Question

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

Answer 1

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

For $x=-0.5$ , the coordinates of point $Q(x,x^2+5x)$ 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$(m)=\frac{y_2-y_1}{x_2-x_1}$

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

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

$=\frac{1.75}{0.5}$

$=\frac{7}{2}$

$=3.5$

Therefore, the slope of the secant line $PQ$ is $m=3.5$