Sorry, but the answer is wrong. Thank you for trying though! This is the explanation for the correct answer:
The vertical asymptote of f(x) is x=2. Hence, d=2.
By long division,f(x)=12x−2−22x−4.
Thus, the oblique asymptote of f(x) is y=12x−2, which passes through (0,−2). Therefore, the oblique asymptote of g(x) is y=−2x−2.
Therefore, g(x)=−2x−2+kx−2 for some constant k.
Finally, f(−2)=(−2)2−6(−2)+62(−6)−4=−114, so g(−2)=−2(−2)−2+k−2−2=−114.
Solving, we find k=19. Hence, g(x)=−2x−2+19x−2=−2x2+2x+23x−2.
We want to solve x2−6x+62x−4=−2x2+2x+23x−2.
Then x2−6x+6=−4x2+4x+46, or 5x2−10x−40=0. This factors as 5(x+2)(x−4)=0, so the other point of intersection occurs at x=4. Since f(4)=42−6⋅4+62(4)−4=−12, the other point of intersection is (4,−12).
Still, thank you for trying, Guest!
