The graph of $f(x)=\frac{(6-4 x)^2}{2 x^2-5 x+12}$ has a horizontal asymptote at $y=$
The function given is \(f(x) = \frac{(6 - 4x)^2}{2x^2 - 5x + 12}\). To find the horizontal asymptote, we need to evaluate the limit of the function as \(x\) approaches positive infinity and negative infinity. This is determined by the behavior of the leading terms of the numerator and the denominator.
First, we expand the numerator \((6 - 4x)^2\):
\[(6 - 4x)^2 = 6^2 - 2(6)(4x) + (4x)^2 = 36 - 48x + 16x^2\]
So, the function can be written as:
\[f(x) = \frac{16x^2 - 48x + 36}{2x^2 - 5x + 12}\]
To evaluate the limit as \(x \to \pm \infty\), we divide every term in the numerator and the denominator by the highest power of \(x\) present, which is \(x^2\):
\[f(x) = \frac{\frac{16x^2}{x^2} - \frac{48x}{x^2} + \frac{36}{x^2}}{\frac{2x^2}{x^2} - \frac{5x}{x^2} + \frac{12}{x^2}} = \frac{16 - \frac{48}{x} + \frac{36}{x^2}}{2 - \frac{5}{x} + \frac{12}{x^2}}\]
Now, we evaluate the limit as \(x \to \infty\) and \(x \to -\infty\). As \(x\) becomes very large, the terms \(\frac{48}{x}\), \(\frac{36}{x^2}\), \(\frac{5}{x}\), and \(\frac{12}{x^2}\) all approach \(0\):
\[\lim_{x \to \pm \infty} f(x) = \frac{16 - 0 + 0}{2 - 0 + 0} = \frac{16}{2} = 8\]
Since the limit is \(8\), the horizontal asymptote of the function is \(y = 8\).
Understanding horizontal asymptotes is essential in engineering and economics. For example, if you are modeling the long-term efficiency of a manufacturing process where the cost per unit \(C(x)\) changes as production quantity \(x\) increases, finding the horizontal asymptote tells you the production cost the system will eventually stabilize at as production scale becomes very large. This helps managers forecast long-term expenses accurately and make informed decisions about scaling operations.