When given the equation of a parabola in the form,
y=ax^2+bx+cy=ax2+bx+c
, the xx coordinate of the vertex, hh, can be found using the formula:
h = -b/(2a)h=−b2a
The yy coordinate of the vertex, kk, is found by evaluating the function at hh:
k = a(h)^2+b(h)+ck=a(h)2+b(h)+c
The focal distance, ff, can be found, using the formula:
f = 1/(4(a))f=14(a)
The equation of the directrix can be found, using the form:
y = k-fy=k−f
Given: y=−1/6x^2+7x−80y=−16x2+7x−80
Please observe that a = -1/6 and b = 7a=−16andb=7, therefore, the xx coordinate of the vertex is:
h = -7/(2(-1/6)) = 42/2h=−72(−16)=422
h = 21h=21
The yy coordinate, kk, of the vertex can be found by evaluating the function at 2121:
k = -1/6(21)^2+7(21)-80k=−16(21)2+7(21)−80
k = -13/2k=−132
The focal distance f is:
f=1/(4(-1/6))= -6/4f=14(−16)=−64
f = -3/2f=−32
The equation of the directrix is:
y = -13/2 - -3/2 = (-13+3)/2 = -10/2y=−132−−32=−13+32=−102
y = -5y=−5
