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 + 2
Next, set y' equal to zero:
0 = -4x + 2
Then solve for x:
4x = 2
x = 2
This is the x-coordinate of the vertex. To find the y-coordinate, plug x = 2 back into the original equation:
y = -2x2 + 2x + 3
y = -8 + 4 + 3
y = -1
So the vertex is at (2, -1).
2x2 + 11x + 12 = 2x2 + 3x + 8x + 12 = x*(2x + 3) + 4*(2x + 3) = (2x + 3)*(x + 4)
2x2 + 5x - 3 (2x - 1)(x + 3)
2x2 + 7x + 3 = (2x + 1)(x + 3)If 2x2 + 7x + 3 = 0 then:(2x + 1)(x + 3) = 0⇒ (2x + 1) = 0 → x = -1/2or (x + 3) = 0 → x = - 3
(2x + 1)(x + 3)
2x2+13x+15 = (2x+3)(x+5)
2x2 + 11x + 12 = 2x2 + 3x + 8x + 12 = x*(2x + 3) + 4*(2x + 3) = (2x + 3)*(x + 4)
2x2 + 5x - 3 (2x - 1)(x + 3)
y=-2x^2+8x+3
2x2 + 7x + 3 = (2x + 1)(x + 3)If 2x2 + 7x + 3 = 0 then:(2x + 1)(x + 3) = 0⇒ (2x + 1) = 0 → x = -1/2or (x + 3) = 0 → x = - 3
2x2+5x+3 = (2x+3)(x+1) when factored
(2x + 1)(x + 3)
2x2+13x+15 = (2x+3)(x+5)
2x2+5x+3 = (2x+3)(x+1) when factored
2x2 + 7x + 6 = 2x2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3)
(2x + 1) (x + 3) = 2x2 + 7x + 3
2x2 + 11x + 15 = (x + 3)(2x + 5).
-2x3 + 2x2 + 12x =(-2x) (x2 - x - 6) =(-2x) (x+2) (x-3)