Help? Very confused on question 8.

Mathematics

1 answer:

Fynjy0 [20]3 years ago

4 0

For part a: you just need to find how far the vertex has been moved from the origin, or the point (0,0). As the vertex is at the point (2,-3), it has been translated right 2 horizontally and down 3 vertically.

For part b: you use the info found in part a to create the equation in the form of y=A(x-h)^2+k. In this case, A =1, so you can ignore it. The h value is the horizontal distance the vertex has been moved. Since it has been moved right 2, this part of the equation would be (x-2). I know it seems like it should be plus 2, but values in parentheses come out opposite. For the k value, find the vertical shift, which is down3, or -3.
Now that you have h and k, substitute them back into the equation.
Your final answer for part b is: y=(x-2)^2 -3.

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As the fellow above said, the particle does make a U-turn at the vertex, and we can find that out by getting the derivative of s(t) and zeroing it out to get the point of the horizontal tangent.

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$\bf s(t)=2t^3-21t^2+60t+3\implies \boxed{\cfrac{ds}{dt}=6t^2-42t+60}\leftarrow v(t)
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\begin{cases}
5\\
2
\end{cases}\impliedby \textit{it makes the first turn at the \underline{2 second}}
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\textit{now let's find the acceleration equation, a(t)}
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\boxed{\cfrac{d^2s}{dt^2}=12t-42}\leftarrow a(t)\\\\
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s(2)=55\qquad \qquad \qquad \qquad a(2)=-18$

the negative acceleration value, simply means a decreasing rate of change, so, it means the particle is slowing down and possibly coming to a stop before changing direction again.

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