Question

the reciprocal of the quadratic function ( f(x) = 20x^{2}+107 ) has the following behaviour near the horizontal asymptote: ( x
ightarrowpminfty,y
ightarrow0^{+} ).
true
false

Explanation:

Step1: Find the reciprocal function

The reciprocal of \(f(x)=20x^{2}+107\) is \(y = \frac{1}{20x^{2}+107}\).

Step2: Analyze the limit as \(x\to\pm\infty\)

As \(x\to\pm\infty\), \(x^{2}\to+\infty\). Then \(20x^{2}+107\to+\infty\).
For \(y=\frac{1}{20x^{2}+107}\), when \(20x^{2}+107\to+\infty\), \(\lim_{x\to\pm\infty}\frac{1}{20x^{2}+107}=0\).
Since \(20x^{2}+107>0\) for all real \(x\) (because \(x^{2}\geq0\) and \(20>0\), \(107 > 0\)), \(\frac{1}{20x^{2}+107}>0\) for all real \(x\). So as \(x\to\pm\infty\), \(y\to0^{+}\).

Answer:

True