You can find the vertex from ANY of the three forms of a parabola: Standard, factored and vertex. Since it is simpler I'm going to converted this into standard form.
y= -3x^2-x-2(3x+5)^2y=−3x2−x−2(3x+5)2
y= -3x^2-x-2*(9x^2+2*5*3*x+25) y=−3x2−x−2⋅(9x2+2⋅5⋅3⋅x+25)
y= -3x^2-x-18x^2-60x-50 y=−3x2−x−18x2−60x−50
y= -21x^2-61x-50 y=−21x2−61x−50
x_{vertex} = {-b}/{2a}=61 /{2*(-21)}=- 61/42~= -1.45 xvertex=−b2a=612⋅(−21)=−6142≅−1.45
(you can prove this by either completing the square in general or averaging the roots found from the quadratic equation)
and then substituted it back into the expression to find y_{vertex}yvertex
y_{vertex}= -21*(-61/42)^2-61*(-61/42)-50 yvertex=−21⋅(−6142)2−61⋅(−6142)−50
y_{vertex}={-21*61*61}/{42*42}+{61*61*42}/{42*42} - {50*42*42}/{42*42}yvertex=−21⋅61⋅6142⋅42+61⋅61⋅4242⋅42−50⋅42⋅4242⋅42
y_{vertex}={ -21*61*61+61*61*42 - 50*42*42}/{42*42}yvertex=−21⋅61⋅61+61⋅61⋅42−50⋅42⋅4242⋅42
y_{vertex}=-10059/1764~=-5.70yvertex=−100591764≅−5.70
The vertex is at (- 61/42, - 10059/1764)(−6142,−100591764) or (-1.45,-5.70)(−1.45,−5.70)
