How do I graph $ x=-2(y+1)^2+4$ I like to know how do I graph this equation? I am a bit confused since the plus sign within the parenthesis I assume means to gown down on the y axis, and the + 4 outside the parenthesis would mean to move right on the x axis? Then I would stretch it by a factor of 2 and flip the graph on the x axis, thus it essentially would stay in the same place. Am I correct? 
 A: You'll get 
$$\begin{align}x=-2(y+1)^2+4&\iff -2(y+1)^2=x-4\\&\iff (y+1)^2=-\frac 12(x-4)\\&\iff \{y-(\color{blue}{-1})\}^2=4\cdot \frac{-1}{8}(x-\color{red}{4}).\end{align}$$
This is a parabola (here you can see a graph) whose vertex is $(\color{red}{4},\color{blue}{-1})$, the axis of symmetry is $y=\color{blue}{-1}$, the focus is $((-1/8)+\color{red}{4},\color{blue}{-1})$, and the directrix is $x=\color{red}{4}-(-1/8)$.
A: No, starting with $x=y^2$


*

*Shift down 1

*Stretch by a factor of 2

*Reflect in the y-axis

*Shift right 4
A: This is a trick question, because x is written alone and y is written with all the stuff around it.  Remember that for a parabola, y is the dependent variable so it should be alone on the left side.
$x=-2(y+1)^2+4 $
$x-4=-2(y+1)^2 $
$ -\frac{1}{2}(x-4)=(y+1)^2$
$y+1=\pm\sqrt{-\frac{1}{2}(x-4)}$
$y=\pm\sqrt{-\frac{1}{2}(x-4)}-1$
This is going to be a "sideways parabola" with vertical shift $-1$, horizontal shift $4$ to the right, and reflected over the $y$-axis so it is pointing to the left, then horizontally stretched by a factor of 2.
A: First thing to note is that this is a function of $y$ rather than $x$:
$$f(y) = -2(y+1)^2 + 4.$$
So this is a parabola that opens to the left because of the $-2$.  The rightmost point on the parabola occurs at $y=-1$ since this is the value for which $y+1 = 0$.  The particular $x$ value at $y=-1$ is $f(-1) = 4$, so the rightmost point is $(4, -1).$
From there, you can find, for example, $f(0), f(1), f(-2), f(-3),$ to sketch it out.
