Erin travels north and south from Main Station. The distance, in km, of the train from Main Station is modeled by the function d(t) = t3 - 9t2 + 6t, where North is positive and South is negative. Time
elapsed after the start of a shift, in hours, is represented by t, where t € (0,12]. 'If the shift starts at
noon, determine at which time(s) the train is more than 16 km south of Main Station.

The times are t = 9, t = 10, t = 11 and t = 12

Step-by-step explanation:

For the train to be more than 16 km South and since south is taken as negative,

d(t) > -16

t³ - 9t² + 6t > -16

t³ - 9t² + 6t + 16 > 0

Since -1 is a factor of 16, inserting t = -1 into the d(t), we have

d(-1) = (-1)³ - 9(-1)² + 6(-1)+ 16 = -1 - 9 - 6 + 16 = -16 + 16 = 0. By the factor theorem, t + 1 is a factor of d(t)

So, d(t)/(t + 1) = (t³ - 9t² + 6t + 16)/(t +1) = t² - 10t + 16

Factorizing  t² - 10t + 16, we have

t² - 2t - 8t + 16

= t(t - 2) - 8(t - 2)

= (t -2)(t - 8)

So t - 2 and t - 8 are factors of d(t)

So (t + 1)(t -2)(t - 8) > 0

when t < -1, example t = -2 ,(t + 1)(t -2)(t - 8) = (-2 + 1)(-2 -2)(-2 - 8) = (-1)(-4)(-10) = -40 < 0

when -1 < t < 2, example t = 0 ,(t + 1)(t -2)(t - 8) = (0 + 1)(0 -2)(0 - 8) = (1)(-2)(-8) = 16 > 0

when 2 < t < 8, example t = 3 ,(t + 1)(t -2)(t - 8) = (3 + 1)(3 -2)(3 - 8) = (4)(1)(-5) = -20 < 0

when t > 8, example t = 9,(t + 1)(t -2)(t - 8) = (9 + 1)(9 -2)(9 - 8) = (10)(7)(1) = 70 > 0

Since t cannot be negative, d(t) is positive in the interval 0 < t < 2 and t > 8

Since t ∈ (0, 12]

In the interval 0 < t < 2 the only value possible for t is t = 1

d(1) = t³ - 9t² + 6t = (1)³ - 9(1)² + 6(1) = 1 - 9 + 6 = -2

Since d(1) < -16 this is invalid

In the interval t > 8 , the only possible values of t are t = 9, t = 10.t = 11 and t = 12.

So,

d(9) = 9³ - 9(9)² + 6(9) = 0 + 54 = 54 km

d(10) = 10³ - 9(10)² + 6(10) = 1000 - 900 + 60 = 100 +60 = 160 km

d(11) = 11³ - 9(11)² + 6(11) = 1331 - 1089 + 66 = 242 + 66 = 308 km

d(12) = 12³ - 9(12)² + 6(12) = 1728 - 1296 + 72 = 432 + 72 = 504 km

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

d

step-by-step explanation: