Question 1199164
<font color=black size=3>
I have a feeling the function should be
h(t) = -16t^2 + 48t + 6
where the 16 is negative up front.


--------------------------------------------------


Part (a)


I'll use x in place of t, and y in place of h(t).


The equation turns into
y = -16x^2 + 48x + 6
Compare this to 
y = ax^2 + bx + c
to find that
a = -16
b = 48
c = 6


We'll use the first two values to compute the following:
h = -b/(2a) 
h = -48/(2*(-16)) 
h = 1.5
This is the x coordinate of the vertex (h,k).



Answer: The ball reaches max height at <font color=red>1.5 seconds</font>


--------------------------------------------------


Part (b)


Use the previous result to find the y coordinate of the vertex.
y = -16x^2 + 48x + 6
y = -16(1.5)^2 + 48(1.5) + 6
y = 42
The vertex is located at (h,k) = (1.5, 42).
This represents the highest point of this parabola.


Answer: The max height of the ball is <font color=red>42 feet</font>


--------------------------------------------------


Part (c)


Replace y with 0 and solve for x.
y = -16x^2 + 48x + 6
0 = -16x^2 + 48x + 6
-16x^2 + 48x + 6 = 0
-2(8x^2 - 24x - 3) = 0
8x^2 - 24x - 3 = 0


We plug these values into the quadratic formula
a = 8
b = -24
c = -3
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-24)+-sqrt((-24)^2-4(8)(-3)))/(2(8))}}}


{{{x = (24+-sqrt(672))/(16)}}}


{{{x = (24+-  25.922963)/(16)}}}


{{{x = (24+25.922963)/(16)}}} or {{{x = (24-25.922963)/(16)}}}


{{{x = (49.922963)/(16)}}} or  {{{x = (-1.922961)/(16)}}}


{{{x = 3.120185}}} or  {{{x = -0.120185}}}
The steps in your homework do not to be as verbose as what I've described above. 
You can skip a few steps.
The decimal values are approximate.
Ignore the negative x solution because a negative time value makes no sense. 


Answer: The ball hits the ground at <font color=red>approximately 3.12 seconds</font>


--------------------------------------------------


I'll let the student handle parts (d) and (e).
</font>