We know that the point is in Q II. Thus, the x-coordinate of this point will be negative and the y-coordinate will be positive.
The y-coordinate is essentially given, and is y=4.
Imagine drawing a vertical line thru x=-2. This line will intersect y=4 at (-2,4).
The coordinates of the point are (-2, -4).
Answer:
so u see
Step-by-step explanation:
Since these are parabolas with y being squared, the standard form of such a parabola with vertex at (h, k) is
<span>x = a(y - k)^2 + h </span>
<span>There are no steps. Just look at what is inside the parentheses and compare it to (y - k) with k = -3. </span>
<span>Then look at what is added and compare it to h = -1. </span>
<span>For instance, A has (y + 1) in parentheses. So k = -1. And it has -3 added, so h = -3. That would be a vertex at (-3, -1).</span>
Simple....
you have:

So...first you know you're trying to find x..
But, you have a square root on one side...to remove this you'd have to square both sides--->>>

Leaving you with....

Move the terms over and set it equal to
zero.
-->>

Factor out a 3...

What multiplies to -12 and adds to 4?
6*-2=-12
6+-2=4
Leaving you with...

Remember you're solving for 0....
3x-2=0
3x-2=0
+2 +2
3x=2

x=
Then-->>
x+2=0
x+2=0
-2 -2
x=-2
Thus, your answer.
................5 FT cubed