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In the x-y plane, what is the slope of the line segment joi

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In the x-y plane, what is the slope of the line segment joi  [#permalink]

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New post 10 Apr 2018, 01:37
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[GMAT math practice question]

In the x-y plane, what is the slope of the line segment joining the y-intercept and the positive x-intercept of the curve \(y=x^2-5x-6\)?

A. 1
B. 2
C. 3
D. 4
E. 5

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In the x-y plane, what is the slope of the line segment joi  [#permalink]

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New post 10 Apr 2018, 06:18
MathRevolution wrote:
[GMAT math practice question]

In the x-y plane, what is the slope of the line segment joining the y-intercept and the positive x-intercept of the curve \(y=x^2-5x-6\)?

A. 1
B. 2
C. 3
D. 4
E. 5


The graph intercepts y-axis (y-intercept) when x=0. Thus, y-intercept is (0;-6).
Similarly, from the equation x^2-5x-6=0 we know that positive x-intercept is (6;0), because x=-1 and x=6.

So, we have two points of the line segment: (0;-6) and (6;0).
The simplest way to look at the slope is rise/run (rise over run).

To get from (0;-6) to (6;0), you rise up 6 and run 6. So the slope is 6/6=1.

Hence, A.
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In the x-y plane, what is the slope of the line segment joi  [#permalink]

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New post 10 Apr 2018, 06:43
MathRevolution wrote:
[GMAT math practice question]

In the x-y plane, what is the slope of the line segment joining the y-intercept and the positive x-intercept of the curve \(y=x^2-5x-6\)?

A. 1
B. 2
C. 3
D. 4
E. 5

Find the x-intercepts of the parabola (the quadratic equation, graphed): Set y=0 and factor
\(y=x^2-5x-6\)
\(x^2-5x-6=0\)
\((x-6)(x+1)\)
\(x = 6, x = -1\)
x-intercepts: \((6,0), (-1,0)\)

Find y-intercept: set x equal to 0
\(y=0^2-5(0)-6\)
\(y=-6\)
y-intercept: \((0,-6)\)

Slope of line segment that joins positive x-intercept \((6,0)\) and y-intercept \((0,-6)\):

The segment's end points are mirror points over
the line y = -x, which has a slope of (-1).
The segment is perpendicular
to that line.
Segment's opposite slope = 1

Or: slope =
\(\frac{rise}{run}=\frac{y_2-y_1}{x_2-x_1}=\frac{0-(-6)}{6-0}=\frac{6}{6}=1\)

Answer A
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Re: In the x-y plane, what is the slope of the line segment joi  [#permalink]

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New post 12 Apr 2018, 01:59
=>

First, we find the x-intercepts of the curve. These occur when y = 0. Now,
\(y=x^2-5x-6\)
\(=> y = (x+1)(x-6).\)

Thus, the x-intercepts are the points \((-1,0)\) and \((6,0),\) and the positive x-intercept is \((6,0).\)
As the y-intercept occurs when \(x = 0,\) it is the point \((0,-6).\)

Thus, the required slope is \(\frac{(6 – 0)}{( 0 – (-6) )} = 1.\)

Therefore, the answer is A.
Answer: A
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Re: In the x-y plane, what is the slope of the line segment joi &nbs [#permalink] 12 Apr 2018, 01:59
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