Re: tricky calculus problem

Re: tricky calculus problem

big thanks earboth! I figured out points 1, 2, 3, and 4.. but I was writing the equation of the line QP as: y-0 = m(x-18), using the point-slope formula and i was substituting m'=-1/2a into m. So I actually had the equation of the line all I was missing was the slope... What I was missing was the other formula for m that you have reminded me of which was (y2-y1)/(x2-x1).. which when equated with the derived m of the line let's us solve for a. But this still doesn't explain what the book was talking about in it's "suggestion"... anyone here who gets it?

Re: tricky calculus problem

Re: tricky calculus problem

I see now, the book is actually trying to explain how to solve for the roots of a cubic equation. I was wondering why it had to say "integral root" rather than just "root" because that just confused me. Also when it said "find the latter" i thought it meant 'find the other roots' but actually it meant "the quadratic polynomial". it makes sense that the roots of the quadratic portion should be imaginary since there is only one possible point that fits the picture..

the hardest part was knowing that you must equate the f'(a) to (y2-y1)/(x2-x1) formula for m, which is the hint i think the book should have given..