Having some trouble factoring. I think I'm using the wrong factoring method, or that it can't be factored. The function I need to factor is f(t) = -16t^2 - 32t + 384.
f(t) = -16t^2 - 32t + 384 = -16(t^2 +2t -24) can you now
Would this be the correct first step to that? -16(t^2 + 2t - 24).
Oh. So that is the correct first step?
yup
= -16(t^2 +2t -24) = = -16(t^2 +6t -4t -24)
It's supposed to be like that, next? It IS asking me to "Factor the function f(t) and use the factors to interpret the meaning of the x-intercept of the function," but is that really how it is then? If so, what does it mean, exactly?
@matricked Where in the world did you get that -4t?
-24 = 6 x-4
@matricked, so where is the t?
If you're trying to get negative twenty-four, why still leave it in the expression?
= -16(t^2 +2t -24) =-16 (t^2 +(6-4)t -24)
Oh, I see. Your parens are messed up. You mean \[-16(t-4)(t+6)\] For the factored product.
...That would make a lot more sense.
@FrostFelon Yeah, I've never seen anyone factor that way. I makes sense if you were used to it. He is basically solving the problem, but instead of factoring, inserts the two terms where they are multiplied.
Then factors simplifies I assume.
=-16 (t^2 +(6-4)t -24) = -16 ( t^2 +6t -4t -24) =-16( t(t+6) -4(t+6)) =-16(t+6)(t-4)
That actually takes longer than the leap of logic I usually use.
That is super messed up. If it works for you, fine I guess, but you have not written what you think you have. According to your last step you have \[f(x)=-16(t^{2}+6t)+(-4t-24)\]
Last step before you give -16(t+6)(t-4).
Opps sorry; \[f(x)=-16((t^{2}+6t)+(-4t-24)\] So, \[f(x)=-16t^{2}+-96t+48t+384\] \[f(x)=-16t^{2}-48t+384\]
There's another part of the problem that I'm not entirely sure about.
Complete the square of the expression for f(x) to determine the vertex of the graph of f(x). Would this be a maximum or minimum on the graph? I can easily tell if it's a maximum or a minimum, but I don't know how to complete the square.
The vertex is (-1, 400), right?
@FrostFelon, Little late, but I believe complete the square means find t and solve \[16t^{2}\]
Oh.
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