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Find the vertex of the parabola f(x)=-4x^(2)+40x-93
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I tried using the quadratic formula but im just getting (-40 plusminus sqrt112)/-8
\[\left( -\frac{ b }{ 2a },f \left( -\frac{ b }{ 2a } \right) \right)\]
Would I make it -40/8 or -40/-8?
\[-\frac{ b }{ 2a }=-\left( \frac{ 40 }{ 2\cdot \left( -4 \right) } \right)=\frac{ 40 }{ 8 }=5\]
That makes more sense. So then I'd have for my y it'd be 27?
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no
(5,7) isn't it
-4(5)^2+40(5)-93 = -4(25)+40(5)-93 = -100 + 200 - 93 = 7 so yes marigirl
Oops sorry. Thanks.
now worries rrey! you never need to apologize for effort... only lack of it!
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no, not now
as in no worries
as in, it's okay
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