Answer to Question #124399 in Calculus for Jade

Question #124399
P(-1,-4) lies on the curve y=x^2+5x.

If Q is the point (x,x^2+ 5x), find the slope of the secant line PQ or x= -0.5
1
Expert's answer
2020-06-29T18:53:25-0400

Consider the curve y=x2+5xy=x^2+5x and a point P(1,4)P(-1,-4) lies on the curve.


For x=0.5x=-0.5 , the coordinates of point Q(x,x2+5x)Q(x,x^2+5x) is,


Q=(0.5,(0.5)2+5(0.5))=(0.5,2.25)Q=(-0.5,(-0.5)^2+5(-0.5))=(-0.5,-2.25)


The slope of the secant line joining the points PP and QQ is evaluated as,


Slope(m)=y2y1x2x1(m)=\frac{y_2-y_1}{x_2-x_1}


=2.25(4)0.5(1)=\frac{-2.25-(-4)}{-0.5-(-1)}


=2.25+40.5+1=\frac{-2.25+4}{-0.5+1}


=1.750.5=\frac{1.75}{0.5}


=72=\frac{7}{2}


=3.5=3.5


Therefore, the slope of the secant line PQPQ is m=3.5m=3.5

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