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Question 192074: Find the exact vertex of the parabola algebraically...
f(x) = -3x^2 + 5x + 1
Could you please provide a detailed explanation of how you achieved the solution. Thanks!
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! The "vertex form" of a parabola is:
y= a(x-h)^2+k
where
(h,k) is the vertex
.
So, the idea is to manipulate the original equation into the "vertex form":
f(x) = -3x^2 + 5x + 1
First, group the x terms:
f(x) = (-3x^2 + 5x) + 1
Factor out a -3:
f(x) = -3(x^2 + (-5/3)x) + 1
Complete the square:
f(x) = -3(x^2 + (-5/3)x + __ ) + 1
f(x) = -3(x^2 + (-5/3)x + (25/36) ) + 1 + 25/12
f(x) = -3(x - 5/6 )^2 + 12/12 + 25/12
f(x) = -3(x - 5/6 )^2 + 37/12 (this is the vertex form)
.
Vertex = (5/6, 37/12)
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