# Math Help - Secant Line

1. ## Secant Line

I would just like to make sure I'm doing this right..thanks
let f(x)=x^2 x=-2, and x=1.9
f(-1.9)^2=-3.61
f(-2)^2=-4
(-3.61)-(-4)
(-1.9)-(-4)
=.39
2.1
=0.185714285

2. Originally Posted by lisa1984wilson
I would just like to make sure I'm doing this right..thanks
let f(x)=x^2 x=-2, and x=1.9 --------------------------> [is it x = 1.9 or x = -1.9 ? ]
f(-1.9)^2=-3.61
f(-2)^2=-4
(-3.61)-(-4)
(-1.9)-(-4)
=.39
2.1
=0.185714285
Is it x = 1.9 or x = -1.9 ?
I assume that it is x = -1.9

$f(-1.9) = (-1.9)^2 = (-1.9)\times (-1.9) = +3.61$

$f(-2)=(-2)^2=+4$

$slope = \frac{3.61-4}{-1.9-(-2)}=\frac{-0.39}{-1.9+2}=\frac{-0.39}{0.1}=-3.9$

3. ## Tangent

I need to compute the slope of the line that is tangent to the graph when x=-2 and compare with the slope of the secant in the first part! Thanks

4. Originally Posted by lisa1984wilson
I need to compute the slope of the line that is tangent to the graph when x=-2 and compare with the slope of the secant in the first part! Thanks
the slope of tangent line to the curve $f(x) = x^2$ at x= -2 is -4

You already have found the slope of secant in first part.