Question: Transformations and Graphs - Year 1 Core 4. Figure 4 Figure 4 shows a sketch of part of the curve $C_1$ with equation y = 2x^3 + 10 \quad x > 0 and part of the curve $C_2$ with equation y = 42x - 15x^2 - 7 \quad x > 0 (a) Verify that the curves intersect at $x = \frac{1}{2}$ (2) The curves intersect again at the point $P$ (b) Using algebra and showing all stages of working, find the exact $x$ coordinate of $P$ (5) Graph shape/position hints: - top-right: a curve labeled $C_1$ rising through the point $(\frac{1}{2}, 9)$ and continuing up to the right. - center: a second curve labeled $C_2$ is a downward-opening parabola, passing through the same intersection near $(\frac{1}{2}, 9)$, rising to a peak above it, then falling and crossing the x-axis to the right. - top-left: the axes are shown with y-axis vertical and x-axis horizontal, origin marked $O$.

1. **State the problem:** We have two curves $C_1$ and $C_2$ with equations: $$C_1: y = 2x^3 + 10, \quad x > 0$$ $$C_2: y = 42x - 15x^2 - 7, \quad x > 0$$ We need to: (a) Verify that they intersect at $x = \frac{1}{2}$. (b) Find the exact $x$ coordinate of the second intersection point $P$. 2. **Part (a) Verify intersection at $x = \frac{1}{2}$:** Calculate $y$ for both curves at $x = \frac{1}{2}$. For $C_1$: $$y = 2\left(\frac{1}{2}\right)^3 + 10 = 2 \times \frac{1}{8} + 10 = \frac{2}{8} + 10 = \frac{1}{4} + 10 = 10.25$$ For $C_2$: $$y = 42 \times \frac{1}{2} - 15 \times \left(\frac{1}{2}\right)^2 - 7 = 21 - 15 \times \frac{1}{4} - 7 = 21 - \frac{15}{4} - 7$$ Simplify: $$21 - 7 = 14$$ $$14 - \frac{15}{4} = \frac{56}{4} - \frac{15}{4} = \frac{41}{4} = 10.25$$ Since both $y$ values are equal at $x = \frac{1}{2}$, the curves intersect there. 3. **Part (b) Find the exact $x$ coordinate of the second intersection $P$:** Set the two equations equal to find all intersection points: $$2x^3 + 10 = 42x - 15x^2 - 7$$ Bring all terms to one side: $$2x^3 + 10 - 42x + 15x^2 + 7 = 0$$ Simplify constants: $$2x^3 + 15x^2 - 42x + 17 = 0$$ 4. **Use the known root $x = \frac{1}{2}$ to factor the cubic:** Multiply both sides by 2 to clear fraction for synthetic division: $$4x^3 + 30x^2 - 84x + 34 = 0$$ Check root $x = \frac{1}{2}$: $$4 \left(\frac{1}{2}\right)^3 + 30 \left(\frac{1}{2}\right)^2 - 84 \left(\frac{1}{2}\right) + 34 = 4 \times \frac{1}{8} + 30 \times \frac{1}{4} - 42 + 34 = \frac{1}{2} + 7.5 - 42 + 34 = 0$$ So $x=\frac{1}{2}$ is a root. Divide polynomial by $(x - \frac{1}{2})$: Use polynomial division or synthetic division: Coefficients: 2 (for $x^3$), 15 (for $x^2$), -42 (for $x$), 17 (constant) Synthetic division with root $\frac{1}{2}$: - Bring down 2 - Multiply 2 by $\frac{1}{2}$ = 1, add to 15 = 16 - Multiply 16 by $\frac{1}{2}$ = 8, add to -42 = -34 - Multiply -34 by $\frac{1}{2}$ = -17, add to 17 = 0 (remainder) So quotient polynomial is: $$2x^2 + 16x - 34$$ 5. **Solve quadratic $2x^2 + 16x - 34 = 0$ for the second intersection:** Divide entire equation by 2: $$x^2 + 8x - 17 = 0$$ Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{64 + 68}}{2} = \frac{-8 \pm \sqrt{132}}{2}$$ Simplify $\sqrt{132}$: $$\sqrt{132} = \sqrt{4 \times 33} = 2\sqrt{33}$$ So: $$x = \frac{-8 \pm 2\sqrt{33}}{2} = -4 \pm \sqrt{33}$$ 6. **Determine which root is valid:** Since $x > 0$, check: - $-4 + \sqrt{33}$: $\sqrt{33} \approx 5.7446$, so $-4 + 5.7446 = 1.7446 > 0$ valid - $-4 - \sqrt{33} < 0$ invalid **Final answer:** The exact $x$ coordinate of $P$ is: $$\boxed{-4 + \sqrt{33}}$$