Show that the tangent to P: y= ax^2+bx+c with gradient m has y intercept c-(m-b)^2/4a. thank you.
Hello, slaypullingcat!
The procedure is simple, but there's a LOT of algebra!
I was lucky to get the answer on the first try . . .
Show that the tangent to: .$\displaystyle y \:=\:ax^2+bx+c$ with gradient $\displaystyle m$
. . has y-intercept: .$\displaystyle c-\frac{(m-b)^2}{4a}$
The derivative of: .$\displaystyle y \:=\:ax^2+bx+c$ [1] .is: .$\displaystyle y' \:=\:2ax + b$
If the gradient is $\displaystyle m\!:\;\;2ax + b \:=\:m \quad\Rightarrow\quad x \:=\:\frac{m-b}{2a}$
Substitute into [1]: .$\displaystyle y \:=\:a\left(\tfrac{m-b}{2a}\right)^2 + b\left(\tfrac{m-b}{2a}\right) + c \quad\Rightarrow\quad y \;=\;\frac{m^2-b^2 + 4ac}{4a}$
The tangent has point $\displaystyle P\left(\tfrac{m-b}{2a},\;\tfrac{m^2-b^2+4ac}{4a}\right)$ . and slope $\displaystyle m.$
Its equation is: .$\displaystyle y - \tfrac{m^2-b^2+4ac}{4a} \;=\;m\left(x - \tfrac{m-b}{2a}\right) $
. . $\displaystyle \text{which simplifies to: }\;y \;=\;mx + \underbrace{c - \tfrac{(m-b)^2}{4a}}_{y\text{-intercept}} $