0
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)
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∙ 14y ago
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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).
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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).
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What are the factors of 4x3 plus 10x2 plus 2x plus 5?
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