1. **State the problem:** Given the function $y=\frac{4}{x^2+x}$ and the corresponding table of $x$ and $y$ values: $x:$ -2, -1.5, -1.2, -1, -0.8, 1, 1.5, 2, 3, 4 $y:$ -1, 0.3, L, m, 5.5, n, 3.3, 3, p, 4.3 We need to find the missing values $L$, $m$, $n$, and $p$. 2. **Calculate each missing $y$ value by substituting the given $x$: ** $$y = \frac{4}{x^2 + x}$$ - For $x = -1.2$ (find $L$): $$L = \frac{4}{(-1.2)^2 + (-1.2)} = \frac{4}{1.44 - 1.2} = \frac{4}{0.24} = 16.67$$ - For $x = -1$ (find $m$): $$m = \frac{4}{(-1)^2 + (-1)} = \frac{4}{1 - 1} = \frac{4}{0}$$ This is undefined because division by zero is not possible. So, \textbf{$m$ is undefined}. - For $x = 1$ (find $n$): $$n = \frac{4}{1^2 + 1} = \frac{4}{1 + 1} = \frac{4}{2} = 2$$ - For $x = 3$ (find $p$): $$p = \frac{4}{3^2 + 3} = \frac{4}{9 + 3} = \frac{4}{12} = 0.333$$ 3. **Summary of missing values:** - $L = 16.67$ - $m$ is undefined (division by zero) - $n = 2$ - $p = 0.333$ Hence, the values of $L$, $m$, $n$, and $p$ in the table are respectively: $16.67$, undefined, $2$, and $0.333$