This was a problem in a math completion I was doing: Find $m−b$ where $y=mx+b$ is the line tangent to the parabola $y=x^2$ at the point $(1,1)$.
After the test (i couldn't get the answer in time), I tried broadening the question by asking what is the line tangent to the equation $y=Ax^2+Bx+C$ at point $(a,\ Aa^2+Ba+C)$ and here's how I solved that.
If the line $y=mx+b$ passes throught the point $(a, Aa^2+Ba+C)$ then $b=Aa^2+a(B-m)+C$
If the line $y=mx+Aa^2+a(B-m)+C$ is tangent to $y=Ax^2+Bx+C$ then $mx+Aa^2+a(B-m)+C=Ax^2+Bx+C$ must have one solution and thus its discriminant must be $0$.
$$mx+Aa^2+a(B-m)+C=Ax^2+Bx+C$$ $$0=Ax^2+x(B-m)-Aa^2-aB+am$$ $${(B-m)^2-4(A)(-Aa^2-aB+am)}=\Delta_2=0$$ $$m^2+M(-2B-4Aa)+B^2+4A^2a^2+4ABa$$ Solving for the slope: $$m=\frac{(2B+4Aa)\pm\sqrt{(2B+4Aa)^2-4(1)(B^2+4A^2a^2+4ABa)}}{2}=B+2Aa$$ So the equation of the tangent line is: $$y=(B+2Aa)x-Aa^2+C$$
This method works fine for quadratics since it has a simple discriminant. I tried to extend this argument for quartics and polynomials in the form $Ax^{2n}+Bx^{2n-1}+...$*. by saying that if it has one real root then it'll be in the form $(rx+q)^{2n}=0$ but that doesn't include the case where you have say $4$ complex roots then a root with a multiplicity of $2n-4$.
Can someone please help me figure out a formula for the line tangent to polynomial $Ax^{2n}+Bx^{2n-1}+...$ at the point $(a, Aa^{2n}+Ba^{2n-1}+...)$
*Has to have an even power for it to have a tangent.
I'm 15 years old so I have no background in higher lever complex maths. I have taken the courses Algebra I and II, Geometry and I have started Pre-Cal this year. Any solution or hint will be greatly appreciated.
Translating $y=Ax^2+Bx+C$ so that $(a,Aa^2+Ba+C)$ goes to the origin: $x\mapsto x+a$ $y\mapsto y+Aa^2+Ba+C$ we get $-2 a A x - A x^2 - B x + y=0$ so taking the tangent cone (keeping only lowest degree terms) we get $-2 a A x - B x + y=0$ or $y=(2Aa+B)x$. Now to translate the tangent back to $(a,Aa^2+Ba+C)$ by $x\mapsto x-a$ $y\mapsto y-(Aa^2+Ba+C)$ we get $a^2 A - 2 a A x - B x - C + y=0$ or $y=(2Aa+B)x-Aa^2+C$ as you say.
Now $y+(Aa^3+Ba^2+Ca+D)-(A(x+a)^3+B(x+a)^2+C(x+a)+D)$ by the same process gives $-3 a^2 A x - 2 a B x - C x + y$ or $y=(3 a^2 A + 2 a B + C)x$ which translates back to $2 a^3 A - 3 a^2 A x + a^2 B - 2 a B x - C x - D + y=0$ or $y=(3 a^2 A + 2 a B + C)x-2 a^3 A - a^2 B + D.$
For the general case let's use $y=a_0 x^n+a_1x^{n-1}+\ldots+a_n.$
$y = (n a^{n-1} a_0 + (n-1) a^{n-2} a_1 + \ldots + 2 a a_2 + a_1)x$ which translates back to $y = (n a^{n-1} a_0 + (n-1) a^{n-2} a_1 + \ldots + 2 a a_2 + a_1)x -(n-1) a^n a_0 - (n-2) a^{n-1} a_1 - \ldots- a^2 a_{n-2} + a_n$