Quad IV x is right side, y is down.
The coordinate or Cartesian plane is divided into four quadrants by the axes. The axes, themselves, do not belong to any quadrant. Assuming the normal x and y-axes, Quadrant I : x > 0, y > 0 Quadrant II : X < 0, y > 0 Quadrant III : X < 0, y < 0 Quadrant IV: X > 0, y < 0 That's it. No special sides, nothing to solve.
If x = 0 then the point is on the y-axis and so it not in any quadrant.
Quadrant I: x > 0, y > 0 Quadrant II: x < 0, y > 0Quadrant I: x < 0, y < 0Quadrant I: x > 0, y < 0
y=6x is in the third quadrant while x is negative and in the first quadrant while x is positive.
X < 0 and y > 0.
Third and fourth
For y - 2y - 3y equals 0, y equals 0.
Points on the x-axis or y-axis are not in any quadrant. Therefore, (-3,0) is not contained in a quadrant.
I can not graph for you, but two points can be found. Zero out X and Y. -X - 3 = 0 -X = 3 X = - 3 Y = - 3, of course Draw a line from the second quadrant into the third quadrant and through the fourth quadrant connecting these two points into a descending line.
It doesn't. Its a matter of interpretation. When drawing the unit circle, we start at x=1, y=0. As we draw, maintaining a radius of 1 from the origin at x=0, y=0, we proceed counter-clockwise. Initially, both x and y are positive. That is quadrant 1. When x becomes negative at x=0, y=1, that is quadrant 2. When y becomes negative at x=-1, y=0, that is quadrant 3. And when x becomes positive again at x=0, y=-1, that is quadrant 4. So you see, its all in the perspective of which comes first, and in trigonometry, the vector where theta = 0 comes first, not where your eye just happens to scan from left to right.