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)
Wiki User
∙ 14y agoSince 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).
4x4 + 1 = (2x2 - 2x + 1)(2x2 + 2x + 1)
The vertex is at (-1,0).
2x2 + 11x + 12 = 2x2 + 3x + 8x + 12 = x*(2x + 3) + 4*(2x + 3) = (2x + 3)*(x + 4)
4x3+10x2+2x+5= (2x+5)(2x2+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).
2x2 + 2x = 2x(x+1)
4x4 + 1 = (2x2 - 2x + 1)(2x2 + 2x + 1)
-5
The vertex is at (-1,0).
2x2 + 11x + 12 = 2x2 + 3x + 8x + 12 = x*(2x + 3) + 4*(2x + 3) = (2x + 3)*(x + 4)
4x3+10x2+2x+5= (2x+5)(2x2+1)
2x2 + 3x + 1 = (x + 1)(2x + 1)
2x2+13x+15 = (2x+3)(x+5)
y=-2x^2+8x+3
-2x3 + 2x2 + 12x =(-2x) (x2 - x - 6) =(-2x) (x+2) (x-3)
2x2 + 21x + 49 = 2x2 +14x +7x + 49 = 2x(x + 7) + 7(x + 7) = (2x + 7)(x + 7)