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wolverine [178]
3 years ago
5

Write an equation in point slope form for the line that passes through (-5,6) and (-3,-9).

Mathematics
1 answer:
Finger [1]3 years ago
8 0

\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{6})\qquad
(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-9})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-9-6}{-3-(-5)}\implies \cfrac{-9-6}{-3+5}\implies -\cfrac{15}{2}


\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-6=-\cfrac{15}{2}[x-(-5)]\implies y-6=-\cfrac{15}{2}(x+5)
\\\\\\
y-6=-\cfrac{15}{2}x-\cfrac{75}{2}\implies y=-\cfrac{15}{2}x-\cfrac{75}{2}-6\implies y=-\cfrac{15}{2}x-\cfrac{87}{2}

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we have a maximum at t = 0, where the maximum is y = 30.

We have a minimum at t = -1 and t = 1, where the minimum is y = 20.

<h3>How to find the maximums and minimums?</h3>

These are given by the zeros of the first derivation.

In this case, the function is:

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The first derivation is:

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We can just evaluate the function in these 3 values to see which ones are maximums and minimums.

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w(0) = 10*0^4 - 20*0^2 + 30    = 30

w(1) =  10*(1)^4 - 20*(1)^2 + 30 =  20

So we have a maximum at x = 0, where the maximum is y = 30.

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If you want to learn more about maximization, you can read:

brainly.com/question/19819849

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