Let f(x)=x^2-2x+2,the straight line L intersects the graph y=f(x) at M and N.The slope and the y-intercept of L are m and 1 respectively. 1)

We can write L as

y = mx + 1

So, let us find the intersection of these two functions

mx + 1 = x^2 - 2x + 2

X^2 - 2x + 2 - mx - 1 = 0

x^2 -(2+m)x + 1 = 0

So, using the quadratic formula, we have

x = [( 2 + m) ± √[(2+m)^2 - 4]/2 = [(2 +m) ±√(m^2 + 2m)] /2

So, the y coordinates of M, N lie on line L and are given by

y₁ = m[(2+m) + √(m^2 + 2m)] /2 + 1    and

y₂ = m[(2+m) - √(m^2 + 2m)] /2 + 1 

Adding these two we have

[m[(2+m) + √(m^2 + 2m)] /2 + 1 ] + [ m[(2+m) - √(m^2 + 2m)] /2 + 1 ] =

[m(2 + m) + 2]

And dividing this by 2, we get the midpoint =

(m/2)(2 + m) + 1 =

(1/2)m^2 + m + 1

That looked like a lot of work Chris :)

Yeah.....it seemed like too much.....I'm a little suspicious about my answer.....there's probably a more direct way to do it.......but my brain is too fried to want to think of another way!!!