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.