The vertex of the positive parabola turns at point (-2, -11)
Need signs between terms.
It is: y = -4x+12
8x + 12 = 4x - 4 8x - 4x = -4 - 12 4x = -16 x = -4
4x-5=7 4x=12 x=3
The vertex of the positive parabola turns at point (-2, -11)
Need signs between terms.
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)
If: 4x-4 = 12 Then: x = 4
If: 4x+(-12) = 8 then 4x = 20 and x = 5
It is: y = -4x+12
8x + 12 = 4x - 4 8x - 4x = -4 - 12 4x = -16 x = -4
48
If: 4x-4 = 12 Then: x = 4
Differentiate and equate to zero Hence y = x^2 + 4x - 12 dy/dx = 2x + 4 = 0 2x = -4 x = -2 Hence y = (-2)^2 +4(-2) - 12 y = 4 - 8 - 12 y = - 16 Hence the vertex is (x,y,) = ( -2,-16)
4x + 6 = 6x - 12-4x -4x6 = 2x - 12+12 +1218 = 2xx=9
4x-5=7 4x=12 x=3