# if someone could give me an explanation for this, that would be great! thank you I dont know how to solve... or at least where to start.

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Given `y=-48t^2+135t` , put into factored form with integer coefficients and the find the coordinates of the horizontal intercepts:

`y=-48t^2+135t`

`=-3t(16t-45)`

One of the problems with computer graded assignments is that the answer must be in a specific form. Ususally factored form looks like (x-p)(x-q) so you might try the following equivalent alternatives:

y=3t(-16t+45)

y=(3t+0)(-16t+45)

y=(-3t+0)(16t-45)

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The coordinates of the intercepts can be found by setting the factors equal to zero.

3t=0 ==> t=0 so the coordinates are (0,0)

16t-45=0 ==> `t=45/16` so the coordinates are `(45/16,0)`

I don't see much that you can do with (0,0).

If the computer program wants exact answers then `(45/16,0)` should be correct -- of course you could find the decimal equivalent which is (2.8125,0). You will have to inquire (or read any instructions) as to how the coordinates are to be represented -- fractional form, decimal form, decimal rounded to 2,3,4,etc... places.

Factor `y=-48t^2+135t` as follows:

`y=3t(-16t+45)`

Determine the horizontal intercepts or points where the curve intercepts the x axis, by finding the values of t that make y=0.

`y=(3t)(-16t+45)=0`

Equal each factor to zero. First factor:

`3t=0rArr t=0`

Second factor =0

`-16t+45=0`

`-16t=-45`

`t=45/16=2.8`

For each value of t, t=0 and t=2.8, we know that the value of y is equal to 0.** Therefore the horizontal intercepts are (0,0) and (2.8,0)**

Thank you for this great explanation.

Unfortunately, every time that I have a wrong answer, the website reboot the page with new numbers... It makes more practice I guess!