# What is the vertex of  y^2+4y+3x-4=0 ?

Mar 8, 2016

$\text{ } x = \frac{1}{3} {\left(y + 2\right)}^{2} - \frac{8}{3}$

#### Explanation:

This is a quadratic expressed in terms of y instead of terms in x. Consequently the graph will be of shape type $\subset$ instead of type $\cap$.
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$\textcolor{b l u e}{\text{Manipulating the equation to give the required format}}$

Given:$\text{ } {y}^{2} + 4 y + 3 x - 4 = 0$

$\textcolor{b r o w n}{\text{Subtract "3x" from both sides}}$

$\text{ } {y}^{2} + 4 y + 0 - 4 = - 3 x$

$\textcolor{b r o w n}{\text{Divide both sides by 3}}$

$\text{ } \frac{1}{3} {y}^{2} + \frac{4}{3} y - \frac{4}{3} = x$

$\text{ } \textcolor{b l u e}{x = \frac{1}{3} {y}^{2} + \frac{4}{3} y - \frac{4}{3}}$ ........................(1)

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$\textcolor{b l u e}{\text{Converting to Vertex Form}}$

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

$\textcolor{b r o w n}{\text{Changing the structure into vertex form and jumping a}}$
$\textcolor{b r o w n}{\text{number of steps.}}$

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

But $k = - \frac{4}{3}$ giving

$\text{ } x = \frac{1}{3} {\left(y + 2\right)}^{2} - \frac{8}{3}$..........................(2)

$\textcolor{red}{\text{If you need further explanation go to my profile page}}$ $\textcolor{red}{\text{and leave me a message. You must also give me a link to this page.}}$

Link $\to$ http://socratic.org/s/aszzseH6
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$\textcolor{g r e e n}{\text{Observe that the plots of equation (1) and equation (2) coincide.}}$
$\textcolor{g r e e n}{\text{This demonstrates that they are the same thing but just look }}$$\textcolor{g r e e n}{\text{different!}}$

Also notice the reversal of where you obtain the vertex coordinates
If the equation form had been y=... then you would have $y = - \frac{8}{3}$ but in this case it is $x = - \frac{8}{3}$ so also in this case $y = \left(- 1\right) \times 2 = - 2$

Vertex$\text{ "->(x,y)" } \to \left(- \frac{8}{3} , - 2\right)$ 