For 1) and 2), Find the equation for the tangent line to the graph of the function . I got another one right in point-slope but these, i am pretty lost on. 1) f(x)= s^3 at x=1/2 ans.3x-4y+1=0. 2)f(x)= 1/(x+3) at x=2. ans. x+25y-7=0 This one is for the normal line. 3) f(x)=1/(x+3) at x=3. ans 216x-6y-647=0 On a side note, I do know the equation for derivatives (f(x+change of x)-f(x))/change of x and the point-slope form is applied somewhere. I just don't know how to get these to that form.