# Thread: slope of secant line

1. ## slope of secant line

Consider function y=f(x)=sqrt of x

The slope of secant line connecting the points (4,2) and (9,3) is?

I get m=x2-x1/y2-y1 = 1/5

Now from that I have to come up with equation of the secant line, and the slope of the tangent line to the curve at point (4,2)... can you guys shed a little light on this? Thanks in advance!

2. Originally Posted by boousaf
Consider function y=f(x)=sqrt of x

The slope of secant line connecting the points (4,2) and (9,3) is?

I get m=x2-x1/y2-y1 = 1/5

Now from that I have to come up with equation of the secant line, and the slope of the tangent line to the curve at point (4,2)... can you guys shed a little light on this? Thanks in advance!
the equation of a line with slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$ is given by the point-slope form:
$\displaystyle y - y_1 = m(x - x_1)$

for the second part of the question, find $\displaystyle f'(4)$, that will give you the slope of the tangent line at $\displaystyle x = 4$