<PRE><B>Question 54474</B>
A)
Well, for a vertical parabola, the domain is always all real numbers (unless acted upon by a different source - not a parabola.) The domain is all real values.
B) Y - Intercept
y = -16(0)^2 + 50(0) + 4
y = 4
(0,4)
C)
0 = -16t^2 + 50t + 4
0 = 8t^2 - 25t - 2
{{{x = (-b +- sqrt( b^2 - 4*a*c ))/(2*a) }}}
{{{x = (25 +- sqrt( -25^2 - 4*8*-2 ))/(2*8) }}}
{{{x = (25 +- sqrt( 625 + 64 ))/(16) }}}
{{{x = (25 +- sqrt( 689 ))/(16) }}} about 3.2 seconds
D)
f(x) = -16t^2 + 50t + 4
v(1.5625,68.56640625)
The highest the ball will go is about 68 feet.
30 = -16t^2 + 50t + 4
0 = -16t^2 + 50t - 26
0 = 8t^2 - 25t + 13
{{{x = (-b +- sqrt( b^2 - 4*a*c ))/(2*a) }}}
{{{x = (25 +- sqrt( -25^2 - 4*8*13 ))/(2*8) }}}
{{{x = (25 +- sqrt( 625 - 416 ))/(16) }}}
{{{x = (25 +- sqrt( 209 ))/(16) }}} About 2.5 seconds
E)
v(1.5625,68.56640625) Vertex explains it all.
You might have to check if that is right (the vertex).</PRE>