posted by Laurie on .
Find equations for the lines that are tangent and normal to the graph of y = sinx + 3 at x = pi
Now, the slope of the tangent line at any point (x,y) on the curve is just y'(x) = cosx
At x=pi, y' = -1 and y=3
So, we want the line through (pi,3) with slope -1.
y-3 = -(x-pi) = pi-x
y = -x + 3+pi
The slope of the normal line is -(1/-1) = 1
y-3 = (x-pi)
y = x + 3-pi