Question

End Behavior Model Consider the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ Show that (a) \(y=\pm \frac{b}{a} \sqrt{x^{2}-a^{2}}\) (b) \(g(x)=(b / a)|x|\) is an end behavior model for $$f(x)=(b / a) \sqrt{x^{2}-a^{2}}$$ (c) \(g(x)=-(b / a)|x|\) is an end behavior model for $$f(x)=-(b / a) \sqrt{x^{2}-a^{2}}$$

Short answer

In the first part, it was shown that \(y=\pm \frac{b}{a} \sqrt{x^{2}-a^{2}}\) can be derived from the given hyperbola’s formula. The function \(g(x)=(b / a)|x|\) is behaving as an end model for \(f(x)=(b / a) \sqrt{x^{2}-a^{2}}\) when \(x >> a\) or \(x << -a\). Similarly, \(g(x)=-(b / a)|x|\) is an end behavior model for \(f(x)=-(b / a) \sqrt{x^{2}-a^{2}}\) at the same large absolute \(x\) values.

Step-by-step solution

Derive \(\pm \frac{b}{a} \sqrt{x^{2}-a^{2}}\) from the hyperbola equation

Start from the equation of hyperbola given, \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\). Rearranging for \(y^2\) gives \(y^2=b^2\left(\frac{x^2}{a^2}-1\right)\) or \(y^{2}=b^{2}\left(\frac{x^{2}-a^{2}}{a^{2}}\right)\). Finally, take the square root to get \\(y=\pm \frac{b}{a} \sqrt{x^{2}-a^{2}}\). This matches the equation from part (a).

Check that \(g(x)=(b / a)|x|\) is an end behavior model for \(f(x)=(b / a) \sqrt{x^{2}-a^{2}}\)

Analyzing the formula for \(f(x)\), when \(x >> a\) (or \(x << -a\)), the \(a^{2}\) inside the square root can be ignored relative to \(x^{2}\). Then, \(f(x)\) simplifies to \(f(x)=(b / a)|x|\), which is the formula for \(g(x)\) given in part (b). So, \(g(x)=(b / a) |x|\) is indeed behaving as an end behavior model for \(f(x)=(b / a) \sqrt{x^{2}-a^{2}}\).

Check that \(g(x)=-(b / a)|x|\) is an end behavior model for \(f(x)=-(b / a) \sqrt{x^{2}-a^{2}}\)

Using a similar approach as in step 2, but now with the negative versions of the functions, it can be seen that \(g(x)=-(b / a)|x|\) behaves like \(f(x)=-(b / a) \sqrt{x^{2}-a^{2}}\) when \(x >> a\) (or \(x << -a\)), thus functioning as an end behavior model. The \(a^{2}\) inside the square root again becomes negligible compared to \(x^{2}\) for larger absolute \(x\) values.