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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