1. **Problem statement:** (a) Verify that the cubic function $f(x)$ with roots at $x=-3$, $x=-1$, and $x=2$, and passing through $(0,-6)$ can be written as $f(x) = x^3 + 2x^2 - 5x - 6$. 2. **Formula and rules:** A cubic function with roots $r_1$, $r_2$, and $r_3$ can be expressed as $$f(x) = a(x - r_1)(x - r_2)(x - r_3)$$ where $a$ is a constant. 3. **Step-by-step verification:** - Given roots: $-3$, $-1$, and $2$, so $$f(x) = a(x + 3)(x + 1)(x - 2)$$ - Expand the factors: $$ (x + 3)(x + 1) = x^2 + 4x + 3 $$ - Then, $$ f(x) = a(x^2 + 4x + 3)(x - 2) $$ - Expand further: $$ f(x) = a\left(x^3 - 2x^2 + 4x^2 - 8x + 3x - 6\right) = a(x^3 + 2x^2 - 5x - 6) $$ - Simplify inside parentheses: $$ f(x) = a(x^3 + 2x^2 - 5x - 6) $$ - Use the point $(0, -6)$ to find $a$: $$ f(0) = a(0 + 0 - 0 - 6) = -6a $$ Since $f(0) = -6$, we have $$ -6a = -6 \implies a = 1 $$ 4. **Conclusion:** Therefore, $$ f(x) = x^3 + 2x^2 - 5x - 6 $$ --- 5. **Problem statement:** (b)(i) Find the points of intersection between $f(x) = x^3 + 2x^2 - 5x - 6$ and $g(x) = -2x - 6$ by solving $f(x) = g(x)$. 6. **Set the equations equal:** $$ x^3 + 2x^2 - 5x - 6 = -2x - 6 $$ 7. **Simplify:** $$ x^3 + 2x^2 - 5x - 6 + 2x + 6 = 0 $$ $$ x^3 + 2x^2 - 3x = 0 $$ 8. **Factor the equation:** $$ x(x^2 + 2x - 3) = 0 $$ 9. **Factor the quadratic:** $$ x(x + 3)(x - 1) = 0 $$ 10. **Solve for roots:** $$ x = 0, x = -3, x = 1 $$ 11. **Find corresponding $y$ values using $g(x) = -2x - 6$:** - For $x=0$: $y = -2(0) - 6 = -6$ - For $x=-3$: $y = -2(-3) - 6 = 6 - 6 = 0$ - For $x=1$: $y = -2(1) - 6 = -2 - 6 = -8$ 12. **Points of intersection:** $$ (0, -6), (-3, 0), (1, -8) $$ --- 13. **Problem statement:** (b)(ii) Draw the graph of $g(x) = -2x - 6$ on the diagram (not shown here). 14. **Graph description:** $g(x)$ is a straight line with slope $-2$ and $y$-intercept $-6$. It passes through points $(0, -6)$, $(-3, 0)$, and $(1, -8)$ as found above. --- **Final answers:** - (a) Verified $f(x) = x^3 + 2x^2 - 5x - 6$ - (b)(i) Intersection points: $(0, -6)$, $(-3, 0)$, $(1, -8)$