1. **State the problem:** We need to model two functions with tables of values and graphs. 2. **Functions given:** - $f(x) = -5x + 5$ - $f(x) = -8x^2 - 9$ 3. **Create tables of values:** Choose values for $x$ and calculate corresponding $f(x)$. 4. **For $f(x) = -5x + 5$:** - When $x = -2$, $f(-2) = -5(-2) + 5 = 10 + 5 = 15$ - When $x = -1$, $f(-1) = -5(-1) + 5 = 5 + 5 = 10$ - When $x = 0$, $f(0) = -5(0) + 5 = 0 + 5 = 5$ - When $x = 1$, $f(1) = -5(1) + 5 = -5 + 5 = 0$ - When $x = 2$, $f(2) = -5(2) + 5 = -10 + 5 = -5$ 5. **Table for $f(x) = -5x + 5$:** | $x$ | $f(x)$ | |-----|--------| | -2 | 15 | | -1 | 10 | | 0 | 5 | | 1 | 0 | | 2 | -5 | 6. **For $f(x) = -8x^2 - 9$:** - When $x = -2$, $f(-2) = -8(-2)^2 - 9 = -8(4) - 9 = -32 - 9 = -41$ - When $x = -1$, $f(-1) = -8(1) - 9 = -8 - 9 = -17$ - When $x = 0$, $f(0) = -8(0) - 9 = 0 - 9 = -9$ - When $x = 1$, $f(1) = -8(1) - 9 = -8 - 9 = -17$ - When $x = 2$, $f(2) = -8(4) - 9 = -32 - 9 = -41$ 7. **Table for $f(x) = -8x^2 - 9$:** | $x$ | $f(x)$ | |-----|--------| | -2 | -41 | | -1 | -17 | | 0 | -9 | | 1 | -17 | | 2 | -41 | 8. **Graphing:** - The first function is linear with slope $-5$ and y-intercept $5$. - The second function is a downward-opening parabola with vertex at $(0, -9)$. 9. **Summary:** Tables provide points to plot the graphs accurately. Final answers: - Table and graph for $f(x) = -5x + 5$ - Table and graph for $f(x) = -8x^2 - 9$