# What is the the vertex of x =-1/2(y-2)^2-4y+12 ?

Aug 30, 2017

Vertex $\to \left(x , y\right) = \left(12 , - 2\right)$

#### Explanation:

$\textcolor{b l u e}{\text{General introduction}}$

Instead of a quadratic in $x$ this is a quadratic in $y$

If the ${y}^{2}$ term is positive then the general shape is $\subset$
If the ${y}^{2}$ term is negative then the general shape is $\supset$

If you expand the brackets we end up with $- \frac{1}{2} {y}^{2}$ which is negative. So the general shape is $\supset$
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$\textcolor{b l u e}{\text{Answering the question}}$

I choose to opt for the 'completed square' form of equation

Expanding the brackets we have:

$x = - \frac{1}{2} \left({y}^{2} - 4 y + 4\right) - 4 y + 12$

$x = - \frac{1}{2} {y}^{2} - 2 y + 10$

$x = - \frac{1}{2} {\left(y + 2\right)}^{2} + 12 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
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Check
$x = - \frac{1}{2} {y}^{2} - 2 y - 2 + 12 \text{ "->" } \textcolor{g r e e n}{x = - \frac{1}{2} {y}^{2} - 2 y + 10}$

Original eqn: $x = - \frac{1}{2} {\left(y - 2\right)}^{2} - 4 y + 12$

$x = - \frac{1}{2} {y}^{2} + 2 y - 2 - 4 y + 12$
$\textcolor{g r e e n}{x = - \frac{1}{2} {y}^{2} - 2 y + 10} \textcolor{red}{\leftarrow \text{ Thay match}}$
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From $E q u a t i o n \left(1\right)$

${y}_{\text{vertex}} = \left(- 1\right) \times 2 = - 2$

${x}_{\text{vertex}} = + 12$

Vertex $\to \left(x , y\right) = \left(12 , - 2\right)$