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

Mar 3, 2016

The vertex form is the following,
$y = a \cdot {\left(x - \left({x}_{v e r t e x}\right)\right)}^{2} + {y}_{v e r t e x}$
for this equation it is given by:
$y = - 3 \cdot {\left(x - \left(- \frac{1}{3}\right)\right)}^{2} + \frac{4}{3}$.

It is found by completing the square, see below.

Explanation:

Completing the square.

We begin with
$y = - 3 \cdot {x}^{2} - 2 x + 1$.

First we factor the $3$ out of ${x}^{2}$ and $x$ terms
$y = - 3 \cdot \left({x}^{2} + \frac{2}{3} x\right) + 1$.
Then we separate out a $2$ from in from of the linear term ($\frac{2}{3} x$)
$y = - 3 \cdot \left({x}^{2} + 2 \cdot \frac{1}{3} x\right) + 1$.

A perfect square is in the form

${x}^{2} + 2 \cdot a \cdot x + {a}^{2}$,

if we take $a = \frac{1}{3}$, we just need $\frac{1}{9}$ (or ${\left(\frac{1}{3}\right)}^{2}$) for a perfect square!

We get our $\frac{1}{9}$, by adding and subtracting $\frac{1}{9}$ 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
$y = - 3 \cdot \left({x}^{2} + 2 \cdot \frac{1}{3} x + \frac{1}{9} - \frac{1}{9}\right) + 1$.

Now we collect the bits of our perfect square
$y = - 3 \cdot \left(\left({x}^{2} + 2 \cdot \frac{1}{3} x + \frac{1}{9}\right) - \left(\frac{1}{9}\right)\right) + 1$
Next we take the (-1/9) out of the bracket.

$y = - 3 \cdot \left({x}^{2} + 2 \cdot \frac{1}{3} x + \frac{1}{9}\right) + \left(- 3\right) \cdot \left(- \frac{1}{9}\right) + 1$

and neaten up a bit

$y = - 3 \cdot \left({x}^{2} + 2 \cdot \frac{1}{3} x + \frac{1}{9}\right) + \left(\frac{3}{9}\right) + 1$
$y = - 3 \cdot {\left(x + \frac{1}{3}\right)}^{2} + \frac{4}{3}$.
Remember the vertex for is
$y = a \cdot {\left(x - \left({x}_{v e r t e x}\right)\right)}^{2} + {y}_{v e r t e x}$
$y = - 3 \cdot {\left(x - \left(- \frac{1}{3}\right)\right)}^{2} + \frac{4}{3}$.
This is the equation in vertex form and the vertex is $\left(- \frac{1}{3} , \frac{4}{3}\right)$.