# What is the vertex form of  y= (5x-9)(3x+4)+x^2-4x?

Sep 11, 2017

See below.

#### Explanation:

First multiply out the brackets and collect like terms:

$15 {x}^{2} - 27 x + 20 x - 36 + {x}^{2} - 4 x \implies 16 {x}^{2} - 11 x - 63$

Bracket terms containing the variable:

$\left(16 {x}^{2} - 11 x\right) - 63$

Factor out the coefficient of ${x}^{2}$:

$16 \left({x}^{2} - \frac{11}{16} x\right) - 63$

Add the square of half the coefficient of $x$ inside the bracket, and subtract the square of half the coefficient of $x$ outside the bracket.

$16 \left({x}^{2} - \frac{11}{16} x + {\left(\frac{11}{32}\right)}^{2}\right) - 63 - {\left(\frac{11}{32}\right)}^{2}$

Rearrange $\left({x}^{2} - \frac{11}{16} x + {\left(\frac{11}{32}\right)}^{2}\right)$ into the square of a binomial.

$16 {\left(x - \frac{11}{32}\right)}^{2} - 63 - {\left(\frac{11}{32}\right)}^{2}$

Collect like terms:

$16 {\left(x - \frac{11}{32}\right)}^{2} - 63 - {\left(\frac{11}{32}\right)}^{2}$

$16 {\left(x - \frac{11}{32}\right)}^{2} - \frac{64633}{1024}$

This is now in vertex form: $a {\left(x - h\right)}^{2} + k$
Where $h$ is the axis of symmetry and $k$ is the maximum or minimum value of the function.

So from example:

$h = \frac{11}{32}$ and $k = - \frac{64633}{1024}$

Sep 11, 2017

$y = 16 {\left(x - \frac{11}{32}\right)}^{2} - \frac{2425}{64}$

#### Explanation:

$\text{the first step is to rearrange the parabola in standard form}$

$\text{that is } y = a {x}^{2} + b x + c \to \left(a \ne 0\right)$

$\text{expand factors using FOIL and collect like terms}$

$y = 15 {x}^{2} - 7 x - 36 + {x}^{2} - 4 x$

$\textcolor{w h i t e}{y} = 16 {x}^{2} - 11 x - 36 \leftarrow \textcolor{red}{\text{ in standard form}}$

$\text{the x-coordinate of vertex in standard form is}$

${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a}$

$y = 16 {x}^{2} - 11 x - 36$

$\text{with } a = 16 , b = - 11 , c = - 36$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{- 11}{32} = \frac{11}{32}$

$\text{substitute this value into the equation for y}$

${y}_{\textcolor{red}{\text{vertex}}} = 16 {\left(\frac{11}{2}\right)}^{2} - 11 \left(\frac{11}{32}\right) - 36 = - \frac{2425}{64}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{11}{32} , - \frac{2425}{64}\right)$

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where )h , k ) are the coordinates of the vertex and a is a multiplier.

$\text{here "(h,k)=(11/32,-2425/64)" and } a = 16$

$\Rightarrow y = 16 {\left(x - \frac{11}{32}\right)}^{2} - \frac{2425}{64} \leftarrow \textcolor{red}{\text{ in vertex form}}$