1. **Problem statement:** We have two curves: Curve C defined by $y=\frac{k}{x}$ where $k\neq0$, and Curve Q defined by $y=3x^2 - 2x$. 2. **Part (a): Sketch curve C for $k=4$** - The equation becomes $y=\frac{4}{x}$. - Vertical asymptote: $x=0$ (since division by zero is undefined). - Horizontal asymptote: $y=0$ (as $x\to\pm\infty$, $y\to0$). - Intersections with axes: - No intersection with $x$-axis because $y=0$ implies $\frac{4}{x}=0$ which is impossible. - No intersection with $y$-axis because $x=0$ is undefined. - The curve is a hyperbola in the first and third quadrants approaching the asymptotes. 3. **Part (b): Show the cubic equation for intersections** - At intersection points, $y$ values are equal: $$\frac{k}{x} = 3x^2 - 2x$$ - Multiply both sides by $x$ (assuming $x\neq0$): $$k = 3x^3 - 2x^2$$ - Rearranged: $$3x^3 - 2x^2 - k = 0$$ - This is the cubic equation satisfied by the $x$-coordinates of intersection points. 4. **Part (c): Find intersection points for $k=1$** - Substitute $k=1$ into the cubic: $$3x^3 - 2x^2 - 1 = 0$$ - Rewrite $-2x^2$ as $-3x^2 + x^2$: $$3x^3 - 3x^2 + x^2 - 1 = 0$$ - Group terms: $$(3x^3 - 3x^2) + (x^2 - 1) = 0$$ - Factor each group: $$3x^2(x - 1) + (x - 1)(x + 1) = 0$$ - Factor out $(x - 1)$: $$(x - 1)(3x^2 + x + 1) = 0$$ - Solve each factor: - $x - 1 = 0 \Rightarrow x=1$ - $3x^2 + x + 1 = 0$ has discriminant $\Delta = 1^2 - 4\times3\times1 = 1 - 12 = -11 < 0$, so no real roots. - So the only real intersection is at $x=1$. - Find $y$ coordinate: $$y = 3(1)^2 - 2(1) = 3 - 2 = 1$$ - Intersection point: $(1,1)$. 5. **Part (d): Range of $k$ for exactly two distinct intersection points** - The cubic is: $$3x^3 - 2x^2 - k = 0$$ - The number of real roots depends on the shape of the cubic. - Find critical points by differentiating: $$f(x) = 3x^3 - 2x^2 - k$$ $$f'(x) = 9x^2 - 4x$$ - Set derivative to zero: $$9x^2 - 4x = 0 \Rightarrow x(9x - 4) = 0$$ $$x=0 \text{ or } x=\frac{4}{9}$$ - Evaluate $f(x)$ at critical points (ignoring $k$ for now): $$f(0) = -k$$ $$f\left(\frac{4}{9}\right) = 3\left(\frac{4}{9}\right)^3 - 2\left(\frac{4}{9}\right)^2 - k$$ - Calculate: $$3\times \frac{64}{729} - 2 \times \frac{16}{81} - k = \frac{192}{729} - \frac{32}{81} - k = \frac{192}{729} - \frac{288}{729} - k = -\frac{96}{729} - k = -\frac{32}{243} - k$$ - For the cubic to have exactly two distinct real roots, the cubic must be tangent to the x-axis at one root (a repeated root) and cross at another. - This happens when the discriminant is zero. - The discriminant $\Delta$ of cubic $3x^3 - 2x^2 - k=0$ is: $$\Delta = -4(-2)^3 - 27(3)^2(-k)^2 = -4(-8) - 27 \times 9 \times k^2 = 32 - 243 k^2$$ - Set $\Delta=0$ for repeated roots: $$32 - 243 k^2 = 0 \Rightarrow k^2 = \frac{32}{243}$$ $$k = \pm \frac{4\sqrt{2}}{9 \sqrt{3}}$$ - For exactly two distinct real roots, $k$ must satisfy: $$|k| = \frac{4\sqrt{2}}{9 \sqrt{3}}$$ - For two distinct roots (one repeated), $k$ equals these values. - For three distinct roots, $|k| < \frac{4\sqrt{2}}{9 \sqrt{3}}$. - For one real root, $|k| > \frac{4\sqrt{2}}{9 \sqrt{3}}$. **Final answers:** - (a) Curve $y=\frac{4}{x}$ with vertical asymptote $x=0$ and horizontal asymptote $y=0$. - (b) Intersection $x$ satisfy $3x^3 - 2x^2 - k=0$. - (c) For $k=1$, intersection point is $(1,1)$. - (d) Exactly two distinct intersection points when $k=\pm \frac{4\sqrt{2}}{9 \sqrt{3}}$.