SOLUTION: I am having trouble finding the Vertex of the quadratic equation of -0.2x^2 + 12x + 11 = 0 I have worked it to find that it has two solutions that being x = -0.90307428 and x =

Algebra ->  Graphs -> SOLUTION: I am having trouble finding the Vertex of the quadratic equation of -0.2x^2 + 12x + 11 = 0 I have worked it to find that it has two solutions that being x = -0.90307428 and x =       Log On


   

Question 333583: I am having trouble finding the Vertex of the quadratic equation of
-0.2x^2 + 12x + 11 = 0
I have worked it to find that it has two solutions that being x = -0.90307428 and x = 60.90307428.
I am confused at how to find the vertex of the curve.
Thanks for any help.
Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
your quadratic equation is y=-0.2x%5E2+%2B+12x+%2B+11
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the vertex can be found where x=-b/(2a) this comes from the quadratic solution for 0=-0.2x%5E2+%2B+12x+%2B+11 and its the front part of the quadratic solution.
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so vertex (x,y) is where x=-b/(2a) and substituting this into y=-0.2x%5E2+%2B+12x+%2B+11 to get the y part.
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from y=-0.2x%5E2+%2B+12x+%2B+11
a=-0.2, b=12, c=11
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x=-b/(2a)= -12/(2*(-0.2))=30
y=-0.2*30^2 +12(30)+11 = -180+360+11=191
vertex = (30,191)
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you could have also worked y=-0.2x%5E2+%2B+12x+%2B+11
into the form y=d%2A%28x-h%29%5E2%2Bk by completing the square and in this form
the vertex would have been (h,k)