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).