# Show that the Tangent has a y intercept of...

• May 3rd 2010, 04:55 AM
Joel
Show that the Tangent has a y intercept of...
Show that the tangent to P: ax^2 + bx + c with gradient m has a y intercept of c - ( (m - b)^2 / 4a )

OK so far my working would be...

find dydx to find gradient of tangent

dydx = 2ax + b

y intercept means x = 0, so when x =0, 2ax + b = b

From here i am stuck.

Any ideas would be appreciated
• May 3rd 2010, 05:12 AM
calypso
I think that answer is wrong

The tangent to the equation will have an equation in the form y = mx + yin

(where yin= yitnercept, havent called it c as you already have c defined)

therefore tangent equation is y = (2ax+b)x + yin

you also have an equation y = ax^2 + bx + c

you know these intercept so you can equate them, ie.

ax^2 + bx + c = (2ax +b)x + yin

then solve for yin

Calypso
• May 3rd 2010, 05:30 AM
Joel
Thanks Calypso

so how does this look...

The tangent to the equation will have an equation in the form y = mx + yin

(where yin= yitnercept, havent called it c as you already have c defined)

therefore tangent equation is y = (2ax+b)x + yin

you also have an equation y = ax^2 + bx + c

you know these intercept so you can equate them, ie.

ax^2 + bx + c = (2ax +b)x + yin

ax^2 + bx + c - (2ax +b)x = yin

ax^2 + bx + c - 2ax^2 - bx = yin

- ax^2 + c = yin

???
• May 3rd 2010, 06:27 AM
calypso
Yes looks correct. However its strange that it is so different to what was stated as the correct answer. Where did you get that answer from of (c - ( (m - b)^2 / 4a ))