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)
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).
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)