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
_________________
SC Butler has resumed! Get
two SC questions to practice, whose links you can find by date,
here.Choose life.