Two boats depart from a port located at (–10, 0) in a coordinate system measured in kilometers, and they travel in a positive x-direction. The first boat follows a path that can be modeled by a quadratic function with a vertex at (0, 5), and the second boat follows a path that can be modeled by a linear function and passes through the point (10, 4). At what point, besides the common starting location of the port, do the paths of the two boats cross?

Question

anonymous · Answer

Possible Answers...
(–6, 0.8)
(–6, 3.2)
(6, 3.2)
(6, 0.8)

anonymous · Answer

:( sorry, those diamond question marks are both negative signs.

anonymous · Answer

@Directrix

directrix · Answer

I can see how to get the equation of boat 2 - the linear path.

Maybe we need a graph.

directrix · Answer

Is this the same question: Two boats depart from a port located at (–10, 0) in a coordinate system measured in kilometers, and they travel in a positive x-direction. The first boat follows a path that can be modeled by a quadratic function with a vertex at (0, 5), and the second boat follows a path that can be modeled by a linear function and passes through the point (10, 4). At what point, besides the common starting location of the port, do the paths of the two boats cross? http://www.softmath.com/algebra-word-problems/show.php?id=20795

directrix · Answer

Do we agree with these imported statements regarding Boat 1?

First boat 

the equation of a quadratic with vertex in (h, k) is given by y = a(x – h)2 + k

in our case the vertex is (0, 5) that is h = 0 and k = 5

y = a(x - 0)^2 + 5

y = ax^2 + 5

also the line goes trough the point (-10, 0)

0 = a(-10)^2 + 5

a = -5/100

a = - 1/20

the path of the first boat is described by the equation y = (-1/20)x^2 + 5

directrix · Answer

Second boat

the equation of the linear function through the points (-10, 0) and (10, 4) has a slope equal to:

m = (4 - 0)/(10 - (-10)) = 4/20 = 1/5

y - y1 = m(x - x1)

y - 0 = (1/5)(x - (-10)

y = (1/5)x + 10/5

y = (1/5)x + 2

the path of the second boat is described by the equation y = (1/5)x + 2

directrix · Answer

At what point, besides the common starting location of the port, do the paths of the two boats cross?

we fill find this by solving the system 

y = (-1/20)x^2 + 5

y = (1/5)x + 2

that is by solving the equation

(-1/20)x^2 + 5 = (1/5)x + 2

we find the 2 roots

x1 = 6 

y1 = (1/5)6 + 2 = 6/5 +2 = 3.2

(6, 3.2)

x2 = -10

y2 =(1/5)(-10) + 2 = 0

(-10, 0)

directrix · Answer

All of the work posted above is that of @Algebrator at http://www.softmath.com/algebra-word-problems/show.php?id=20795

mathmath333 · Answer

https://www.desmos.com/calculator/w45arwv7eq

anonymous · Answer

Ok, so (6, 3.2) is answer, because, when graphed, the equations intersect at that point(and another, but that one wouldn't make sense).

mathmath333 · Answer

they meet two times 
1.) when they start at the staring point 
2.) when they intersect

anonymous · Answer

They made it clear they didn't want #1. That is why I said it didn't make sense.

directrix · Answer

@GodGirl360  Look back at the question:

 At what point, besides the common starting location of the port, do the paths of the two boats cross?

mathmath333 · Answer

yes ^

directrix · Answer

>>>They made it clear they didn't want #1. That is why I said it didn't make sense.

You are correct.

mathmath333 · Answer

if u dont wanna graph u can solve the two equations http://www.wolframalpha.com/input/?i=y%3D%28-1%2F20%29x%5E2%2B5%2C10y%3D2x%2B20

anonymous · Answer

Ok, thank you.

directrix · Answer

We are happy to help.