5.4 Tangents and normalsIB Maths: Analysis and Approaches HL: Revision notes

The tangent to $y=f(x)$ at the point $(a,\,f(a))$ has gradient $f'(a)$. Its equation is $y-f(a)=f'(a)(x-a).$ Method:

1. Find the point: $y$-coordinate from $f(a)$.
2. Find the gradient: $m=f'(a)$.
3. Substitute into $y-y_{1}=m(x-x_{1})$.

Example: $y=x^{2}-4x+1$ at $x=3$: point $(3,-2)$, gradient $2(3)-4=2$, tangent $y=2x-8$.

The normal at a point is the line through that point perpendicular to the tangent. Perpendicular gradients multiply to $-1$, so $m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}}.$ Example: $y=\frac{2}{x}$ at $(2,1)$: tangent gradient $-\frac{2}{4}=-\frac{1}{2}$, normal gradient $2$, normal $y=2x-3$.

If the tangent is horizontal ($f'(a)=0$), the normal is the vertical line $x=a$.

To find where the tangent has gradient $m$, solve $f'(x)=m$. Parallel tangents have equal gradients, so to find another point with a tangent parallel to the one at $x=a$, solve $f'(x)=f'(a)$ and discard $x=a$.

Example: on $y=\frac{2}{x}$, $-\frac{2}{x^{2}}=-\frac{1}{2}$ gives $x=\pm2$, so the tangent at $(-2,-1)$ is parallel to the tangent at $(2,1)$.

A tangent can cross the curve somewhere else. Set the curve equal to the tangent line and solve. Because the line touches the curve at $x=a$, $(x-a)^{2}$ is always a factor of the resulting polynomial, which makes factorising easier.

Example: $y=x^{3}-2x^{2}+1$ with tangent $y=4x-7$ at $x=2$: $x^{3}-2x^{2}-4x+8=(x-2)^{2}(x+2)=0$, so the tangent meets the curve again at $(-2,\,-15)$.

On Paper 2 you can use a GDC to find a derivative at a point (numerical derivative) and to draw a tangent, and to find where lines meet the axes or the curve. Always write down the mathematics you used: the gradient value, the equation, and the equation you solved.

Work with unrounded values and give the final answer to 3 significant figures, e.g. a strut along the normal from $(8,\,8.96)$ meeting the ground at $x=16.6016$ has length $\sqrt{8.6016^{2}+8.96^{2}}=12.4$ m.