1. **State the problem:** We have an account with an initial investment of $1 that pays 150% interest per month but also charges a fixed fee of 3 per month. We want to find the value of the account $y$ after $x$ months for $x = -2, -1, 0, 1, 2$. 2. **Write the formula:** The account grows by 150% per month, meaning it multiplies by $1 + 1.5 = 2.5$ each month, but we subtract 3 each month as a fee. The recursive formula is: $$y_{x} = 2.5 y_{x-1} - 3$$ with initial value $y_0 = 1$. 3. **Calculate values step-by-step:** - For $x=0$, $y_0 = 1$ (given). - For $x=1$: $$y_1 = 2.5 \times y_0 - 3 = 2.5 \times 1 - 3 = 2.5 - 3 = -0.5$$ - For $x=2$: $$y_2 = 2.5 \times y_1 - 3 = 2.5 \times (-0.5) - 3 = -1.25 - 3 = -4.25$$ 4. **Calculate for negative $x$ (going backwards):** We solve for $y_{x-1}$: $$y_x = 2.5 y_{x-1} - 3 \implies y_{x-1} = \frac{y_x + 3}{2.5}$$ - For $x=0$, $y_0=1$ (given). - For $x=-1$: $$y_{-1} = \frac{y_0 + 3}{2.5} = \frac{1 + 3}{2.5} = \frac{4}{2.5} = 1.6$$ - For $x=-2$: $$y_{-2} = \frac{y_{-1} + 3}{2.5} = \frac{1.6 + 3}{2.5} = \frac{4.6}{2.5} = 1.84$$ 5. **Complete the table:** | x | y | |----|------| | -2 | 1.84 | | -1 | 1.6 | | 0 | 1 | | 1 | -0.5 | | 2 | -4.25| 6. **Type of function:** Because the account value decreases after $x=0$ and becomes negative, this is not a pure exponential growth or decay. The fixed fee causes the value to decrease despite the growth factor. So it is neither pure growth nor decay exponential. 7. **Asymptote:** Find steady state $y$ where $y = 2.5 y - 3$: $$y = 2.5 y - 3 \implies 3 = 2.5 y - y = 1.5 y \implies y = \frac{3}{1.5} = 2$$ So the horizontal asymptote is $y=2$. 8. **Y-intercept:** At $x=0$, $y=1$. 9. **Graph description:** The graph crosses $y=1$ at $x=0$, rises to about 1.6 and 1.84 for negative $x$, and then falls below zero for positive $x$. The horizontal asymptote is $y=2$.