finding the tangent line is equivalent to finding the derivative and one point you our point is x=0 y=arccos(0)=(pi/2) so our point is(0,pi/2) derivative arccos(x)=-1/(sqrt(1-x^2) substituting 0 .the derivative at x=0 is -1 so using the equation of line y-y1=m(x-x1) y-(pi/2)=-1(x-0) y=pi/2-x 2)the horizontal asymptote is the value of the function as it approaches positive and negative infinity = 11/9 for both positive and negative infinity so the equation of the tangent line is y=11/9
