If x2 + y2 = 1, find the maximum and minimum of y = 3x2 + 4y2-2x-5 Specific steps should be taken

If x2 + y2 = 1, find the maximum and minimum of y = 3x2 + 4y2-2x-5 Specific steps should be taken

Because x2 + y2 = 1
So, - 1 ≤ x ≤ 1
3x2+4y2-2x-5
=4(x^2+y^2)-x^2-2x-5
=4-(x+1)^2-4
=-(x+1)^2
When x = - 1, there is a maximum value = 0
When x = 1, there is a minimum value = - 4