algebra 2

The maximum can be found by setting the first derivative of the function equal to zero.

What Brandon said:

y=f(x)
f(x)=-0.073x(x-33)
f(x)=-0.073x^2+2.409x
f'(x)=-0.146x+2.409
0=-0.146x+2.409
0.146x=2.409
x=16.5
f(x)=-0.073(16.5)(16.5-33)
f(x)=19.87425

Which is how high the flea the jumped.

To get the maximum horizontal range we set y=0 (since the flea has to come to ground after covering the horizontal range i.e. finishing the jump )
and x not 0 (since the flea jumped!) , so we get -0.073x(x-33)=0 and x not 0
i.e. x=33 is the horizontal range

Algebra 2 isn't the Calculus

Using Algebra 2 methods you would walk along the ordinate to the maximum.

y=-0.073x(x-33)

Change this to.

y=o.o73x(33-x)


set y=0 to determine range (x=0 & x=33), as this is a downward quadratic (parabola) there is one maximum between x=0 to 33

lets see what it is at the midpoint (x=33/2)

y(33/2)=0.073(33/2)^2

pick two points equal distance from this midpoint (33/2) say 16 and 17

y(16)=0.073(16)(17)

y(17)=0.073(17)(16)

so y(16)=y(17)