Complete the square to find the transformation.
-2(x-2)^2-4
the y value is -4.
so the vertex is at (2,-4)
OR
Use the vertex formula:
-b/2a
a=-2
b=8
x=-8/(2)-2
x=2
Run 2 through the original equation:
y=-(2)^2+8(2)-12
y=4
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2x2 + 4 + 1 = 2x2 + 5 So, the vertex is (0, 5)
It is (1, 1).
Since this question is in the calculus section, I'm assuming you know how to take the derivative. We know that y = -2x2 + 2x + 3 is a parabola, so it has one vertex, which is a minimum. We can use the first derivative test to find this extreme point.First, take the derivative:y' = -4x + 2Next, set y' equal to zero:0 = -4x + 2Then solve for x:4x = 2x = 2This is the x-coordinate of the vertex. To find the y-coordinate, plug x = 2 back into the original equation:y = -2x2 + 2x + 3y = -8 + 4 + 3y = -1So the vertex is at (2, -1).
8x12 plus the extra 3 is 99in
The vertex is (-9, -62).