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

Jan 23, 2018

$y = 3 {\left(x - \frac{1}{3}\right)}^{2} - \frac{4}{3}$

#### Explanation:

Given a quadratic of the form $y = a {x}^{2} + b x + c$ the vertex, $\left(h , k\right)$ is of the form $h = - \frac{b}{2 a}$ and $k$ is found by substituting $h$.

$y = 3 {x}^{2} - 2 x - 1$ gives $h = - \frac{- 2}{2 \cdot 3} = \frac{1}{3}$.

To find $k$ we substitute this value back in:

$k = 3 {\left(\frac{1}{3}\right)}^{2} - 2 \left(\frac{1}{3}\right) - 1 = \frac{1}{3} - \frac{2}{3} - \frac{3}{3} = - \frac{4}{3}$.

So the vertex is $\left(\frac{1}{3} , - \frac{4}{3}\right)$.

Vertex form is $y = a \cdot {\left(x - h\right)}^{2} + k$, so for this problem:

$y = 3 {\left(x - \frac{1}{3}\right)}^{2} - \frac{4}{3}$

Jan 23, 2018

$y = 3 {\left(x - \frac{1}{3}\right)}^{2} - \frac{4}{3}$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to obtain this form use "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1"

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

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} - \frac{2}{3} x$

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

$\textcolor{w h i t e}{y} = 3 {\left(x - \frac{1}{3}\right)}^{2} + 3 \left(- \frac{1}{9} - \frac{3}{9}\right)$

$\Rightarrow y = 3 {\left(x - \frac{1}{3}\right)}^{2} - \frac{4}{3} \leftarrow \textcolor{red}{\text{in vertex form}}$

Jan 23, 2018

$y = 3 {\left(x - \frac{1}{3}\right)}^{2} - \frac{4}{3}$

#### Explanation:

You must complete the square to put this quadratic into turning point form.

First, factorise out the ${x}^{2}$ coefficient to get:

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

Then halve the $x$ coefficient, square it, and add it and subtract it from the equation:

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

Note that the polynomial inside the brackets is a perfect square. The extra $- \frac{1}{3}$ has been added to maintain equality (this is equivalent to adding and subtracting $\frac{1}{9}$, multiplying by $3$ when removing it from the brackets).

Hence:

$y = 3 {\left(x - \frac{1}{3}\right)}^{2} - \frac{4}{3}$

From this the turning point can be found to be located at $\left(\frac{1}{3} , - \frac{4}{3}\right)$