# derivative application

• Apr 25th 2009, 07:37 PM
slaypullingcat
derivative application
Show that the tangent to P: y= ax^2+bx+c with gradient m has y intercept c-(m-b)^2/4a. thank you.
• Apr 25th 2009, 07:44 PM
The Second Solution
Quote:

Originally Posted by slaypullingcat
Show that the tangent to P: y= ax^2+bx+c with gradient m has y intercept c-(m-b)^2/4a. thank you.

m = 2ax + b => x = ....

Substitute that expression for x into y= ax^2+bx+c to get an expression for y. Now you know a point (x1, y1) on the tangent.

Then the tangent is y - y1 = m(x - x1).

Now substitute x = 0 and solve for y.
• Apr 25th 2009, 08:07 PM
Soroban
Hello, slaypullingcat!

The procedure is simple, but there's a LOT of algebra!
I was lucky to get the answer on the first try . . .

Quote:

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: .$\displaystyle y - \tfrac{m^2-b^2+4ac}{4a} \;=\;m\left(x - \tfrac{m-b}{2a}\right)$

. . $\displaystyle \text{which simplifies to: }\;y \;=\;mx + \underbrace{c - \tfrac{(m-b)^2}{4a}}_{y\text{-intercept}}$

• Apr 25th 2009, 08:31 PM
slaypullingcat
That was a quality post, thank you very much. I understand it now.