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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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10 Apr 2018, 00:37
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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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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 28 Nov 2017 Posts: 145 Location: Uzbekistan In the x-y plane, what is the slope of the line segment joi [#permalink] Show Tags 10 Apr 2018, 05:18 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. _________________ Kindest Regards! Tulkin. Senior SC Moderator Joined: 22 May 2016 Posts: 2204 In the x-y plane, what is the slope of the line segment joi [#permalink] Show Tags 10 Apr 2018, 05:43 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6620 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: In the x-y plane, what is the slope of the line segment joi [#permalink] Show Tags 12 Apr 2018, 00:59 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: In the x-y plane, what is the slope of the line segment joi &nbs [#permalink] 12 Apr 2018, 00:59
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