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What is the vertex of the quadratic y -2x2 plus 4x-1?
It is (1, 1).
What is the vertex of -2x2 plus 2x plus 3?
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).
Find the vertex of -2x2 plus 2x plus 8?
f(x) = -2x2 + 2x + 8 ∴ f'(x) = -4x + 2 Let f'(x) = 0: 0 = -4x + 2 ∴ 4x = 2 ∴ x = 0.5 Now find the corresponding value: f(0.5) = -2(0.5)2 + 2(0.5) + 8 = -0.5 + 1 + 8 = 8.5 So the vertex of this parabola occurs at the point (0.5, 8.5)
What is the range of y equals 2x2 plus 4x-3?
y = 2x2 + 4x - 3 This equation describes a parabola. Because it's first term is positive, we know that it goes infinitely upward, and has a minimum that occurs at it's vertex. You can find it's vertex by taking it's derivative and solving for zero: y' = 4x + 4 0 = 4x + 4 0 = x + 1 x = -1 y = (-1)2 + 4(-1) - 3 y = 1 - 4 - 3 y = -6 So the vertex is at (-1, -6), which means that y ≥ -6
Factor 4x4 plus 1?
4x4 + 1 = (2x2 - 2x + 1)(2x2 + 2x + 1)
