Question: Find the equation of the normal to the curve $y = x^2 - 4x + 5$ at the point where $x = 3$.

Step 1. Find $y$ at $x = 3$: $y = 9 - 12 + 5 = 2$. Point: $(3, 2)$.

Step 2. Differentiate and evaluate:

$$\dfrac{dy}{dx} = 2x - 4 \quad \Rightarrow \quad \dfrac{dy}{dx}\bigg|_{x=3} = 2$$

Gradient of tangent at $(3,2)$ is $2$.

Step 3. Gradient of normal $= -\dfrac{1}{2}$ (negative reciprocal).

Step 4. Normal equation via $y - y_1 = m_n(x - x_1)$:

$$y - 2 = -\dfrac{1}{2}(x - 3) \;\Longrightarrow\; y = -\dfrac{1}{2}x + \dfrac{3}{2} + 2 = -\dfrac{1}{2}x + \dfrac{7}{2}$$

Or: $2y + x = 7$.

Tip: Tangent uses $m$; normal uses $-1/m$. Always verify the point lies on the curve before substituting into the line formula.

Question: Find the coordinates of the point on $y = x^3 - 3x$ at which the tangent has gradient $9$.

Step 1. Differentiate:

$$\dfrac{dy}{dx} = 3x^2 - 3$$

Step 2. Set equal to the given gradient:

$$3x^2 - 3 = 9 \;\Longrightarrow\; 3x^2 = 12 \;\Longrightarrow\; x^2 = 4 \;\Longrightarrow\; x = \pm 2$$

Step 3. Find the corresponding $y$-values:

• $x = 2$: $y = 8 - 6 = 2$. Point $(2, 2)$.
• $x = -2$: $y = -8 + 6 = -2$. Point $(-2, -2)$.

Answer: Two points: $(2, 2)$ and $(-2, -2)$.

Tip: Constraint problems on tangents = solving $f'(x) = (\text{given value})$. Expect multiple solutions (one for each tangent direction). If the question specifies "the point on the curve in the first quadrant", pick the right one.

Question: Curves $y = x^2$ and $y = -x^2 + 4x - 2$ have a common tangent at a shared point. Find the shared point.

Step 1. "Common tangent at a shared point" means the curves both pass through the point AND have the same gradient there.

Set the curves equal: $x^2 = -x^2 + 4x - 2 \Rightarrow 2x^2 - 4x + 2 = 0 \Rightarrow (x-1)^2 = 0 \Rightarrow x = 1$.

At $x = 1$: $y = 1$. Shared point: $(1, 1)$.

Step 2. Verify same gradient:

• Curve 1: $\dfrac{dy}{dx} = 2x$, at $x=1$: gradient $= 2$.
• Curve 2: $\dfrac{dy}{dx} = -2x + 4$, at $x=1$: gradient $= 2$. $\checkmark$

Step 3. Tangent line at $(1, 1)$ with $m = 2$:

$$y - 1 = 2(x - 1) \;\Longrightarrow\; y = 2x - 1$$

Answer: Shared point $(1, 1)$; common tangent $y = 2x - 1$.

Tip: Tangent contact between two curves $\Leftrightarrow$ they intersect with a repeated root (discriminant = 0 of $f - g$).

Question: Find the points on $y = x^3 - 3x^2 + 2x$ where the tangent is parallel to the line $y = 2x + 5$.

Step 1. Parallel tangents share gradient. Line gradient $= 2$.

Step 2. Differentiate:

$$\dfrac{dy}{dx} = 3x^2 - 6x + 2$$

Step 3. Solve $\dfrac{dy}{dx} = 2$:

$$3x^2 - 6x + 2 = 2 \;\Longrightarrow\; 3x^2 - 6x = 0 \;\Longrightarrow\; 3x(x - 2) = 0$$ $$x = 0 \text{ or } x = 2$$

Step 4. Corresponding $y$-values:

• $x = 0$: $y = 0$. Point $(0, 0)$.
• $x = 2$: $y = 8 - 12 + 4 = 0$. Point $(2, 0)$.

Answer: Tangent is parallel to $y = 2x + 5$ at $(0, 0)$ and $(2, 0)$.

Tip: Parallel $\Leftrightarrow$ same $m$. Perpendicular $\Leftrightarrow$ $m_1 \cdot m_2 = -1$, so $m_2 = -\dfrac{1}{m_1}$. Always check whether multiple points exist before stating "the" answer.

Question: The tangent to $y = \dfrac{4}{x}$ at $P(2, 2)$ meets the $x$-axis at $A$ and the $y$-axis at $B$. Find the area of triangle $OAB$.

Step 1. Differentiate $y = 4x^{-1}$: $\dfrac{dy}{dx} = -4x^{-2} = -\dfrac{4}{x^2}$.

At $x = 2$: $\dfrac{dy}{dx} = -\dfrac{4}{4} = -1$. Tangent gradient $= -1$.

Step 2. Tangent equation:

$$y - 2 = -1(x - 2) \;\Longrightarrow\; y = -x + 4$$

Step 3. Intercepts:

• $x$-axis ($y = 0$): $x = 4$. So $A = (4, 0)$.
• $y$-axis ($x = 0$): $y = 4$. So $B = (0, 4)$.

Step 4. Triangle $OAB$ is right-angled at $O$ with legs $OA = 4$ and $OB = 4$:

$$\text{Area} = \dfrac{1}{2} \times 4 \times 4 = 8 \text{ square units.}$$

Tip: For tangent-meets-axes triangles, find the intercepts of the tangent line — the triangle's vertices are $(0,0)$, $(x\text{-intercept}, 0)$, $(0, y\text{-intercept})$. Use $\tfrac{1}{2} \times \text{base} \times \text{height}$.

Question: Find the equation of the tangent to $y = \ln(2x + 1)$ at the point where $x = 0$.

Step 1. $y$ at $x = 0$: $y = \ln(1) = 0$. Point: $(0, 0)$.

Step 2. Differentiate using chain rule:

$$\dfrac{dy}{dx} = \dfrac{1}{2x+1} \cdot 2 = \dfrac{2}{2x+1}$$

At $x = 0$: $\dfrac{dy}{dx} = \dfrac{2}{1} = 2$.

Step 3. Tangent equation through $(0, 0)$ with $m = 2$:

$$y - 0 = 2(x - 0) \;\Longrightarrow\; y = 2x$$

Tip: For tangent/normal on special functions, the heavy lifting is correct differentiation:

• $\dfrac{d}{dx}\ln(f(x)) = \dfrac{f'(x)}{f(x)}$
• $\dfrac{d}{dx}e^{f(x)} = f'(x)e^{f(x)}$
• $\dfrac{d}{dx}\sin(f(x)) = f'(x)\cos(f(x))$

The line construction is identical to polynomial cases.