calculus
posted by Laurie .
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.
y3 = (xpi) = pix
y = x + 3+pi
The slope of the normal line is (1/1) = 1
y3 = (xpi)
y = x + 3pi