If P = (x, y) = ∞ is on a monic cubic polynomial, then −P is the other finite point of intersection of the curve and the vertical line through P. Show that −P = (x, −$a_{1}x$ − $a_3$ − y). (Hint: This involves solving a quadratic in y. Note that the sum of the roots of a monic quadratic polynomial equals the negative of the coefficient of the linear term.)
Attempt: Plug y = −$a_{1}x$ − $a_3$ − y directly in to get a quadratic in y equal to a cubic in x. Actually to be honest Idk where to start on this problem.
Let $f(x,y)=y^2+a_1xy+a_3y-x^3-a_2x^2-a_4x-a_6$. Suppose $P=(x_0,y_0)$ is s.t. $f(P)=0$. To find $-P$ we need to intersect the vertical line through $P$ with $f$. Such a line has equation $x=x_0$, so of course the $x$-coordinate of $-P$ is $x_0$ again. Now plug this into $f$ and get $f(x_0,y)=y^2+(a_1x_0+a_3)y-x_0^3-a_2x_0^2-a_4x_0-a_6$. Seen as a polynomial in the variable $y$, $f(x_0,y)$ has $2$ roots: one is $y_0$ by hypothesis and the second one is the $y$-coordinate of $-P$, call it $y_0'$. Now the opposite of the sum of the roots of a quadratic polynomial equals the coefficient of the term of degree $1$, namely $-(y_0+y_0')=a_1x_0+a_3$, which is your claim.