The point besides the common starting location of the port, the paths of the two boats cross when vertex is given would be (6, 3.2).
We have,
Two boats depart from a port located at (–10, 0) measured in kilometers, and they travel in a positive x-direction,
i.e.
x = -10 and y = 0
And,
The first boat follows a path of quadratic function with a vertex at (0, 5),
i.e. h = 0 and k = 5,
And,
The second boat follows a path of a linear function and passes through the point (10, 4),
So,
Now,
According to the question,
Vertex form of the equation,
f(x) = a(x - h)² + k
Here,
(h, k) is the vertex,
Now,
Putting values,
i.e.
f(x) = a(x - 0)² + 5
f(x) = ax² + 5
i.e.
y = ax² + 5
Now,
We have,
x = -10, y = 0
So,
Putting values we get,
0 = a(10)² + 5
0 = 100a + 5
On solving we get,
a = [tex]-\frac{1}{20}[/tex]
So,
y = [tex]-\frac{1}{20}[/tex] x² + 5 .....(i)
i.e.
f(x) = [tex]-\frac{1}{20}[/tex] x² + 5
Now,
For Boat B,
Slope for linear function,
Slope = [tex]\frac{(4-0)}{(10-(-10))} =\frac{4}{20}=\frac{1}{5}[/tex]
So,
Now,
The linear function is,
[tex]\frac{1}{5}[/tex](x - (-10)) = y - 4
On solving we get,
y = [tex]\frac{1}{5}[/tex]x + 2 .....(ii)
Now,
The point where the boats will meet,
From equation (i) and (ii),
We get,
[tex]-\frac{1}{20}[/tex] x² + 5 = [tex]\frac{1}{5}[/tex]x + 2
On solving we get,
5(-x² + 100) = 20(x + 10)
i.e.
-x² + 100 = 4x + 40
x² - 100 + 4x + 40 = 0
We get,
x² + 4x - 60 = 0
On solving using mid term splitting method,
x² + 10x - 6x - 60 = 0
x(x + 10) - 6(x + 10) = 0
On solving we get,
x = -10 or x = 6
Now,
For x = -10,
y = 0
And,
For x = 6,
y = 3.2
So,
The point will be (6, 3.2).
Hence we can say that the point besides the common starting location of the port, the paths of the two boats cross when vertex is given would be (6, 3.2).
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