Curve Intersections E9Fd0A

1. **State the problem:** We have two curves: $$y = 2x^3 + 10$$ for curve C1 and $$y = 42x - 15x^2 - 7$$ for curve C2, both defined for $$x > 0$$. We need to: (a) Verify that the curves 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 C1: $$y = 2\left(\frac{1}{2}\right)^3 + 10 = 2 \times \frac{1}{8} + 10 = \frac{2}{8} + 10 = \frac{1}{4} + 10 = 10.25$$ For C2: $$y = 42 \times \frac{1}{2} - 15 \times \left(\frac{1}{2}\right)^2 - 7 = 21 - 15 \times \frac{1}{4} - 7 = 21 - \frac{15}{4} - 7 = 21 - 3.75 - 7 = 10.25$$ Since both curves have $$y = 10.25$$ at $$x = \frac{1}{2}$$, they intersect there. 3. **Part (b): Find the exact $$x$$ coordinate of the second intersection point $$P$$** Set the two equations equal to find intersection points: $$2x^3 + 10 = 42x - 15x^2 - 7$$ Bring all terms to one side: $$2x^3 + 10 - 42x + 15x^2 + 7 = 0$$ Simplify: $$2x^3 + 15x^2 - 42x + 17 = 0$$ 4. **Use the known root $$x = \frac{1}{2}$$ to factor the cubic:** Divide the cubic by $$\left(x - \frac{1}{2}\right)$$ using polynomial division or synthetic division. Multiply divisor by 2 to clear fraction: divide by $$2x - 1$$. Perform polynomial division: Divide $$2x^3 + 15x^2 - 42x + 17$$ by $$2x - 1$$. - Leading term: $$2x^3 \div 2x = x^2$$ - Multiply divisor by $$x^2$$: $$2x^3 - x^2$$ - Subtract: $$ (2x^3 + 15x^2) - (2x^3 - x^2) = 16x^2$$ Bring down $$-42x$$: - Divide $$16x^2 \div 2x = 8x$$ - Multiply divisor by $$8x$$: $$16x^2 - 8x$$ - Subtract: $$ (16x^2 - 42x) - (16x^2 - 8x) = -34x$$ Bring down $$+17$$: - Divide $$-34x \div 2x = -17$$ - Multiply divisor by $$-17$$: $$-34x + 17$$ - Subtract: $$(-34x + 17) - (-34x + 17) = 0$$ Remainder is zero, so factorization is: $$2x^3 + 15x^2 - 42x + 17 = (2x - 1)(x^2 + 8x - 17)$$ 5. **Solve quadratic $$x^2 + 8x - 17 = 0$$ for other intersection points:** Use quadratic formula: $$x = \frac{-8 \pm \sqrt{8^2 - 4 \times 1 \times (-17)}}{2 \times 1} = \frac{-8 \pm \sqrt{64 + 68}}{2} = \frac{-8 \pm \sqrt{132}}{2}$$ 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 $$x > 0$$, discard $$x = -4 - \sqrt{33}$$ (negative). The second intersection point $$P$$ has: $$x = -4 + \sqrt{33}$$ **Final answer:** The curves intersect at $$x = \frac{1}{2}$$ and again at $$x = -4 + \sqrt{33}$$.