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Ax+by-2=0

In the function above, a and b are constants. If the graph of the function has a negative slope and a negative y-intercept, which is the following is true?
A, a=0
B, a>0
C, a<0

User Chinasaur
by
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2 Answers

5 votes

Answer:

C, a < 0

Step-by-step explanation:

We have the equation,

ax + by - 2 = 0

Since the graph of the function has a negative slope and a negative y-intercept, we first write the equation in slope-intercept form,

ax + by - 2 = 0

ax + by = 2

by = -ax + 2

y = (-a/b)x + 2/b

Now, here, -a/b is the slope and 2/b is the y-intercept.

Since the y-intercept is negative, 2/b must be negative and hence b must be negative.

Also, the slope is negative as well,

Now since b is negative we get, (-a/b), to make this whole expression negative, a must also be negative.

Hence a < 0

User Schlangi
by
8.6k points
7 votes

Final answer:

In the equation ax+by-2=0, for a graph to have a negative slope, the coefficient 'a' must be negative (a<0). This means that option C, a<0, is correct.

Step-by-step explanation:

The equation ax+by-2=0 is a linear equation, and we are given that its graph has a negative slope and a negative y-intercept. To have a negative slope, in the slope-intercept form y = mx + b, the coefficient of x, which is m, must be negative.

Therefore, in our equation, 'a' (the coefficient of x in ax+by-2=0) must be negative to have a negative slope. For a negative y-intercept, when x=0, y should also be negative. The equation can be rewritten as y = -a/bx + 2/b. Here, we see that for y to be negative when x is 0, 2/b must be negative, which means b must be positive since the '2' is already negative in the original equation.

However, since we are only concerned with 'a', we conclude 'a' must be less than zero (a<0), which makes answer C correct.

User Alex Fu
by
8.1k points

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