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How do you find the VERTEX of a parabola #y=-x^2+4x+12#?

Answer 1

I found:
#x_v=2#
#y_v=16#

The #x# coordinate of the vertex can be found by using the derivative (setting it equal to zero), but since I'm not sure if you know the derivative, you can use the relationship: #x_v=-b/(2a)# using your equation in the form #y=ax^2+bx+c# where: #a=-1# #b=4# #c=12#. This gives you: #x_v=-4/(-2)=2#. Subtracting this value back into your equation yields: #y_v=-4+8+12=16#.

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Answer 2

Jace Anderson

To find the vertex of a parabola in the form y = ax^2 + bx + c, you can use the formula x = -b/(2a) to find the x-coordinate of the vertex. Then, substitute this x-value into the equation to find the corresponding y-coordinate.

For the given equation y = -x^2 + 4x + 12: a = -1 b = 4

Substitute these values into the formula: x = -b/(2a) = -4/(2*(-1)) = -4/(-2) = 2

Now, substitute x = 2 into the equation to find the y-coordinate: y = -(2)^2 + 4(2) + 12 = -4 + 8 + 12 = 16

Therefore, the vertex of the parabola y = -x^2 + 4x + 12 is (2, 16).

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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.