Question: Determine the equation of the graph
Graph: increasing S-shaped curve with x-intercept at $(-1, 0)$ and marked point at $(2, 5)$, centered in the lower-middle of the page. position_hint: center
Determine the equation for the following graph
1. **State the problem:** We need to find the equation of an increasing S-shaped curve (sigmoid-like) with an x-intercept at $(-1, 0)$ and passing through the point $(2, 5)$.
2. **Identify the type of curve:** An S-shaped curve is often modeled by a cubic function or a logistic function. Since the curve passes through $(-1,0)$ and $(2,5)$ and is increasing, a cubic polynomial is a good candidate:
$$y = ax^3 + bx^2 + cx + d$$
3. **Use the x-intercept:** At $x = -1$, $y = 0$, so:
$$0 = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d$$
4. **Use the point $(2,5)$:**
$$5 = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d$$
5. **Assume the curve crosses the x-axis only once at $x=-1$ and is increasing, so the root at $x=-1$ is simple. For simplicity, assume $d=0$ (since $y=0$ at $x=-1$ and the curve passes through origin shifted by $-1$). Alternatively, we can express $d$ in terms of $a,b,c$ from step 3:**
$$d = a - b + c$$
6. **Substitute $d$ into the equation from step 4:**
$$5 = 8a + 4b + 2c + (a - b + c) = 9a + 3b + 3c$$
7. **We have one equation with three unknowns. To find a unique solution, we need more conditions. Since the curve is S-shaped and increasing, the derivative changes sign once. Let's impose the derivative at $x=-1$ is positive (increasing):**
$$y' = 3ax^2 + 2bx + c$$
At $x=-1$:
$$y'(-1) = 3a - 2b + c > 0$$
8. **For simplicity, choose $a=1$ to reduce variables:**
From step 6:
$$5 = 9(1) + 3b + 3c \\ 5 = 9 + 3b + 3c \\ 3b + 3c = 5 - 9 = -4 \\ b + c = -\frac{4}{3}$$
From step 7:
$$y'(-1) = 3(1) - 2b + c > 0 \\ 3 - 2b + c > 0$$
Substitute $c = -\frac{4}{3} - b$:
$$3 - 2b - \frac{4}{3} - b > 0 \\ 3 - 3b - \frac{4}{3} > 0 \\ \frac{9}{3} - 3b - \frac{4}{3} > 0 \\ \frac{5}{3} - 3b > 0 \\ -3b > -\frac{5}{3} \\ b < \frac{5}{9}$$
9. **Choose $b=0$ (a simple value less than $\frac{5}{9}$), then $c = -\frac{4}{3} - 0 = -\frac{4}{3}$ and $d = a - b + c = 1 - 0 - \frac{4}{3} = -\frac{1}{3}$**
10. **Final equation:**
$$y = x^3 + 0 \cdot x^2 - \frac{4}{3}x - \frac{1}{3} = x^3 - \frac{4}{3}x - \frac{1}{3}$$
11. **Check points:**
At $x=-1$:
$$y = (-1)^3 - \frac{4}{3}(-1) - \frac{1}{3} = -1 + \frac{4}{3} - \frac{1}{3} = 0$$
At $x=2$:
$$y = 8 - \frac{8}{3} - \frac{1}{3} = 8 - 3 = 5$$
This matches the given points.
**Answer:** The equation of the graph is
$$y = x^3 - \frac{4}{3}x - \frac{1}{3}$$