Find the laplace transform y(s) of the solution of the given initial value problem. then invert to find y(t) . write uc for the heaviside function that turns on at c , not uc(t) .

y'' + 16y = e^(? 2t)u2

y(0) = 0 y'(0) = 0

y(s) =

y(t) =

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

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answer: it's d

step-by-step explanation: