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How do you write #f(x) = 3x^2 - 6x + 9# into vertex form?

Answer 1

#y = 3(x - 1)^2 + 6#

x-coordinate of vertex: #x = -b/(2a) = 6/6 = 1# y-coordinate of vertex: y (1) = 3 - 6 + 9 = 6 Vertex (1, 6) Vertex form #y = 3(x - 1)^2 + 6#

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Answer 2

Hazel Anglin

To write the function ( f(x) = 3x^2 - 6x + 9 ) into vertex form, complete the square:

[ f(x) = 3(x^2 - 2x) + 9 ] [ f(x) = 3(x^2 - 2x + 1) + 9 - 3 ] [ f(x) = 3(x - 1)^2 + 6 ]

So, the vertex form of the function is ( f(x) = 3(x - 1)^2 + 6 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.