1. **State the problem:** We are given the function $y = x^3 - 3x^2 + x$ and asked to: (b) Draw the graph for $-0.75 \leq x \leq 2.75$. (c) Use the graph to find intervals where $y > -1$. (d) Solve $x^3 - 3x^2 + 2x - 1 = 0$ by drawing a line. (e) Estimate the gradient of $y = x^3 - 3x^2 + x$ at $x = -0.25$. 2. **Graphing $y = x^3 - 3x^2 + x$ (b):** Plot the points from the table: $x$: -0.75, -0.5, -0.25, 0, 0.5, 1, 1.5, 2, 2.5, 2.75 $y$: -2.9, -1.4, -0.5, (missing), -0.1, -1, -1.9, (missing), -0.6 Connect these points smoothly to form the cubic curve. 3. **Finding where $y > -1$ (c):** From the table and graph: - At $x = -0.5$, $y = -1.4 < -1$; at $x = -0.25$, $y = -0.5 > -1$. - At $x = 0.5$, $y = -0.1 > -1$; at $x = 1$, $y = -1$. - At $x = 1.5$, $y = -1.9 < -1$; at $x = 2.5$, $y = -0.6 > -1$. So $y > -1$ between approximately $-0.25 < x < 1$ and for $x > 2.25$ (estimated from graph). 4. **Equation of the line for (d)(i):** Rewrite $x^3 - 3x^2 + 2x - 1 = 0$ as $$x^3 - 3x^2 + x = 1 - x$$ The line is therefore: $$y = 1 - x$$ 5. **Solving $x^3 - 3x^2 + 2x - 1 = 0$ graphically (d)(ii):** Draw the line $y = 1 - x$ on the same graph as $y = x^3 - 3x^2 + x$. The solutions are the $x$-values where the graphs intersect. From the graph, intersections occur approximately at: $$x \approx 0.35, 1, 2.65$$ 6. **Estimate gradient at $x = -0.25$ (e):** Gradient is the derivative: $$\frac{dy}{dx} = 3x^2 - 6x + 1$$ Calculate at $x = -0.25$: $$3(-0.25)^2 - 6(-0.25) + 1 = 3(0.0625) + 1.5 + 1 = 0.1875 + 1.5 + 1 = 2.6875$$ Estimate gradient is approximately $2.7$. --- 7. **Problem 5(a): Estimate mean height** Class intervals and frequencies: - 60 < h ≤ 70: 8 - 70 < h ≤ 90: 26 - 90 < h ≤ 100: 35 - 100 < h ≤ 110: 67 - 110 < h ≤ 115: 28 - 115 < h ≤ 125: 16 Calculate midpoints: - 65, 80, 95, 105, 112.5, 120 Calculate weighted sum: $$\sum f \times x = 8\times65 + 26\times80 + 35\times95 + 67\times105 + 28\times112.5 + 16\times120$$ $$= 520 + 2080 + 3325 + 7035 + 3150 + 1920 = 18030$$ Total frequency = 180 Mean estimate: $$\bar{x} = \frac{18030}{180} = 100.17$$ Rounded to 1 decimal place: $$100.2$$ cm 8. **Problem 5(b): Calculate histogram bar heights** Height of bar = frequency / class width Class widths: - 60 < h ≤ 70: width = 10, bar height = 0.4 cm given Calculate bar heights for: - 115 < h ≤ 125: width = 10, frequency = 16 $$\text{height} = \frac{16}{10} = 1.6 \text{ cm}$$ - 110 < h ≤ 115: width = 5, frequency = 28 $$\text{height} = \frac{28}{5} = 5.6 \text{ cm}$$ - 70 < h ≤ 90: width = 20, frequency = 26 $$\text{height} = \frac{26}{20} = 1.3 \text{ cm}$$