describe the graph of xy = -(1/2)?

Answer 1

It's a diagonal hyperbola. The center (which is NOT part of the graph) is the origin, and asymptotes are the x-axis and y-axis. The graph occupies only Quadrants II and IV.

Answer 2

y = -1/2x, So this graph has a shape like the graph of y = 1/x, but with the curves reflected to the other side of the x-axis.

Imagine the graph of y = 1/x, this has a curve in the 1st and 3rd quadrants, with an asymptote line at x=0, and another one at y=0

the graph of y = -1 /2x is similar, but the curves are in the 2nd and 4th quadrants, and each graph is stretched by a factor of
1/2. If this is a sketch, then no worry about the stretch thing.
Note that the asymptote lines are still in the same positions.

Answer 3

plot these points then join them as a curve, not a straight line:

(1,-1/2) (-1,1/2) (2,-1/4) (-2,1/4) (4,-1/8) (-4,1/8)

the curve passes through the origin. make sure that the curve does not intersect the X-axis at the sides (left and right of the graph), however it will keep getting closer

Answer 4

down 1, right 2

Answer 5

xy = (-1/2)
y = (-1/2)/x
y = (-1/2)/(x/1)
y = (-1/2)*(1/x)
y = (-1/(2x))

y < 0
y > 0
x > 0
x < 0

for a graph, go to http://www.calculator.com/calcs/GCalc.html

type in (-1/(2x))

Answer 6

St Line passing through origin (0,0) and slope 26.56deg and leying in IInd and IVth Qud. passing through pts. (2,-1),(4,-2),(1,-2),(2,-4) and so on

Answer 7

It's a hyperbola. It has the x-axis and y-axis for asymptotes.
It lies in the 2nd and 4th quadrants.

Answer 8

It's a hyperbola.