Sketch the Graph $x = {(y + 4)^2} - 8$ My vertex for this parabola that opens towards the right is (4,8). Is that correct? This program is giving me a totally different graph. Am I wrong or is the program wrong? https://www.desmos.com/calculator
 A: As Jyrki stated, the minimum x value will occur when $$\left(y+4\right)^2$$ is minimized, or when $$y=-4$$
At this point, $$x=(-4+4)^2-8=0-8=-8$$
Thus, the vertex will be at (-8,-4).
A: I think you need to revise the transformations:
$x=\pm(y+a)^2+b$


*

*$+a \rightarrow $ shifts $a$ down

*$-a \rightarrow $ shifts $a$ up

*$+b \rightarrow $ shifts $b$ to right

*$-b \rightarrow $ shifts $b$ to left


In this particular example: 
$x=(y+4)^2-8$ is just the parabola $x=y^2$ shifted 4 units down and 8 units to the left, so the vertex becomes $(-8,-4)$.
A: No, that is not correct: (4,8) is not the vertex of the parabola. If you move the constant term from the right to the left side and change to subtraction you get
$$x - - 8 = {(y -  - 4)^2}$$
This compares to 
$$x - a = c{(y -  b)^2}$$
where the vertex is $(a,b)$. We now see that the vertex is at $(-8, -4)$.
A: I use calculus the find the relative extrema:
$$
x'(y) = 2(y + 4) = 0 \Rightarrow y = -4
$$
Because of $x''(y) = 2 > 0$ that is at least a local minimum at $y = -4$. So the shape is pretty clear.
Further $x(-4) = -8$, so the coordinates of your vertex point are $(y, x) = (-4, -8)$. 
Please check the labels of your coordinate system, the usual plots are for $(x,y)$ pairs, $x$ on the horizontal, $y$ on the vertical axis. That way read the result as $(x,y) = (-8, -4)$.
