How do you write the vertex form equation of the parabola # y=-x^2+4x+12#?

Answer 1

#y=-(x-2)^2+16#

Vertex form of equation is of the type #y=a(x-h)^2+k#
Note that here coefficient of #x^2# is #a#, hence in #y=-x^2+4x+12#
we take #a=-1#. Then we go on to form a square of form #(x-h)^2#
#:.y=-x^2+4x+12#
= #-(x^2-2xx2xx x+2^2-2^2)+12#
#-((x-2)^2-4)+12#
= #-(x-2)^2+4+12#
= #-(x-2)^2+16#
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Answer 2

To write the vertex form equation of the parabola (y = -x^2 + 4x + 12), complete the square on the quadratic expression. This involves rearranging the equation to isolate the squared term and then adding or subtracting a constant term to complete the square. The vertex form equation is (y = a(x - h)^2 + k), where ((h, k)) represents the vertex of the parabola.

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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.

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.

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.

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.

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