How do you find the vertex and intercepts for x = –2(y– 3)^2 - 2?

Mar 29, 2016

Vertex at $\left(- 2 , 3\right)$

${x}_{\text{intercept}} = - 20$

Explanation:

Given $x = - 2 {\left(y \textcolor{b l u e}{- 3}\right)}^{2} \textcolor{m a \ge n t a}{- 2}$

$\textcolor{b l u e}{\text{Solving for vertex}}$

As in completing the square for x we have the same scenario but this time in y

$\text{y-vertex} = \left(- 1\right) \times \textcolor{b l u e}{- 3} = + 3$

$\text{x-vertex} = \textcolor{m a \ge n t a}{- 2}$

$\implies \text{Vertex} \to \left(x , y\right) \to \left(- 2 , 3\right)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for x intercept}}$

x intercept @ y=0 giving

$x = - 2 {\left(y - 3\right)}^{2} - 2 \to x = - 2 \left(9\right) - 2 = - 20$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for y intercept}}$

The determinant for $y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ Must be positive

$x = - 2 {\left(y - 3\right)}^{2} - 2 \text{ "->" } x = - 2 {y}^{2} + 12 y - 20$

Thus the determinant is: $\sqrt{{12}^{2} - 4 \left(- 2\right) \left(- 20\right)} = \sqrt{- 16}$

Thus there are only Complex number roots for x=0

color(magenta)(y_("intercept") " Does not exist")
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~