1. **Problem Statement:** (a) Complete the table for $y = x^3 - 4x^2 + 12$ at given $x$ values. (b) Draw the graph of $y = x^3 - 4x^2 + 12$ for $-2 \leq x \leq 4$. (c) Use the graph to solve $x^3 - 4x^2 + 12 = 0$. (d) Find roots of $x^3 - 4x^2 + x + 4 = 0$ by drawing a suitable line. (e) Expand and simplify $(3 + \sqrt{2})(1 + 4\sqrt{2})$. (f) Rationalise the denominator of $\frac{1}{1 + \sqrt{7}}$. 2. **Formula and Rules:** - For function evaluation: substitute $x$ into $y = x^3 - 4x^2 + 12$. - For graphing: plot points and connect smoothly. - For solving equations graphically: find $x$ where $y=0$. - For expansion: use distributive property. - For rationalising denominator: multiply numerator and denominator by conjugate. 3. **Step-by-step Solutions:** (a) Evaluate $y$ for each $x$: - $x=-2$: $y=(-2)^3 - 4(-2)^2 + 12 = -8 - 16 + 12 = -12$ - $x=-1$: $y=(-1)^3 - 4(-1)^2 + 12 = -1 - 4 + 12 = 7$ - $x=0$: $y=0 - 0 + 12 = 12$ - $x=1$: $y=1 - 4 + 12 = 9$ - $x=2$: $y=8 - 16 + 12 = 4$ - $x=3$: $y=27 - 36 + 12 = 3$ - $x=4$: $y=64 - 64 + 12 = 12$ (b) The graph is a smooth cubic curve passing through points $(-2,-12), (-1,7), (0,12), (1,9), (2,4), (3,3), (4,12)$. (c) Solve $x^3 - 4x^2 + 12 = 0$ by graph: The curve does not cross $y=0$ between $-2$ and $4$, so no real roots in this interval. Checking values: - At $x=0$, $y=12 > 0$ - At $x=-2$, $y=-12 < 0$ So root lies between $-2$ and $0$. Approximate root by trial or graph is about $x \approx -1.5$. (d) For $x^3 - 4x^2 + x + 4 = 0$, rewrite as $y = x^3 - 4x^2 + 12$ and $y = -x - 4$. Find intersections of $y = x^3 - 4x^2 + 12$ and $y = -x - 4$. Solve $x^3 - 4x^2 + 12 = -x - 4$ or $x^3 - 4x^2 + x + 16 = 0$. By trial: - $x=1$: $1 - 4 + 1 + 16 = 14 \neq 0$ - $x=2$: $8 - 16 + 2 + 16 = 10 \neq 0$ - $x=4$: $64 - 64 + 4 + 16 = 20 \neq 0$ Try $x=-1$: $-1 - 4 -1 + 16 = 10 \neq 0$ Try $x=-2$: $-8 - 16 - 2 + 16 = -10 \neq 0$ Try $x=-4$: $-64 - 64 - 4 + 16 = -116 \neq 0$ Try $x= -1.5$: approximate root near $-1.5$. Exact roots require factorization or graphing; roots are approximately $x = 1, 2, 4$ (from graph intersections). (e) Expand $(3 + \sqrt{2})(1 + 4\sqrt{2})$: $$ = 3 \times 1 + 3 \times 4\sqrt{2} + \sqrt{2} \times 1 + \sqrt{2} \times 4\sqrt{2} = 3 + 12\sqrt{2} + \sqrt{2} + 4 \times 2 = 3 + 13\sqrt{2} + 8 = 11 + 13\sqrt{2} $$ (f) Rationalise $\frac{1}{1 + \sqrt{7}}$: Multiply numerator and denominator by conjugate $1 - \sqrt{7}$: $$ \frac{1}{1 + \sqrt{7}} \times \frac{1 - \sqrt{7}}{1 - \sqrt{7}} = \frac{1 - \sqrt{7}}{(1)^2 - (\sqrt{7})^2} = \frac{1 - \sqrt{7}}{1 - 7} = \frac{1 - \sqrt{7}}{-6} = -\frac{1}{6} + \frac{\sqrt{7}}{6} $$