Consider the curve y=x2+5x and a point P(−1,−4) lies on the curve.
For x=−0.5 , the coordinates of point Q(x,x2+5x) is,
Q=(−0.5,(−0.5)2+5(−0.5))=(−0.5,−2.25)
The slope of the secant line joining the points P and Q is evaluated as,
Slope(m)=x2−x1y2−y1
=−0.5−(−1)−2.25−(−4)
=−0.5+1−2.25+4
=0.51.75
=27
=3.5
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