An object travels along the x-axis so that its position after t seconds is given by x(t) = 2t2 – 5t – 18 for all times t such that t ≥ 0. Which inequality describes all times t for which the object is traveling toward the right?

the function is given, and it's value is where the object is ("how far to the right").

so as long as it rises (going more right), this will be apply.

in the screenshot I graphed the function. of course t is graphed as x and "along the x-axis" is graphed as y, but the pattern is the same anyways.

for the first 1.25 seconds the object goes to the left, and after that always to the right.

since we look at t to calculate x, t effectively takes the role of the important variable that is normally given to x. the calculation pattern are just the same. so let's find the lowest point of this function by calculating it out.

x(t) = 2t² – 5t – 18

x'(t) = 4t -5

x'(t) = 0

0 = 4t -5

5 = 4t

1.25 = t

plugging it into the second derivative

x''(t) = 4

x''(1.25) = 4

it's positive, so at t=1.25 there is a low point

(of course the second derivative is constant anyways.)

the object is traveling toward the right

the object is traveling toward the rightfor t > 1.25

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y