# Thread: Show that the tangent.....

1. ## Show that the tangent.....

Show that the tangent to P: y= ax˛ + bx + c with gradient of m has a y-intercept
C – ( (m – b)˛/4a )

My thoughts is that the firstly I would differentiate to get

y' = 2ax + b
then well I am not sure of the best way to tackle this...
help appreciated

2. I assume gradient is your term for slope. If so, then the derivative you have taken is defined as the slope of the curve at some point x. Thus you can write your derivative as:

$m=2ax+b \Rightarrow \frac{m-b}{2a}$

See if you can take it from there.

3. Originally Posted by Joel
Show that the tangent to P: y= ax˛ + bx + c with gradient of m has a y-intercept
C – ( (m – b)˛/4a )

My thoughts is that the firstly I would differentiate to get

y' = 2ax + b
then well I am not sure of the best way to tackle this...
help appreciated
At any point $(x_1, ax_1^2+bx_1+c)$ the slope of the tangent line is

$m=2ax_1+b$

The equation of the tangent line is

$y-(ax_1^2+bx_1+c)=(2ax_1+b)(x-x_1)$

$y-(ax_1^2+bx_1+c)=(2ax_1+b)x-2ax_1^2-bx_1$

$y=(2ax_1+b)x-2ax_1^2-bx_1+ax_1^2+bx_1+c$

$y=mx+c-ax_1^2=mx+c-\frac{4a^2x_1^2}{4a}$

$=mx+c-\frac{(2ax_1+b-b)^2}{4a}$

$y=mx+c-\frac{(m-b)^2}{4a}$