what is the equationfor a line that passes through the point (-1,8)and only quadrants I,II,and IV on a graph?

Answer 1

There is no single solution to this problem. Several lines can pass through the point (-1,8) and not go through the third quadrant.

Since you specified that the line must go through quadrant I then passing through the origin (0,0) is out, but using one point as (-1,8) and any other point (x,0) where x is positive, would satisfy your conditions.

One example would be Y = -X +7

Answer 2

This answer is not unique. Pick any positive number you want for the x-intercept. This will give you a line that passes only through quadrants 1, 2, and 4. Say you pick (10,0).
Find the slope between the two points.
m = (8-0)/(-1-10) = 8/-11 = -8/11
Now use the equation y = mx + b
y = (-8/11)x + b
Replace x with 10 and y with 0 to find b.
0 = (-8/11)(10) + b
0 = -80/11 + b
80/11 = b
Answer: y = (-8/11)x + 80/11
This is just one answer. You could choose (1,0) as your x-intercept and get another equation.

Answer 3

There are many lines that will satisfy that criteria.

One one side (Or "bound") there is the line that passes through
(0,0) and (-1, 8) You usually want the equation in the form

y = mx + b

so m = (y2 - y1)/(x2 - x1) = (8-0)/(-1 -0) = 8/-1 = -8

at (-1, 8) y = mx + b or 8 = -8(-1) + b

so solve for b

(If the line doesn't pass through (0,0) it does pass extremely close to that point, so we say that the line described by that expression is one bound)


On the other side is the line that passes through the points (-1,8) and (Q,-e) where Q = positive infinity and e = the smallest possible number greater than zero) (My keyboard isn't allowing me to type special characters)

m = (y2-y1)/(x2-x1) = (-e-8) / (Q - -1)

since the denominator is large, this expression simplifies to zero

y = mx + b => 8 = 0x + b

Solve for b

This is the other bound.


Extra Credit:

Show
y = mx + (8+m)
where -8
(You don't expect me to do all your homework, do you?)

Answer 4

data insufficiency