Posted by Nicole on Tuesday, October 18, 2011 at 2:43pm.
A person standing close to the edge on the top of a 160foot building throws a baseball vertically upward. The quadratic function
s(t) = 16t^2 +64t + 160
models the ball's height above the ground, s(t), in feet, t seconds after it was thrown.
After how many seconds does the ball reach its maximum height? What is the maximum height?
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A person standing close to the edge on the top of a 160foot building throws a baseball vertically upward. The quadratic function
s(t) = 16t^2 +64t + 160
models the ball's height above the ground, s(t), in feet, t seconds after it was thrown.
How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.
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The equation shown below models the head circumference of severely autistic children, y, in centimeters, at age x months.
y = 4(sqrt x) + 35
According to the model, what is the head circumference at birth?
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Write an equation in vertex form of the parabola that has the same shape as the graph of f(x) = 2x^2, but with the point (7, 4) as its vertex.
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Find the coordinates of the vertex for the parabola defined by the functionf(x) = 3(x2)^2 + 12.
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Find the coordinates of the vertex for the parabola defined by the functionf(x) = 3(x2)^2 + 12.

Math  Henry, Wednesday, October 19, 2011 at 5:17pm
1. Vf = Vo + gt,
t = (Vf  Vo) / g,
t(up) = (0  64) / 32 = 2s. = Time to reach max. ht.
2. h(2) = 16*2^2 + 64*2 + 160 = 224Ft.
above ground.
3. h = Vo*t + 0.5g*t^2 = 224Ft.,
0*t + 16t^2 = 224,
t^2 = 14,
t(dn) = 3.7s.
T = t(up) + t(dn) = 2 + 3.7 = 5.7s. = Time to hit ground.
4. Y = 4(sqrtx) + 35,
Y = 4*o + 35 = 35cm.
5. Eq1: F(x) = Y = 2x^2.
Let X = 1, Y = 2*1^2 = 2
Eq1:Y=2x^2Eq2:
V(0,0)V(7,4)
P(1,2)P(8,6)
Y = a(xh)^2 + k,
a(87)^2 + 4 = 6,
a = 2.
Eq2: Y = 2(x7)^2 + 4.
Notice that the vertex is shifed 7 units to the right and 4units upward.Therefore, each point on a graph
of Eq2 should be shifted likewise.
6. F(x) = Y = 3(x2)^2 + 12.
Y = a(xh)^2 + k.
V(h,k),
V(2,12).