Two boats depart from a port located at in a coordinate system measured in kilometers, and they fravel in a
positive -direction. The first boat follows a path that can be modeled by a quadratic function with a vertex at , and
the second boat follows a path that can be modeled by a linear function and passes through the point . At what
point, besides the common starting location of the port, do the paths of the two boats cross?

Ask by Hampton Vaughn. in the United States

Mar 25,2025

Question

Two boats depart from a port located at in a coordinate system measured in kilometers, and they fravel in a
positive -direction. The first boat follows a path that can be modeled by a quadratic function with a vertex at , and
the second boat follows a path that can be modeled by a linear function and passes through the point . At what
point, besides the common starting location of the port, do the paths of the two boats cross?

Ask by Hampton Vaughn. in the United States

Mar 25,2025

UpStudy AI · Accepted Answer

We first model the quadratic path. Since its vertex is at $$ (0,5) $$, its equation can be written in vertex form as
$$
y = a(x - 0)^2 + 5 = ax^2 + 5.
$$
Because the boat starts at the port $$ (-10,0) $$, this point lies on the quadratic. Substituting $$ x = -10 $$ and $$ y = 0 $$ gives:
$$
0 = a(-10)^2 + 5 \quad \Longrightarrow \quad 0 = 100a + 5.
$$
Solving for $$ a $$ we have:
$$
100a = -5 \quad \Longrightarrow \quad a = -\frac{5}{100} = -\frac{1}{20}.
$$
Thus, the quadratic function becomes:
$$
y = -\frac{1}{20} x^2 + 5.
$$

Next, we model the linear path. The line passes through the points $$ (-10,0) $$ and $$ (10,4) $$. The slope $$ m $$ is calculated as:
$$
m = \frac{4 - 0}{10 - (-10)} = \frac{4}{20} = \frac{1}{5}.
$$
Using point-slope form with the point $$ (-10,0) $$, the linear function is:
$$
y = \frac{1}{5}(x + 10).
$$
Expanding, we have:
$$
y = \frac{1}{5} x + 2.
$$

To find the intersection of the quadratic and the linear paths (besides the port $$ (-10,0) $$), set the functions equal:
$$
-\frac{1}{20} x^2 + 5 = \frac{1}{5} x + 2.
$$
Multiply through by $$ 20 $$ to eliminate fractions:
$$
-x^2 + 100 = 4x + 40.
$$
Rearrange the equation:
$$
-x^2 - 4x + 60 = 0.
$$
Multiply through by $$ -1 $$ to simplify:
$$
x^2 + 4x - 60 = 0.
$$
Solve the quadratic using the quadratic formula:
$$
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-60)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 240}}{2} = \frac{-4 \pm \sqrt{256}}{2} = \frac{-4 \pm 16}{2}.
$$
This gives:
$$
x = \frac{-4 + 16}{2} = \frac{12}{2} = 6 \quad \text{or} \quad x = \frac{-4 - 16}{2} = \frac{-20}{2} = -10.
$$
The solution $$ x = -10 $$ represents the starting point minus the port. Thus, the other intersection occurs at $$ x = 6 $$.

Substitute $$ x = 6 $$ into the linear function to determine $$ y $$:
$$
y = \frac{1}{5}(6) + 2 = \frac{6}{5} + 2 = \frac{6}{5} + \frac{10}{5} = \frac{16}{5} = 3.2.
$$
The two boat paths cross at the point:
$$
(6, 3.2).
$$
The two boats cross at the point $$ (6, 3.2) $$.

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