how do i find the value of p for which the line y=6 is tangent to curve y=x^2+(1-p)x+2p?

Answer 1

This might help

http://www.swan.ac.uk/education/pgcemaths/mk/tests/core_a1.html

Answer 2

Let y=6 be tgt to the given curve at x1,y1
But for line ,y1=6

Hence,P =(x1,6)
P satisfies eq of curve

hence,
6 =x1 ^ 2 +(1-p)x +2p---------------1

equation of curve is y =xsq +(1-p)x +2p

Tangent to curve is given by dy/dx=0

Hence,2x +1-p =0------------------------2

Solve 2 eq in 2 variables to get X1 and p

I hope this is right.

Answer 3

4p^2=x^2+x

Answer 4

Dude.................... your teacher needs psychiatric help if he gives you that for your homework!