1. **Problem Statement:** Sketch the graph with the given asymptotes and points. 2. **Given information:** - Horizontal asymptote: $y=1$ - Vertical asymptotes: $x=-2$ and $x=2$ - $y$-intercept: $(0,0)$ - $x$-intercept: $(0,0)$ - Symmetric to the $y$-axis - Passes through the point $(4,4)$ 3. **Analyze the asymptotes and symmetry:** - The vertical asymptotes at $x=-2$ and $x=2$ suggest the function is undefined there and the function has factors causing division by zero at these points. - Symmetry about the $y$-axis means the function $f(x)$ satisfies $f(-x)=f(x)$. 4. **Investigate the function form:** - A rational function with vertical asymptotes at $x=\pm2$ has denominator factors $(x-2)(x+2)=x^2-4$. - Since the function is symmetric about $y$-axis, it depends on $x^2$ terms. 5. **Horizontal asymptote $y=1$ and point $(4,4)$:** - The horizontal asymptote $y=1$ implies numerator and denominator have the same degree and leading coefficients ratio is 1. - Assume function is $f(x) = \frac{ax^2 + b}{x^2 - 4}$. - As $x \to \pm \infty$, $f(x) \to a$ because leading terms dominate, so $a=1$. 6. **Use $y$-intercept at $(0,0)$:** - $f(0) = 0 = \frac{a\cdot0 + b}{0 - 4} = \frac{b}{ -4 }$ implies $b=0$. 7. **Use point $(4,4)$ to find consistency:** - $f(4) = 4 = \frac{1\cdot16 + 0}{16 - 4} = \frac{16}{12} = \frac{4}{3}$ - This does not equal 4, so try adding a linear numerator term zeroed by symmetry. 8. **Consider numerator $x^2 + c$:** - $f(x) = \frac{x^2 + c}{x^2 - 4}$ - At $x=0$: $0 = \frac{0 + c}{-4} \Rightarrow c=0$ - At $x=4$: $4 = f(4) = \frac{16 + 0}{16 - 4} = \frac{16}{12} = \frac{4}{3}$ mismatch again. 9. **Try adding a constant term in numerator:** - Let numerator be $kx^2 + m$. Since $f(0)=0$, $m=0$. - Horizontal asymptote $= \frac{k}{1} = 1 \Rightarrow k=1$. - At $x=4$: $f(4) = \frac{1\cdot16 + 0}{16 - 4} = \frac{16}{12} = \frac{4}{3}$ mismatch. 10. **Add a multiplicative factor:** - Suggest function $f(x) = \frac{A x^2}{x^2 - 4}$. - Horizontal asymptote is $A$; set $A=1$. - At $x=0$: $f(0)=0$ correct. - At $x=4$: $f(4)= \frac{16}{12} = \frac{4}{3}$ less than 4. 11. **Adjust with vertical stretch:** - We need $f(4)=4$ so let $f(x) = B \frac{x^2}{x^2 - 4}$. - At $x=4: 4 = B \frac{16}{12} = B \frac{4}{3} \Rightarrow B = 3$. - Horizontal asymptote = $\lim_{x\to \infty} f(x) = B = 3$ not 1 as required. 12. **Conclusion:** Given conditions conflict for simple rationals with vertical asymptotes at $\pm2$, intercept at zero, symmetry, and passing point $(4,4)$ with horizontal asymptote $y=1$. 13. **Sketch instructions:** - Draw vertical dashed lines at $x=-2$, $x=2$ for vertical asymptotes. - Draw a horizontal dashed line at $y=1$ for horizontal asymptote. - Mark the point $(0,0)$ for both intercepts. - Mark point $(4,4)$. - Indicate symmetry about $y$-axis. **Final answer:** The function cannot be uniquely determined from the data but the asymptotes and points for sketching are correctly identified.