Question 1181301
h(t)=-2t^2+8t+6.2.
a) This is the constant 6.2 m
b)-2t^2+8t+6.2=h(t)
-2(t^2-4t)+6.2=h(t)
complete the square by taking half of -4 and squaring it, then adding 8 to the constant to balance the equation.
-2(t^2-4t+4)+14.2
-2(t-2)^2+14.2
The vertex is at (2, 14.2), changing the sign of h and keeping the same sign of k.
This is 2 seconds and 14.2 m
c) Not sure what optimal height means in the context of this question but 14.2 m is maximum height
d)Maximum value given the negative quadratic, and that is 14.2 m after 2 seconds
e) The range needs the time it takes to hit or be 0.
so-2t^2+8t+6.2=0
ad t^2-4t-3.1=0, dividing by -2
t=(1/2)(4+/- sqrt (16+12.4)) and sqrt (28.4)=5.33
positive root is 4.66 m. The range of the function is [0, 4.66] units seconds
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{{{graph(300,300,-2,6,-5,20,-2x^2+8x+6.2)}}}
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Can check this by vertex = -b/2a which is -8/-4 or 2 seconds
h(2)=-8+16+6.2=14.2
Note: the graph is plotting height with time, not distance with time.