For an object whose velocity in ft/sec is given by v(t) = −t^2 + 4, what is its displacement, in feet, on the interval t = 0 to t = 3 secs? 7.67 −3.00 −0.33 3.00

The answer is 3 feet

Step-by-step explanation:

The displacement  function is the integration of the velocity formula, so:

if we evaluate the displacement for the upper and lower limit it would be

D = total displacement = d(3) - d(0)

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.