Someone i am so desperate like actually crying 1. the equation of a parabola is given. y=−1/6x2+7x−80 what is the equation of the directrix of the parabola? 2. (y+2)^2= 129x-5) what is the equation of the directrix of the parabola?

y= (−1/6)x^2+7x−80      multiply both sides by -6

-6y  =    x^2   - 42x  +  480       subtract 480 from both sides

 -6y  - 480   =  x^2  -  42x    take (1/2)  of 42  = 21square this  = 441  and add to both sides

 -6y - 480 + 441  =   x^2  - 42x  +  441       simplify the left, factor the right

 -6y - 39   =  (x  - 21)^2      factor the left side as

 -6 (y  +  39/6)   =  ( x - 21)^2       (1)

 Using the form

 4p (y - k)  =  ( x  - h)^2        we  can   write (1)  as

 4 (-3/2)(y - (-39/6) )   =  ( x  - 21)^2

 The vertex  = ( h, k)  =  ( 21, -39/6)   and    p  = -3/2

 And the directrix   is given by 

 y  = k - p  →   y  = -39/6 - (-3/2)  =  -39/6 + 3/2  = -39/6 + 9/6  =

-30/6  = - 5> y=-5

the answer Part 1) is

y=-5

Part 2) (y+2)^2= 129X-5 

Using the form

 4p (x - h)  =  ( y  - k)^2        we  can   write (1)  as

 4*(129/4)*(x-5/129)=(y+2)²

 The vertex  = ( h, k)  =  ( 5/129, -2)   and    p  = 129/4

 And the directrix   is given by 

x  = h - p  →   x = 5/129 - (129/4)  =  (20-16641)/516> -16621/516

x=-32.21

the answer part 2) is 

you have to put the graph on here with the question

step-by-step explanation: