Question: The diagram shows the curve with equation $y = -x^3 + 9x^2 - 15x - 3$ and a straight line. The straight line passes through the two stationary points of the curve.
(a) Show that the equation of the straight line is
$y = 8x - 18$.
(b) Verify that the curve also meets the straight line when $x = 3$.
1. **State the problem:** We are given a cubic curve with equation $$y = -x^3 + 9x^2 - 15x - 3$$ and a straight line passing through its two stationary points. We need to find the equation of this straight line and verify if the curve meets the line at $x=3$.
2. **Find stationary points:** Stationary points occur where the derivative $\frac{dy}{dx} = 0$.
Calculate the derivative:
$$\frac{dy}{dx} = \frac{d}{dx}(-x^3 + 9x^2 - 15x - 3) = -3x^2 + 18x - 15$$
3. **Solve for stationary points:** Set derivative equal to zero:
$$-3x^2 + 18x - 15 = 0$$
Divide both sides by $-3$:
$$\cancel{-3}x^2 + \cancel{-3} \cdot (-6)x + \cancel{-3} \cdot 5 = 0 \implies x^2 - 6x + 5 = 0$$
4. **Factorize quadratic:**
$$x^2 - 6x + 5 = (x - 1)(x - 5) = 0$$
So stationary points are at $x=1$ and $x=5$.
5. **Find corresponding $y$ values:**
At $x=1$:
$$y = -(1)^3 + 9(1)^2 - 15(1) - 3 = -1 + 9 - 15 - 3 = -10$$
At $x=5$:
$$y = -(5)^3 + 9(5)^2 - 15(5) - 3 = -125 + 225 - 75 - 3 = 22$$
6. **Find equation of line through points $(1, -10)$ and $(5, 22)$:**
Slope $m$:
$$m = \frac{22 - (-10)}{5 - 1} = \frac{32}{4} = 8$$
Use point-slope form with point $(1, -10)$:
$$y - (-10) = 8(x - 1)$$
$$y + 10 = 8x - 8$$
$$y = 8x - 18$$
This matches the required equation.
7. **Verify curve meets line at $x=3$:**
Calculate $y$ on curve at $x=3$:
$$y = -(3)^3 + 9(3)^2 - 15(3) - 3 = -27 + 81 - 45 - 3 = 6$$
Calculate $y$ on line at $x=3$:
$$y = 8(3) - 18 = 24 - 18 = 6$$
Since both values are equal, the curve meets the line at $x=3$.
**Final answers:**
(a) Equation of the straight line is $$y = 8x - 18$$.
(b) The curve meets the line at $x=3$ with $y=6$.