Persm garis singgung 2y+4x-3=0 menyinggung parabola y=ax^2+bx+6 dititik (1,-1/2) maka 4a+b=?

y = ax^2 + bx + 6 ===> (1, -1/2)
-1/2 = a(1)^2 + b(1) + 6
-1/2 = a + b + 6
a + b = -1/2 - 6
a = -13/2 - b

2y + 4x - 3 = 0
2y = 3 - 4x
y = (3 - 4x)/2

y = y
ax^2 + bx + 6 = (3 - 4x)/2
2ax^2 + 2bx + 12 = 3 - 4x
2(-13/2 - b)x^2 + 2bx + 4x + 12 - 3 = 0
(-13 - 2b)x^2 + (2b + 4)x + 9 = 0
menyinggung : D = 0
b^2 - 4ac = 0
(2b + 4)^2 - 4(-13 - 2b)(9) = 0
4b^2 + 16b + 16 + 468 + 72b = 0
4b^2 + 88b + 484 = 0
b^2 + 22b + 121 = 0
(b + 11)(b + 11) = 0
b = -11

a = -13/2 - b = -13/2 - (-11) = -13/2 + 11 = 9/2

4a + b = 4(9/2) + (-11) = 18 - 11 = 7