Question 1192012
<pre>
Part 1:  correct

Part 2:   {{{ h(x) = -0.5x^2 + 0.75x + 3.5 = 3.6 }}}
        {{{ -0.5x^2 + 0.75x - 0.1 = 0 }}}

    Multiply by 100: 
        {{{  -50x^2 + 75x - 10 = 0 }}} 
      
    Which has solutions (thanks to WolframAlpha, or you can complete the square):
        {{{ x = (1/4)(3 +- sqrt(29/5)) }}}

     so at (approx)  x = 0.15ft  and  x = 1.35ft  the frog will be 3.6ft above the ground.


Part 3:  For this, we can use Calculus (take dh/dx, set to zero, solve for x) or we can note that for a parabola  {{{ax^2 + bx + c }}} the max/min will be on the axis of symmetry which is at x = -b/(2a):

           {{{ x = -75 / (2*(-50)) = 75/100 = 0.75}}} ft

[ Calculus method:  dh/dx = -100x + 75 
                   -100x + 75 = 0   -->  x = -75/-100 = 0.75ft ]


Part 4:  Plug in x = 0.75 into h(x):
         {{{ h(0.75) = -0.5(0.75^2) + 0.75(0.75) + 3.5 = 3.78 }}}ft (approx)

Some/most tutors will ignore your post if you post multiple problems.  I can understand why you did it :-)