SOLUTION: Please help me to solve this problem.
A fountain shoots out water that follows a parabolic curve. Water comes out of a port on the ground and goes into another port also on the
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Question 1122315: Please help me to solve this problem.
A fountain shoots out water that follows a parabolic curve. Water comes out of a port on the ground and goes into another port also on the ground. The ports are five meters apart; and the water peaks at four meters from the ground. Consider the midpoint of the distance between the two ports as the origin. (a) Write the equation that models this parabola. (b) What is the horizontal distance of the water arc two meters from the ground? (c) If the horizontal distance of the rising and fall water is two meters, how far from the ground is the water at those point?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The standard form of the equation of a parabola with at vertical axis of symmetry is:
\ =\ ax^2\ +\ bx\ +\ c)
Since the 
-intercept of this parabola is at the point )
, we know that:
^2\ +\ b(0)\ +\ c\ =\ 4)
in other words, 
Also, since the 
-coordinate of the vertex of a parabola is given by 
, and the -coordinate of this particular vertex is 
, we know that 
Now we are left with:
\ =\ ax^2\ +\ 4)
We also know that the roots of this equation are 
and 
.
Substituting either root:
^2\ =\ -4)
\ =\ -4)
Putting it all together is one sock:
\ =\ -\frac{16}{25}x^2\ +\ 4)
John
My calculator said it, I believe it, that settles it
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