$$\begin{gathered}y - y_1 = m(x - x_1)\\[3pt] m = \frac{dy}{dx} \ \text{at } (x_1, y_1)\end{gathered}$$

Differentiate → sub $x_1$ for gradient → find $y_1$ from curve if missing → write the line. Reflex: dy/dx at the point IS the tangent gradient.

$$y - y_1 = -\frac{1}{m}(x - x_1)$$

Here $m$ is the TANGENT gradient — normal is perpendicular to it, so gradients multiply to $-1$. Reflex: Never use $\frac{dy}{dx}$ directly for a normal — flip and negate it.

$$\frac{dy}{dx} = m_L \quad \text{or} \quad \frac{dy}{dx} = -\frac{1}{m_L}$$

Tangent ∥ line → $m_L$; ⊥ → neg. reciprocal; Normal ∥ line → neg. reciprocal (flip twice: line→normal→tangent). Solve for $x$, get $y$, write the line.

Reflex: Count the flips: every perpendicular and every normal is one negative reciprocal.

$$b^2 - 4ac = 0$$

Sub line into curve, get quadratic in one var. Practice 2 Q5: $9kx^2+3x+(7k-1)=0$ → $k=-0.131$ or $0.273$.

Reflex: Tangent line + unknown constant → substitute and set $b^2-4ac=0$.