f'(x)=2ax+b,∴f'(3)=6a+b=5...①

∫y (when x=3) =5x-8=5*3-8=7, ∴ f (3) = 9a+3b+ 1 = 7. ...

①×3-②:9a=9

∴ a= 1，b=5-6a=- 1

( 1)f(x)=x？ -x+ 1

(2) let g (x) = ke x, ∫f(x)= x? -x+ 1=[x-( 1/2)]？ +(3/4), that is, the image of f(x) is: the opening is upward and the vertex is (1/2,3/4).

And ∵ g' (x) = ke x,

When k=0, g(x)=0 that does not intersect with f(x) is irrelevant;

When k < 0, g(x) decreases monotonously; When k>0, g(x) monotonically increases;

..... I can't figure out what to do next for the time being.