How do you find the nature of stationary points for y = 8x3 + 15x2 − 18x + 12?

$$24x^2 + 30x - 18 = 0$$

Divide the entire equation by 6 to simplify:

$$4x^2 + 5x - 3 = 0$$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=4$, $b=5$, $c=-3$:

$$x = \frac{-5 \pm \sqrt{5^2 - 4(4)(-3)}}{2(4)}$$ $$x = \frac{-5 \pm \sqrt{25 + 48}}{8}$$ $$x = \frac{-5 \pm \sqrt{73}}{8}$$

The stationary points occur at $x_1 = \frac{-5 + \sqrt{73}}{8}$ and $x_2 = \frac{-5 - \sqrt{73}}{8}$.

For $x_1 = \frac{-5 + \sqrt{73}}{8}$:
Substitute $x_1$ into the second derivative:

$$\frac{d^2y}{dx^2} \Big|_{x=\frac{-5 + \sqrt{73}}{8}} = 48\left(\frac{-5 + \sqrt{73}}{8}\right) + 30$$ $$= 6(-5 + \sqrt{73}) + 30$$ $$= -30 + 6\sqrt{73} + 30$$ $$= 6\sqrt{73}$$

Since $6\sqrt{73} > 0$, the stationary point at $x = \frac{-5 + \sqrt{73}}{8}$ is a local minimum.

For $x_2 = \frac{-5 - \sqrt{73}}{8}$:
Substitute $x_2$ into the second derivative:

$$\frac{d^2y}{dx^2} \Big|_{x=\frac{-5 - \sqrt{73}}{8}} = 48\left(\frac{-5 - \sqrt{73}}{8}\right) + 30$$ $$= 6(-5 - \sqrt{73}) + 30$$ $$= -30 - 6\sqrt{73} + 30$$ $$= -6\sqrt{73}$$

Since $-6\sqrt{73} < 0$, the stationary point at $x = \frac{-5 - \sqrt{73}}{8}$ is a local maximum.