Question 1209303
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I'll use x in place of t.
{{{-x^2 + 8x -4 - 5x^2 + 14x + 55}}}
Combine like terms to get
{{{-6x^2 + 22x + 51}}}


We could follow the process of completing the square, but I'll use a shortcut.
The vertex of {{{y = ax^2+bx+c}}} is at (h,k) where {{{h = -b/(2a)}}}


Compare {{{-6x^2 + 22x + 51}}} with {{{ax^2+bx+c}}} to see that a = -6, b = 22, c = 51


So,
{{{h = -b/(2a)}}}
{{{h = -22/(2*(-6))}}}
{{{h = 11/6}}}
This is the x coordinate of the vertex.


Plug it into the previous expression to find that
{{{-6x^2 + 22x + 51}}}
= {{{-6(11/6)^2 + 22(11/6) + 51}}}
= {{{427/6}}}
This is the y coordinate of the vertex and it's largest output possible for {{{-6x^2 + 22x + 51}}}


The vertex is located at (11/6, 427/6) which is the highest point of this parabola.


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Answer: <font color=red>427/6</font>
This is approximately equal to 71.166667 where the 6's go on forever but we have to round at some point.


Verification using <a href="https://www.wolframalpha.com/input?i=max+-t%5E2+%2B+8t+-4+-+5t%5E2+%2B+14t+%2B+55">WolframAlpha</a>
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