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

1. ## 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

2. 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

3. 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

???

4. 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 ))