SOLUTION: Hello I have a problem solving this parabola question.The path of a cliff diver as he dives into a lake ,is given by this eqaution,y=-(x-10)(SQAURED)+75,where y metres is the diver
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Question 310688: Hello I have a problem solving this parabola question.The path of a cliff diver as he dives into a lake ,is given by this eqaution,y=-(x-10)(SQAURED)+75,where y metres is the diver's height above the water and,x metres is the horisontal distance travelled by the diver.What is the maximum height the diver is above the water?
Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Since the first term is always negative (since it's squared and multiplied by -1), the max value it reaches is zero (when x=10), which then only leaves the constant term.
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.
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Not sure whether you just need an intuitive approach, a formal algebraic approach, or a calculus approach.
Intuitive
The largest y can be is 75 and that is when ^2\ =\ 0)
. That is because ^2)
is always a positive number, unless it is zero, so ^2)
is always a negative number unless it is zero. Therefore you are always subtracting something from 75 unless , which happens when 
While we are at it -- 75 meters? That is 246 feet. Hitting the water from that height would be like hitting concrete. I rather think your cliff diver is only going to perform this particular stunt once.
Algebraic
^2\ +\ 75)
Expand the binomial expression:
A parabola with a negative lead coefficient opens downward, hence the vertex is a maximum. The vertex of a parabola expressed in 
form has a vertex that occurs at an 
value of 
, so:
And the value of the function at is:
\ =\ -(10)^2\ +\ 20(10)\ -\ 25\ =\ 75)
Calculus
Using the function definition from the Algebraic discussion:
The function will have a maximum where the first derivative is zero and the second derivative is negative.
Set it equal to zero
which is negative for all in the domain of 
So there is a maximum at
John
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