Ok, I've found the answer to these questions, but I did so in more of a trial and error way. The question is: f(x) = 2x^2 + x Find the two tangent lines to the curve, which both pass through the point (2,-3). So, I tried using y=mx+b = 2x^2 + x, where m = f'(x) = 2x + 1, thus: (2x+1)x + b = 2x^2 + x. I worked this out to find an extraneous answer, and one correct tangent of y = -x - 1. I know the other answer has a slope of 11, through trial and error, but there's got to be an easier way. Any thoughts on how to do this algebraically?