Question

What is the vertex of `y= -3x^2-x-2(3x+5)^2`?

Answer 1

The vertex is at `(- 61/42, - 10059/1764)` or `(-1.45,-5.70)`

Explanation:

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)^2`

`y= -3x^2-x-2*(9x^2+2*5*3*x+25)`
`y= -3x^2-x-18x^2-60x-50`
`y= -21x^2-61x-50`

`x_{vertex} = {-b}/{2a}=61 /{2*(-21)}=- 61/42~= -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}`

`y_{vertex}= -21*(-61/42)^2-61*(-61/42)-50`

`y_{vertex}={-21*61*61}/{42*42}+{61*61*42}/{42*42} - {50*42*42}/{42*42}`

`y_{vertex}={ -21*61*61+61*61*42 - 50*42*42}/{42*42}`

`y_{vertex}=-10059/1764~=-5.70`

The vertex is at `(- 61/42, - 10059/1764)` or `(-1.45,-5.70)`