I don't understand this.
An archer shoots an arrow into the air such that its height at any time, t, is given by the function h(t)=-16t^+kt+3. If the maximum height of the arrow occurs at time t=4, what is the value of k?
i plugged it in and got h=-253+4k ???
the answer choices are
a. 128
b. 64
c. 8
d. 4
what am i doing wrong??
An archer shoots an arrow into the air such that its height at any time, t, is given by the function h(t)=-16t^+kt+3. If the maximum height of the arrow occurs at time t=4, what is the value of k?
i plugged it in and got h=-253+4k ???
the answer choices are
a. 128
b. 64
c. 8
d. 4
what am i doing wrong??
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or without calculus:
the vertex of a parabola is at t = -b / (2a)
here, a = -16 and b = k
-b / (2a) = -k / (2 * -16) = 4 (given)
-k = -32(4) = -128
-k = -128 ==> k = 128 (a)
****
or, another way:
transform to vertex form:
h = a(t - 4)^2 + b
h = a(t^2 - 8t + 16) + b
and we know a = -16 ==>
h = -16t^2 + 128t - 256 + b
from the given, k = 128 (and the maximum height will be at b = 259 (since -256 + b = 3)
the vertex of a parabola is at t = -b / (2a)
here, a = -16 and b = k
-b / (2a) = -k / (2 * -16) = 4 (given)
-k = -32(4) = -128
-k = -128 ==> k = 128 (a)
****
or, another way:
transform to vertex form:
h = a(t - 4)^2 + b
h = a(t^2 - 8t + 16) + b
and we know a = -16 ==>
h = -16t^2 + 128t - 256 + b
from the given, k = 128 (and the maximum height will be at b = 259 (since -256 + b = 3)
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the maximum height occurs at the derivative
h(t) = -16t^2 + kt + 3
h'(t) = -32t + k
0 = -32*4 + k
k = 32*4 = 128
A
h(t) = -16t^2 + kt + 3
h'(t) = -32t + k
0 = -32*4 + k
k = 32*4 = 128
A
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You have to differentiate it, and get the answer a. 128