Question

A sailboat with a sick passenger abroad is following a parabolic path given by the equation y =-x^2+ 4x + 3. The boat's captain sent out a distress signal. A speedboat is trying to catch the sailboat

to provide medical aid and is on a path given by the linear equation y = x + 3. Besides the y-intercept
(starting location) where do the paths of the two boats cross?

Please show your work

Answer 1

To find the point where the paths of the sailboat and the speedboat intersect, we need to solve the system of equations formed by the given equations:

y = -x^2 + 4x + 3 (equation for the sailboat's path)

y = x + 3 (equation for the speedboat's path)

To find the x-coordinate of the intersection point, we can set the y-values of the two equations equal to each other:

-x^2 + 4x + 3 = x + 3

Now, let's solve this equation step by step:

First, let's move all the terms to one side to get a quadratic equation:

-x^2 + 4x + 3 - x - 3 = 0

Simplifying further:

-x^2 + 3x = 0

To solve this equation, we can factor out an x:

x(-x + 3) = 0

From this, we can see that either x = 0 or -x + 3 = 0. Solving each equation:

For x = 0, substituting this value into the equation for the sailboat's path:

y = -(0)^2 + 4(0) + 3

y = 3

So one point of intersection is (0, 3).

For -x + 3 = 0, solving for x:

-x = -3

x = 3

Substituting this value into the equation for the sailboat's path:

y = -(3)^2 + 4(3) + 3

y = -9 + 12 + 3

y = 6

So another point of intersection is (3, 6).

Therefore, the paths of the sailboat and the speedboat cross at two points: (0, 3) and (3, 6).

Explanation:

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