The graph of the function \(y=f(x)\) is shown below. For all \(x > 4\), it is true that \(f(x) > 0.4\). If \(f(x) = \frac{x^2}{Ax^2 + Bx + C}\), where \(A,B\) and \(C\) are integers, then find \(A+B+C\).
Goddamn you ROM! If you leave, the forum’s mean IQ will drop by 30 bloody points.
Just take a bleedin’ break, and visit when you have a mind too.
Take your boat to Gilligan’s Island; after a few weeks there this place won’t seem so annoying.
I’m NOT the only one who feels this way!
GA
For the past 5 hours, I have pondered why Rom posted his “goodbye” by editing and replacing a fine teaching post.
I didn’t save a copy, but the post described how to correlate the graphical locations of the asymptotes to an equation. This was an excellent, wonderful post –not a bloody thing a matter with it. Now it’s gone. Replaced with a “goodbye...”
This post was chosen deliberately (it wasn’t his latest post). I know why too...
.
Because it appears that (0,0) is on the graph, the numerator has the form Dx^2
Because we have vertical asymptotes at x = -2 and x = 3.....the denominator has the form
A (x + 2) (x - 3)
And the line y = 0.4 = 2/5 is an asymptote for x > 4......so we might guess that the ratio of D/ A = 2/5
So....it appears that we have
2x^2 2x^2 2x^2
f(x) = ______________ = ______________ = _________________
5 ( x + 2) (x - 3) 5 (x^2 - x - 6) 5x^2 - 5x - 30
So A + B + C = 5 - 5 - 30 = -30
Note that when x < -6, the graph will cross the asymptote of y = 2/5
Here's a graph : https://www.desmos.com/calculator/7i9wnrvuuc