# What is the vertex form of y=(3x-5)(6x-2) ?

May 6, 2018

The vertex form of $y = \left(3 x - 5\right) \left(6 x - 2\right) = 30 {\left(x - 0.6\right)}^{2} - 0.8$

#### Explanation:

First we must know what is meant by the vertex form of a quadratic function, which is
$y = a {\left(x - h\right)}^{2} + k$ (https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html)
We, therefore, want $\left(3 x - 5\right) \left(6 x - 2\right)$ on the above form.
We have $\left(3 x - 5\right) \left(6 x - 2\right) = 30 {x}^{2} - 36 x + 10$

Therefore $a = 30$

30(x-h)^2+k=30(x^2-2hx+h^2)+k= 30x^2-36x+10=30(x^2-1,2x)+10
Therefore $2 h = 1 , 2$

$30 {\left(x - 0.6\right)}^{2} = 30 \left({x}^{2} - 1.2 x + 0.36\right) = 30 {x}^{2} - 36 x + 10.8$
This gives
$30 {x}^{2} - 36 x + 10 = \left(30 {x}^{2} - 36 x + 10.8\right) - 0.8$
Therefore,
$\left(3 x - 5\right) \left(6 x - 2\right) = 30 {\left(x - 0.6\right)}^{2} - 0.8$

May 6, 2018

$y = 18 {\left(x - 1\right)}^{2} - 8$

#### Explanation:

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

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

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

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

$\text{expand the factors}$

$\Rightarrow y = 18 {x}^{2} - 36 x + 10$

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

$\text{factor out 18}$

$y = 18 \left({x}^{2} - 2 x + \frac{5}{9}\right)$

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

$y = 18 \left({x}^{2} + 2 \left(- 1\right) x \textcolor{red}{+ 1} \textcolor{red}{- 1} + \frac{5}{9}\right)$

$\textcolor{w h i t e}{y} = 18 {\left(x - 1\right)}^{2} + 18 \left(- 1 + \frac{5}{9}\right)$

$\textcolor{w h i t e}{y} = 18 {\left(x - 1\right)}^{2} - 8 \leftarrow \textcolor{red}{\text{in vertex form}}$