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4 years ago

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A particle starts at time t=0 and moves along a number line so that its position, at time t is greater than or equal to zero, is

given by x(t)= (t-2)(t-6)^3. The particle is moving to the left for...

1 answer:

olga2289 [7]4 years ago

4 0

<span>at t=0
x(0) = (-2)* (-6)^3= 432.
nitial position. At t=2, x(2)=0.
We can see that our particle is moving towards the left on the axis (by getting from 432 to 0
At t>2 and t<6, x(t) <0.
Our particle is still going left - now beyond the y axis
. At t=6, x(6) = 0
since we were left from the y axis (because we were going negative earlier) that means the particle took a right turn this time.
. At t>6, x(t)>0 the particle is still going right and will keep on going to infinity.
6 seconds.is answer hope this helps</span>

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Shkiper50 [21]

Answer:

Part 1) $y+8=3(x+2)^{2}$ -----> $y=-8.08$

Part 2) $y-14=-(x-3)^{2}$ ----> $y=14.25$

Part 3) $y+7.5=2(x+2.5)^{2}$ ----> $y=-7.625$

Part 4) $y-17=-(x-3)^{2}$ -----> $y=17.25$

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Step-by-step explanation:

we know that

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where

(h,k) is the vertex

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case 1) we have

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the vertex is the point (-2,-8)

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$p=\frac{1}{12}$

The directrix is equal to

$y=-8-\frac{1}{12}=-8.08$

case 2) we have

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The directrix is equal to

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case 3) we have

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The directrix is equal to

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The directrix is equal to

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$y-6=-(x-1)^{2}$

the vertex is the point (1,6)

$\frac{1}{4p}=-1$

$p=-\frac{1}{4}$

The directrix is equal to

$y=6+\frac{1}{4}=6.25$

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