation:
[tex]\[
4 = a(5 - 4)^2 + 2
\][/tex]
[tex]\[
4 = a(1)^2 + 2
\][/tex]
[tex]\[
4 = a + 2
\][/tex]
Solving for [tex]\(a\)[/tex], we get:
[tex]\[
a = 2
\][/tex]
Hence, the quadratic equation is:
[tex]\[
y = 2(x - 4)^2 + 2
\][/tex]
Rearranging it to standard form, we get:
[tex]\[
y - 2 = 2(x - 4)^2
\][/tex]
[tex]\[
y - 2(x - 4)^2 = 2
\][/tex]
2. Linear Equation for the Road:
The linear road passes through points [tex]\((-3, 7)\)[/tex] and [tex]\((8, 2)\)[/tex]. We can find the equation of the line by using the two-point form of the linear equation:
[tex]\[
y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)
\][/tex]
Here, [tex]\((x_1, y_1) = (-3, 7)\)[/tex] and [tex]\((x_2, y_2) = (8, 2)\)[/tex].
The slope [tex]\(\frac{y_2 - y_1}{x_2 - x_1}\)[/tex] is:
[tex]\[
\frac{2 - 7}{8 - (-3)} = \frac{-5}{11}
\][/tex]
So, the equation becomes:
[tex]\[
y - 7 = \frac{-5}{11}(x + 3)
\][/tex]
Multiplying through by 11 to clear the fraction:
[tex]\[
11(y - 7) = -5(x + 3)
\][/tex]
[tex]\[
11y - 77 = -5x - 15
\][/tex]
[tex]\[
11y + 5x = 62
\][/tex]
Rewriting in the standard form:
[tex]\[
5x + 11y = 62
\][/tex]
Now, we have our system of equations:
1. [tex]\(y - (x - 4)^2 = 2\)[/tex]
2. [tex]\(5x + 11y = 62\)[/tex]
Therefore, the system of equations that can be used to determine whether the road intersects the boundary of the tower's signal is:
[tex]\[
\left\{
\begin{aligned}
y - (x - 4)^2 & = 2 \\
5x + 11y & = 62
\end{aligned}
\right.
\][/tex]
Thus, the correct answer is:
[tex]\[
\left\{
\begin{aligned}
y-(x-4)^2 & =2 \\
5 x+11 y & =62
\end{aligned}
\right.
\][/tex]
