When given the equation of a parabola in the form,
#y=ax^2+bx+c#
, the #x# coordinate of the vertex, #h#, can be found using the formula:
#h = -b/(2a)#
The #y# coordinate of the vertex, #k#, is found by evaluating the function at #h#:
#k = a(h)^2+b(h)+c#
The focal distance, #f#, can be found, using the formula:
#f = 1/(4(a))#
The equation of the directrix can be found, using the form:
#y = k-f#
Given: #y=−1/6x^2+7x−80#
Please observe that #a = -1/6 and b = 7#, therefore, the #x# coordinate of the vertex is:
#h = -7/(2(-1/6)) = 42/2#
#h = 21#
The #y# coordinate, #k#, of the vertex can be found by evaluating the function at #21#:
#k = -1/6(21)^2+7(21)-80#
#k = -13/2#
The focal distance f is:
#f=1/(4(-1/6))= -6/4#
#f = -3/2#
The equation of the directrix is:
#y = -13/2 - -3/2 = (-13+3)/2 = -10/2#
#y = -5#
