What is the vertex form of #y=-3x^2-2x+1#?

1 Answer
Mar 3, 2016

The vertex form is the following,
#y=a*(x-(x_{vertex}))^2+y_{vertex}#
for this equation it is given by:
#y=-3*(x-(-1/3))^2+4/3#.

It is found by completing the square, see below.

Explanation:

Completing the square.

We begin with
#y=-3*x^2-2x+1#.

First we factor the #3# out of #x^2# and #x# terms
#y=-3*(x^2+2/3 x)+1#.
Then we separate out a #2# from in from of the linear term (#2/3x#)
#y=-3*(x^2+2*1/3 x)+1#.

A perfect square is in the form

#x^2 + 2*a*x+a^2#,

if we take #a=1/3#, we just need #1/9# (or #(1/3)^2#) for a perfect square!

We get our #1/9#, by adding and subtracting #1/9# so we don't change the value of the left hand side of the equation (because we really just added zero in a very odd way).

This leaves us with
#y=-3*(x^2+2*1/3 x+1/9-1/9)+1#.

Now we collect the bits of our perfect square
#y=-3*((x^2+2*1/3 x+1/9)-(1/9))+1#
Next we take the (-1/9) out of the bracket.

#y=-3*(x^2+2*1/3 x+1/9) + (-3)*(-1/9)+1#

and neaten up a bit

#y=-3*(x^2+2*1/3 x+1/9)+(3/9)+1#
#y=-3*(x+1/3)^2+4/3#.
Remember the vertex for is
#y=a*(x-(x_{vertex}))^2+y_{vertex}#
or we turn the plus sign into two minus signs producing,
#y=-3*(x-(-1/3))^2+4/3#.
This is the equation in vertex form and the vertex is #(-1/3,4/3)#.