What is the the vertex of #y = -3x^2+6x-1 #?

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What is the the vertex of #y = -3x^2+6x-1 #?

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Answer 1 · 2016-01-28T23:39:58

The #v(-1, 2) #
#x = 0; f(0) = -1#

Explanation:

Given #f(x) = y = ax^2 + bx + c " "# form of the equation
The vertex, #v(h, k)#
# h = -b/(2a); and k = f(h)#
Now # f(x) = -3x^2 + 6x - 1 #
#h = - 6/(2*3) = -1; f(-1) = 2 #
Thus #v(-1, 2) #

Intercept is simply -1, to find simply set #x = 0; f(0) = -1#

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What is the the vertex of #y = -3x^2+6x-1 #?

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