P1 4.3 - Tangents and normals | Edexcel A-Level International Mathematics

Turn a derivative into the equation of a line. Find the tangent gradient at a chosen point, use the point to locate the tangent, and find the perpendicular normal, including the case of a horizontal tangent.

Gradient at a point

A curve and its tangent share a point and a direction there. Is knowing the curve's height enough to find that direction? No: you need the derivative evaluated at the point's input.

For $y=x^2$ at $x=2$ , the point is $(2,4)$ , but the tangent gradient comes from $\mathrm{d}y/\mathrm{d}x=2x$ , giving $4$ . Those two values happen to agree here; they generally do not.

Tangent gradient

For $y=f(x)$ at $P=(a,f(a))$ , the tangent gradient is $m=f'(a)$ , provided this derivative exists.

Differentiate before substituting. Substituting into the original equation finds the point's height; substituting into the derivative finds the tangent gradient.

Worked example 1: point and gradient

Question. Find the point and tangent gradient on $y=x^3-4x+2$ where $x=1$ .

Plan. Use the original function for $y$ , and the derivative for the gradient.

\begin{aligned} y&=1^3-4(1)+2=-1\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-4\\ m&=3(1)^2-4=-1 \end{aligned}

Answer. The point is $(1,-1)$ and the tangent gradient is $-1$ .

Check. The symmetric chord gradient using $x=1\pm0.001$ is $-0.999999$ , agreeing with the negative tangent gradient.

Change only the chosen input from $1$ to $2$ . The curve is the same, but its direction can change.

Worked example 2: a new point on the same curve

Question. Find the point and tangent gradient on $y=x^3-4x+2$ where $x=2$ .

Plan. Evaluate the same original function and gradient function at the new input.

\begin{aligned} y&=2^3-4(2)+2=2\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-4\\ m&=3(2)^2-4=8 \end{aligned}

Answer. The point is $(2,2)$ and the tangent gradient is $8$ .

Check. The symmetric chord gradient using $x=2\pm0.001$ is $8.000001$ , agreeing with $8$ . The height $2$ is not the gradient.

A gradient is one number at a specified point. Leaving $3x^2-4$ as the final answer has found a gradient function, but has not evaluated the requested gradient.

Watch out

The wrong step is $y(2)=2$ , therefore gradient $=2$ . The correct step is $f'(2)=8$ , therefore gradient $=8$ . Label the two calculations so they cannot be confused.

Find both pieces of information for a different cubic.

Tangent equations

A gradient fixes a line's direction, and one point fixes which parallel line you need.

If a line has gradient $3$ and passes through $(1,2)$ , then moving from $(1,2)$ to any other point $(x,y)$ on it gives $y-2=3(x-1)$ . This is the point-gradient form you already use for straight lines.

Tangent equation

At $P=(a,f(a))$ on $y=f(x)$ , with finite derivative $f'(a)$ ,

y-f(a)=f'(a)(x-a).

Find the point and tangent gradient first, then use them in the straight-line equation.

Not in the formula booklet: learn it.

The line must pass through the actual point on the curve; using the derivative's value as the point's height would usually place it elsewhere.

Worked example 3: locate the tangent

Question. Find an equation of the tangent to $y=x^2+x$ at the point where $x=1$ .

Plan. Find the height and derivative at $1$ , then use point-gradient form.

\begin{aligned} y_P&=1^2+1=2\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=2x+1 \quad\Longrightarrow\quad m=3\\ y-2&=3(x-1)\\ y&=3x-1 \end{aligned}

Answer. $y=3x-1$ .

Check. The line gives $y=2$ at $x=1$ , and its gradient $3$ equals the curve's derivative there. Both required properties hold.

Change only the point's input from $1$ to $2$ . Recalculate the height as well as the gradient.

Worked example 4: locate a second tangent

Question. Find an equation of the tangent to $y=x^2+x$ at the point where $x=2$ .

Plan. Evaluate the curve and derivative at $2$ , then form the line.

\begin{aligned} y_P&=2^2+2=6\\ m&=2(2)+1=5\\ y-6&=5(x-2)\\ y&=5x-4 \end{aligned}

Answer. $y=5x-4$ .

Check. Substituting $x=2$ gives $6$ , and the line's gradient $5$ equals $f'(2)$ . Also, curve minus line is $x^2+x-(5x-4)=(x-2)^2$ , which is zero at the contact point.

Either point-gradient form or a simplified equivalent equation can describe the line. Keep exact fractions if the gradient is fractional.

Exam tip

Find an equation, not just the gradient. Once you have $m$ and $(a,b)$ , write $y-b=m(x-a)$ ; this makes the line-setting method visible even if later arithmetic slips.

Use the same point-and-gradient sequence on a new quadratic.

Normal equations including horizontal tangents

The normal uses the same point as the tangent but runs at right angles to it.

Normal

A normal to a curve at a point is the straight line through that point perpendicular to the tangent there.

For two perpendicular lines with finite, nonzero gradients, their gradients multiply to $-1$ . Thus a tangent of gradient $2$ has a normal of gradient $-1/2$ : one line rises steeply while the other falls more gently.

Normal gradient

If the tangent has finite nonzero gradient $m$ , the normal gradient is

m_{\mathrm{normal}}=-\frac1m.

A horizontal tangent instead has a vertical normal. Through $(a,b)$ , that normal is $x=a$ , with no finite gradient.

Not in the formula booklet: learn it.

The two cases look different because a vertical line cannot be written as $y=mx+c$ with a finite $m$ .

[DIAGRAM: asset_name: Tangents and perpendicular normals; asset_slug: tangent-normal-cases; caption: A sloping tangent has a negative-reciprocal normal gradient; a horizontal tangent has a vertical normal.; description: Equal-scale coordinate panels show the parabola y equals x squared. Left: at one comma one, tangent y equals two x minus one and normal y equals negative one half x plus three halves meet at right angles. Right: at zero comma zero, tangent y equals zero is horizontal and normal x equals zero is vertical.]
A sloping tangent has a negative-reciprocal normal gradient

Start with a nonzero tangent gradient.

Worked example 5: a sloping normal

Question. Find an equation of the normal to $y=x^2$ at the point where $x=1$ .

Plan. Find the tangent gradient, take its negative reciprocal, and use the curve's point.

\begin{aligned} P&=(1,1)\\ m_{\mathrm{tangent}}&=2(1)=2\\ m_{\mathrm{normal}}&=-\frac12\\ y-1&=-\frac12(x-1)\\ y&=-\frac12x+\frac32 \end{aligned}

Answer. $y=-\frac12x+\frac32$ .

Check. At $x=1$ , the line gives $y=1$ . Its gradient multiplied by the tangent gradient is $(-1/2)(2)=-1$ , confirming perpendicularity.

Change only the point's input from $1$ to $0$ . The tangent is now horizontal, so a negative-reciprocal calculation would involve division by zero.

Worked example 6: a vertical normal

Question. Find an equation of the normal to $y=x^2$ at the point where $x=0$ .

Plan. Evaluate the tangent gradient, then choose the vertical line through the point.

P=(0,0),\qquad m_{\mathrm{tangent}}=2(0)=0.

The tangent is $y=0$ . A perpendicular line is vertical, and the vertical line through $(0,0)$ is

x=0.

Answer. $x=0$ .

Check. This is the y-axis: it passes through the origin and meets the horizontal tangent at a right angle.

You can now answer the opening question completely. The curve's height locates the point; its derivative fixes the tangent's direction; perpendicularity then fixes the normal's direction.

Watch out

If the tangent gradient is zero, do not write a normal gradient of zero or “infinity”. Write the vertical line's equation $x=a$ . If the gradient is nonzero, change its sign and take its reciprocal; changing only the sign is usually wrong.

Check the horizontal-tangent case on a shifted parabola.

Quick Recap

Use the original function to find a point's height and the derivative to find its tangent gradient.
The tangent through $(a,b)$ is $y-b=m(x-a)$ , where $m=f'(a)$ .
A normal passes through the same point and is perpendicular to the tangent.
Use normal gradient $-1/m$ for $m\ne0$ ; a horizontal tangent at $(a,b)$ has normal $x=a$ .

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P1 4.3 - Tangents and normals

Gradient at a point

Tangent equations

Normal equations including horizontal tangents

Quick Recap

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