We know that the x-coordinate of the vertex is x = -b/2a.
We identify a, b, and c in f(x) = ax^2 + bx + c.
y = 2x^2 + 4x or y = 2x^2 + 4x + 0
a = 2, b = 4, c = 0
Substitute 2 for a and 4 for b into the x-coordinate equation:
x = -b/2a = -4/2(2) = -4/4 = -1
So the x-coordinate of the vertex is -1.
To find the y-coordinate, substitute -1 for x in the equation of the function, y = 2x^2 + 4x.
y = 2x^2 + 4x = 2(-1)^2 + 4(-1) = 2 - 4 = -2.
Thus, the vertex is (-1, -2).
It is (1, 1).
2x2= 4
2x2 + 4 + 1 = 2x2 + 5 So, the vertex is (0, 5)
Assuming the missing symbol there is an equals sign, then we have: y - 2x2 - 4x = 4 We can find it's vertex very easily by solving for y, and finding where it's derivative equals zero: y = 2x2 + 4x + 4 y' = 4x + 4 0 = 4x + 4 x = -1 So the vertex occurs Where x = -1. Now we can plug that back into the original equation to find y: y = 2x2 + 4x + 4 y = 2 - 4 + 4 y = 2 So the vertex is at the point (-1, 2)
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).
The derivative of 2x2 + 4x + 8 is 4x+4.
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)
2x2 + (5x-4x)2 + (x-x)2 = 2x2 + x2 + 02 = 2x2 + x2 = 3x2
4x
d/dx 2x2+3x+7=4x+3
The discriminant is 88.
2x2-4x-3x+6 2x2-7x+6 (2x+3)(x+2)