A person standing close to the edge on the top of a 160-foot 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 160-foot 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(x-2)^2 + 12.
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Find the coordinates of the vertex for the parabola defined by the functionf(x) = -3(x-2)^2 + 12.
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^2------Eq2:
V(0,0)----------V(7,4)
P(1,2)----------P(8,6)
Y = a(x-h)^2 + k,
a(8-7)^2 + 4 = 6,
a = 2.
Eq2: Y = 2(x-7)^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(x-2)^2 + 12.
Y = a(x-h)^2 + k.
V(h,k),
V(2,12).
To find the coordinates of the vertex for the parabola defined by the function f(x) = -3(x-2)^2 + 12, you can use the vertex form of a parabola, which is given by y = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.
In this case, the vertex form of the given function is y = -3(x-2)^2 + 12. Comparing this equation to the vertex form equation, we can see that the vertex is located at the point (2, 12).
Therefore, the coordinates of the vertex for the parabola defined by the function f(x) = -3(x-2)^2 + 12 are (2, 12).
To find the maximum height of the ball, we need to determine the vertex of the quadratic function s(t) = -16t^2 + 64t + 160. The vertex form of a quadratic function is given by:
s(t) = a(t - h)^2 + k
where (h, k) represents the coordinates of the vertex.
To find the maximum height, we need to find the value of t when the ball reaches its highest point, which is the vertex of the function.
The formula for finding the x-coordinate of the vertex is given by:
h = -b / (2a)
In our case, a = -16 and b = 64, so substituting these values into the formula, we have:
h = -64 / (2 * -16)
= -64 / -32
= 2
So, the ball reaches its maximum height at t = 2 seconds.
To find the maximum height, we substitute this value of t into the quadratic function:
s(2) = -16(2)^2 + 64(2) + 160
= -64 + 128 + 160
= 224
Therefore, the maximum height of the ball is 224 feet.
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To determine when the ball hits the ground, we need to find the value of t when the height, s(t), is equal to zero, since s(t) represents the height above the ground.
Setting s(t) = 0 in the quadratic function:
-16t^2 + 64t + 160 = 0
To solve this quadratic equation, we can either factor or use the quadratic formula. In this case, the equation does not factor easily, so we'll use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -16, b = 64, and c = 160. Substitute these values into the quadratic formula:
t = (-64 ± √(64^2 - 4 * -16 * 160)) / (2 * -16)
= (-64 ± √(4096 + 10240)) / (-32)
= (-64 ± √14336) / (-32)
= (-64 ± 120) / (-32)
Simplifying further, we have:
t = (-64 - 120) / (-32) or t = (-64 + 120) / (-32)
= -184 / -32 or t = 56 / -32
= 5.75 or t = -1.75
Since time cannot be negative in this context, we can disregard -1.75 as a solution.
Therefore, it takes approximately 5.75 seconds for the ball to hit the ground. Rounded to the nearest tenth of a second, it takes 5.8 seconds.
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To find the head circumference at birth according to the given model equation, we need to substitute x = 0 into the equation and solve for y.
y = 4(sqrt(0)) + 35
= 4 * 0 + 35
= 35
Therefore, according to the model, the head circumference at birth is 35 centimeters.
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To write the equation of a parabola in vertex form, we use the formula:
f(x) = a(x - h)^2 + k
where (h, k) represents the vertex of the parabola.
Since we want to find the equation of a parabola that has the same shape as f(x) = 2x^2 but with the vertex at (7, 4), we can simply substitute these values into the vertex form equation:
f(x) = a(x - 7)^2 + 4
To find the value of a, we can use another point on the parabola. Since the given parabola f(x) = 2x^2 passes through the point (0, 0), we can substitute these values into the equation to solve for a:
0 = a(0 - 7)^2 + 4
-4 = 49a
a = -4 / 49
Substituting this value of a back into the equation, we have:
f(x) = (-4 / 49)(x - 7)^2 + 4
Therefore, the equation in vertex form of the parabola that has the same shape as f(x) = 2x^2, but with the vertex at (7, 4), is f(x) = (-4 / 49)(x - 7)^2 + 4.
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To find the coordinates of the vertex for the given parabola f(x) = -3(x - 2)^2 + 12, we observe that the equation is already in vertex form, given by f(x) = a(x - h)^2 + k.
In this case, the vertex of the parabola is located at (h, k). Therefore, the coordinates of the vertex for this parabola are (2, 12).