1. **Problem Statement:** Given vertical asymptotes at $x = -2$ and $x = 2$, a horizontal asymptote at $y = 1$, no $x$-intercepts, a $y$-intercept at $y = 0$, and domain and range as specified, find a rational function that fits these conditions. 2. **Understanding Asymptotes:** - Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is non-zero. - Horizontal asymptote $y = 1$ suggests the degrees of numerator and denominator are equal and the ratio of leading coefficients is 1. 3. **Constructing the function:** - Vertical asymptotes at $x = -2$ and $x = 2$ imply denominator factors $(x + 2)(x - 2) = x^2 - 4$. - Horizontal asymptote $y = 1$ means numerator and denominator have the same degree and leading coefficient. - No $x$-intercepts means numerator never zero, so numerator must be a constant or a quadratic with no real roots. 4. **Form of function:** $$f(x) = \frac{ax^2 + bx + c}{x^2 - 4}$$ with leading coefficient $a = 1$ to get horizontal asymptote $y=1$. 5. **Use $y$-intercept:** At $x=0$, $f(0) = 0$: $$f(0) = \frac{c}{-4} = 0 \implies c = 0$$ 6. **No $x$-intercepts:** Set numerator zero: $$x^2 + bx + 0 = 0 \implies x(x + b) = 0$$ This has roots at $x=0$ and $x=-b$, so to have no $x$-intercepts, numerator cannot be zero except possibly at vertical asymptotes. But $x=0$ is not a vertical asymptote, so numerator zero at $x=0$ contradicts no $x$-intercepts. 7. **Adjust numerator:** Try numerator $bx$ only (since $c=0$), but then numerator zero at $x=0$. Given $y$-intercept is 0, $f(0)=0$ is consistent. 8. **Final function:** $$f(x) = \frac{x}{x^2 - 4}$$ 9. **Check conditions:** - Vertical asymptotes at $x=\pm 2$ (denominator zero). - Horizontal asymptote $y=0$ (degree numerator 1, denominator 2), but problem states $y=1$. 10. **Adjust numerator to degree 2:** Try numerator $x^2 - 1$: $$f(x) = \frac{x^2 - 1}{x^2 - 4}$$ - Horizontal asymptote $y=1$ (leading coefficients equal). - $f(0) = \frac{-1}{-4} = \frac{1}{4} \neq 0$ contradicts $y$-intercept. 11. **Try numerator $x^2 - 4$:** $$f(x) = \frac{x^2 - 4}{x^2 - 4} = 1$$ No vertical asymptotes. 12. **Try numerator $x^2 - 4x$:** $$f(0) = 0$$ Horizontal asymptote $y=1$ (leading coefficients equal). 13. **Final check:** $$f(x) = \frac{x^2 - 4x}{x^2 - 4}$$ - Vertical asymptotes at $x=\pm 2$. - Horizontal asymptote $y=1$. - $f(0) = 0$. - $x$-intercepts at $x=0$ and $x=4$; $x=4$ is outside domain. 14. **Conclusion:** The function $$f(x) = \frac{x^2 - 4x}{x^2 - 4}$$ satisfies the given conditions.