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Thanks so much Bunuel. I cannot believe how easily you explained it. I have some doubts in the graphs solution but totally understood the algebraic approach.

one question that I have though is:

In the graphs, I see the functions: f(x) = -x-2 and f(x) = -3x-1. How did you derive these formulas and how did you use them towards the question?
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seekmba
Thanks so much Bunuel. I cannot believe how easily you explained it. I have some doubts in the graphs solution but totally understood the algebraic approach.

one question that I have though is:

In the graphs, I see the functions: f(x) = -x-2 and f(x) = -3x-1. How did you derive these formulas and how did you use them towards the question?

These lines are just examples of what actual line k could be look like, \(f(x) = -x-2\) and \(f(x) = -3x-1\) do not represent the exact equations of the line k. I just needed to draw lines for statements 1 and 2 (each line satisfying the given statement) to illustrate the point that x-intercept must be negative.

\(f(x)=-x-2\) is just one of the infinite lines with negative slope (\(m=-1<0\)) and with y-intercept equal to -2 (line passes through the pint \((0,-2)\)).

\(f(x)=-3x-1\) is just one of the infinite lines with negative slope (\(m=-3<0\)) and with negative y-intercept (\(y_{intercept}=b=-1<0\)).

Hope it's clear.
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In the xy-plane, if line k has negative slope and passes through the point (s;-2), is the x-intercept of line k positive?

(1) s = 0
(2) The y-intercept of line k is negative.
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In the xy-plane, if line k has negative slope and passes through the point (s;-2), is the x-intercept of line k positive?

Look at the diagram below:
Attachment:
Intercept.png
Intercept.png [ 11.68 KiB | Viewed 7099 times ]
(1) s = 0. Directly gives the y-intercept of line k: (0, -2). Since the y-intercept of line k is negative and the slope of line k is negative too then the x-intercept of line k must also be negative. Sufficient.

(2) The y-intercept of line k is negative. The same here: since the y-intercept of line k is negative and the slope of line k is negative too then the x-intercept of line k must also be negative.

Answer: D.
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seekmba
In the xy-plane, if line k has negative slope and passes through the point (s, -2), is the x-intercept of the line k positive?

(1) s = 0
(2) The y-intercept of line k is negative

took a minute to solve...but definitely can be cracked in under 30s...

1. if s=0,then -2 is the y interceptor. we know for sure that line k doesn't pass through 1st quadrant, because the slope is negative. therefore, it can't be possible to have an x-intercept of the line positive.
B, C, and E - out.

2. y intercept of line is negative. it means that the line doesn't pass through the quadrant 1. it can't be possible to have an x intercept that is positive.
sufficient.

D is the answer.
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enigma123
In the xy-plane, if line k has negative slope and passes through the point (s;-2), is the x-intercept of line k positive?

(1) s = 0
(2) The y-intercept of line k is negative.

The general equation of a line is y = mx+b. There are two things to consider here:

    If the line passes through \((s,-2)\), then \(-2 = ms+b.\)
    The x-intercept of a line is given by \(x = \frac{-b}{m}\).


Now, since m is negative, the question is asking is b > 0?

(1) s = 0. From the first consideration above, this tells us that b = -2. Sufficient.

(2) The y-intercept is negative (b < 0). Sufficient.

Therefore, either statement is sufficient.

Answer: D
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took some extra seconds to think of the possible options...
1 - x intercept is 0! statement 1 is sufficient.
2 - there is no possible way for x intercept to be positive - sufficient.

D must be the answer
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