f(x)=(dx2+6cx)/x=dx+6c when x<0.
When x=0, f(x)=5, and f(2)=4d+2c+5.
As x→0- (the left of zero), f(x)→6c, so for continuity 6c=5, c=⅚.
Therefore f(2)=4d+5/3+5.
When x→2+ (the right of 2), f(x)→2d-⅚.
For continuity, 2d-⅚=4d+5/3+5, -15/2=2d, d=-15/4.
So f(x)=
{ -15x/4+5, | x<0
{ -15x2/4+5x/6+5 | 0≤x≤2
{ -15x/4-⅚ | x>2
So what does this graph look like? All negative values of x lie on a straight line which reaches the y-axis at y=5. Points then take a ride down the right arm of an inverted parabola, crossing the x-axis and plummeting downwards until the parabola meets a line parallel to the first line at x=-25/3. Points continue indefinitely along this linear track.