Graph the Inequality

xyz_1965

See attachment for question and picture of graph.

Cervesa

$xy > 0$ for all points in quad IV ... why?

joshuaa

I understand why you are confused!

you think
xy >= 4 as same as y >= 4/x

They are not the same

here
y >= 4/x

any points on the curves or above will satisfy the condition

here
xy >= 4

any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition

for example

x = -2
y = -1 (above the left curve)

xy = (-2)(-1) = 3 will not satisfy the condition xy >= 4 cuz 3 is less than 4. so we need values equal or below the left curve

Debsta

MHF Helper
If x is pos, xy>4 implies y>4/× (dividing by a positive keeps the inequality sign).
But, if x is neg, xy>4 implies y<4/x (dividing by a negative reverses inequality sign).

So, in the first quadrant, where x is pos, you shade the area above.
In the third quadrant, where x is neg, you shade the area below.

xyz_1965

$xy > 0$ for all points in quad IV ... why?
The product of xy must be positive for xy > 0 to be a true statement.

xyz_1965

I understand why you are confused!

you think
xy >= 4 as same as y >= 4/x

They are not the same

here
y >= 4/x

any points on the curves or above will satisfy the condition

here
xy >= 4

any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition

for example

x = -2
y = -1 (above the left curve)

xy = (-2)(-1) = 3 will not satisfy the condition xy >= 4 cuz 3 is less than 4. so we need values equal or below the left curve
You said:

"any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition"

I assume that by "satisfy the condition" you meant to say the condition of the given inequality or
xy greater than or equal to 4.

xyz_1965

If x is pos, xy>4 implies y>4/× (dividing by a positive keeps the inequality sign).
But, if x is neg, xy>4 implies y<4/x (dividing by a negative reverses inequality sign).

So, in the first quadrant, where x is pos, you shade the area above.
In the third quadrant, where x is neg, you shade the area below.
As joshuaa said:

"any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition"

joshuaa

You said:

"any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition"

I assume that by "satisfy the condition" you meant to say the condition of the given inequality or
xy greater than or equal to 4.
the condition is
xy greater than or equal to 4

i advise you to always Test your final solution whenever you have a problem with inequalities

xyz_1965

xyz_1965

the condition is
xy greater than or equal to 4

i advise you to always Test your final solution whenever you have a problem with inequalities

joshuaa

what you have said correct except

let x = -4, and y = -3

(-4)(-3) >= 4

12 >= 4

True statement again proving that we must shade below the graph of the function in quadrant 3

xyz_1965