-0.01x^2+0.6x+6.5 distance of the throw

User Generated

parjzna12512

Mathematics

Question Description

an athlete whose event in shot put releases a shot. when the shot is released at an angel of 30 degrees, it's height, f(x), in feet, can be modeled by f(x)=-0.01x^2+0.6x+6.5, where x is the shot's horizontal distance, in feet, from its point of release. 

what is the show's maximum horizontal distance to the nearest tenth of a foot, or distance of the throw?

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Explanation & Answer

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to find the maximum we have to convert the polynomial into

following form

f(x)=-0.01(x^2-60x-650)

f(x)=-0.01[(x-30)^2-900-650]

f(x)=-0.01[(x-30)^2-1550]

to find max distance (x-30)^2 should goes to zero

then 

f(x)max=-0.01(0-1550)

=15.5 feet

so max height =15.5 feet



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fnpuvagun (127)
University of Maryland

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