# What is the the vertex of y =3x^2+4x-18 ?

Feb 26, 2016

x_("vertex")=-2/3" "I will let the reader find ""y_("vertex")

#### Explanation:

Given:$\text{ "y=3x^2+4x-18" }$..................................(1)

Write as:$\text{ } y = 3 \left({x}^{2} + \frac{4}{3} x\right) - 18$

Using the $+ \frac{4}{3} \text{ from } \left({x}^{2} + \frac{4}{3} x\right)$

$\left(- \frac{1}{2}\right) \times \frac{4}{3} = - \frac{4}{6} = - \frac{2}{3}$

$\textcolor{b l u e}{{x}_{\text{vertex}} = - \frac{2}{3}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$- \frac{2}{3} \text{ " =" " -0.6666bar6" "=" } - 0.6667$ to 4 decimal places

$\textcolor{b r o w n}{\text{All you have to do now is substitute "x=-2/3" into}}$color(brown)("equation (1) to find "y_("vertex"))

Feb 26, 2016

May be done as follows

#### Explanation:

The given equation is
$y = 3 {x}^{2} + 4 x - 18$
$\implies y = 3 \left[{x}^{2} + 2 x \left(\frac{2}{3}\right) + {\left(\frac{2}{3}\right)}^{2} - {\left(\frac{2}{3}\right)}^{2} - 6\right]$

$\implies y = 3 \left[{\left(x + \frac{2}{3}\right)}^{2} - {\left(\frac{2}{3}\right)}^{2} - 6\right]$
$\implies y = 3 \left[{\left(x + \frac{2}{3}\right)}^{2} - \frac{4}{9} - 6\right]$
$\implies y = 3 \left[{\left(x + \frac{2}{3}\right)}^{2} - \frac{58}{9}\right]$
$\implies y = 3 {\left(x + \frac{2}{3}\right)}^{2} - \frac{58}{9} \cdot 3$
$\implies y + \frac{58}{3} = 3 {\left(x + \frac{2}{3}\right)}^{2}$
putting ,$y + \frac{58}{3} = Y \mathmr{and} x + \frac{2}{3} = X$ we have
new equation
$Y = 3 {X}^{2}$,which has coordinate of vertex (0,0)
So putting X=0 and Y=0 in the above relation
we get
$x = - \frac{2}{3}$
and $y = - \frac{58}{3} = - 19 \frac{1}{3}$

so the actual coordinate of vertex is $\left(- \frac{2}{3} , - 19 \frac{1}{3}\right)$