Tangents and Normals

The derivative gives the gradient of a curve at a point, which is the gradient of the tangent line there. Put that gradient together with the point and you have the equation of a straight line that just touches the curve. The normal is the line through the same point at right angles to the tangent. Tangent and normal questions come up throughout IB calculus, often as the first step in a longer problem.

Key ideas

The tangent at a point

The tangent to $y = f(x)$ at the point where $x = a$:

• passes through the point $\big(a, f(a)\big)$;
• has gradient $m = f'(a)$, the derivative at $x = a$.

Using the point-gradient form of a line:

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

The normal at a point

The normal passes through the same point and is perpendicular to the tangent. Perpendicular gradients multiply to $-1$, so if the tangent has gradient $m \ne 0$, the normal has gradient

$$m_{\text{normal}} = -\frac{1}{m}$$

Flip the fraction and change the sign: a tangent gradient of $\dfrac{2}{3}$ gives a normal gradient of $-\dfrac{3}{2}$.

If the tangent is horizontal ($m = 0$), the normal is vertical, with equation $x = a$.

The tangent (orange) and normal (green) to $y = x^2 - 4x + 5$ at $(3, 2)$, from Example 1.

Three forms of the answer

IB questions may ask for the equation in a particular form:

Form	Example	Notes
$y - y_1 = m(x - x_1)$	$y - 2 = 2(x - 3)$	quickest to write down; often accepted
$y = mx + c$	$y = 2x - 4$	expand and simplify the form above
$ax + by + d = 0$	$2x - y - 4 = 0$	usually with integer $a$, $b$, $d$

If a question just says “find the equation”, any correct form is fine. If it names a form, give that form.

The method

1. Find the $y$-coordinate of the point, if you aren’t given it: $y_1 = f(a)$.
2. Differentiate (see the power rule) and substitute $x = a$ to get the tangent gradient $m = f'(a)$.
3. For the normal, use the gradient $-\dfrac{1}{m}$ instead.
4. Substitute into $y - y_1 = m(x - x_1)$ and rearrange into the form asked for.

Working backwards: a given gradient

To find where the tangent has a given gradient $k$, solve $f'(x) = k$. A tangent parallel to the line $y = kx + c$ also has gradient $k$, and a horizontal tangent has gradient $0$. There may be more than one answer, so find every solution.

Tangents through a point not on the curve (AA)

Sometimes you’re given a point $P$ that isn’t on the curve and asked for the tangents that pass through it. You don’t know where they touch, so call the point of contact $x = a$:

1. Write the tangent at $x = a$: $y = f(a) + f'(a)(x - a)$.
2. Substitute the coordinates of $P$ for $x$ and $y$.
3. Solve the resulting equation for $a$. Each solution gives one tangent.

Using technology

Your GDC can find the gradient at a point directly: graph the function and use its $\dfrac{dy}{dx}$ (numerical derivative) feature at the $x$-value you want, and many models will also draw the tangent and show its equation. This is especially useful for functions you can’t yet differentiate by hand. Both courses expect you to use analytic methods and technology: differentiate by hand when you can, and use your GDC when you can’t (or to check). Keep the full calculator value for the gradient and round only at the end.

Worked examples

Example 1: Tangent and normal at a point

Let $f(x) = x^2 - 4x + 5$. Find the equations of the tangent and the normal to the graph of $f$ at the point where $x = 3$. Give the normal in the form $ax + by + d = 0$, where $a, b, d \in \mathbb{Z}$.

Solution.

The point. $f(3) = 9 - 12 + 5 = 2$, so the point is $(3, 2)$.

The gradient. $f'(x) = 2x - 4$, so $f'(3) = 2$.

Tangent.

$$y - 2 = 2(x - 3) \quad\Rightarrow\quad y = 2x - 4$$

Normal. The gradient is $-\dfrac{1}{2}$:

$$\begin{aligned} y - 2 &= -\tfrac{1}{2}(x - 3) \\ 2y - 4 &= -(x - 3) && \text{multiply by } 2 \\ 2y - 4 &= -x + 3 \\ x + 2y - 7 &= 0 \end{aligned}$$

Check: $(3, 2)$ is on both lines: $2(3) - 4 = 2$ ✓ and $3 + 2(2) - 7 = 0$ ✓. The gradients $2$ and $-\dfrac{1}{2}$ multiply to $-1$, so the lines are perpendicular. ✓

Example 2: Tangents with a given gradient

Find the points on the curve $y = x^3 - 3x^2 + 2$ where the tangent is parallel to the line $y = 9x + 1$, and find the equations of those tangents.

Solution. Parallel lines have equal gradients, so solve $\dfrac{dy}{dx} = 9$:

$$\begin{aligned} 3x^2 - 6x &= 9 \\ x^2 - 2x - 3 &= 0 && \text{divide by } 3 \\ (x - 3)(x + 1) &= 0 \end{aligned}$$

So $x = 3$ or $x = -1$.

• At $x = 3$: $y = 27 - 27 + 2 = 2$. Tangent: $y - 2 = 9(x - 3)$, so $y = 9x - 25$.
• At $x = -1$: $y = -1 - 3 + 2 = -2$. Tangent: $y + 2 = 9(x + 1)$, so $y = 9x + 7$.

There are two tangents, both parallel to $y = 9x + 1$.

Example 3: Using technology

Let $f(x) = \dfrac{10}{1 + x^2}$. Use technology to find the gradient of the curve at $x = 1$. Hence find the equations of the tangent and normal at this point.

Solution.

The point. $f(1) = \dfrac{10}{1 + 1} = 5$, so the point is $(1, 5)$.

The gradient. Graph $y = \dfrac{10}{1 + x^2}$ on your GDC and use the $\dfrac{dy}{dx}$ feature at $x = 1$. It gives $f'(1) = -5$.

Tangent. $y - 5 = -5(x - 1)$, so $y = -5x + 10$.

Normal. The gradient is $-\dfrac{1}{-5} = \dfrac{1}{5}$:

$$y - 5 = \tfrac{1}{5}(x - 1) \quad\Rightarrow\quad y = 0.2x + 4.8$$

or, with integer coefficients, $x - 5y + 24 = 0$. Check at $(1, 5)$: $1 - 25 + 24 = 0$. ✓

(Once you can differentiate this function by hand, you can confirm $f'(1) = -5$ exactly.)

Example 4: Tangents through an outside point (AA)

Find the equations of the two tangents to $y = x^2$ that pass through the point $P(1, -3)$.

Solution. $P$ isn’t on the curve ($1^2 \ne -3$), so let the tangent touch the curve at $(a, a^2)$. The gradient there is $2a$, so the tangent is

$$y - a^2 = 2a(x - a) \quad\Rightarrow\quad y = 2ax - a^2$$

It passes through $P(1, -3)$, so substitute $x = 1$ and $y = -3$:

$$\begin{aligned} -3 &= 2a - a^2 \\ a^2 - 2a - 3 &= 0 \\ (a - 3)(a + 1) &= 0 \end{aligned}$$

So $a = 3$ or $a = -1$.

• $a = 3$: the tangent touches at $(3, 9)$ and is $y = 6x - 9$.
• $a = -1$: the tangent touches at $(-1, 1)$ and is $y = -2x - 1$.

Check that both pass through $P$: $6(1) - 9 = -3$ ✓ and $-2(1) - 1 = -3$ ✓.

Common mistakes

Using the derivative function as the gradient. The gradient of the tangent is a number, $f'(a)$. Writing $y - 2 = (2x - 4)(x - 3)$ gives a curve, not a line. Substitute the $x$-value into $f'(x)$ first.

Using the derivative’s value as the y-coordinate. The point on the curve is $\big(a, f(a)\big)$, found from the original function. In Example 1, the point is $(3, 2)$ because $f(3) = 2$; it is a coincidence that $f'(3)$ is also $2$.

Getting the normal gradient wrong. The normal gradient is the negative reciprocal, $-\dfrac{1}{m}$. Common slips are using $\dfrac{1}{m}$ (forgetting the sign change) or $-m$ (forgetting to flip). Check that the two gradients multiply to $-1$.

Mishandling a horizontal tangent. If $f'(a) = 0$, the tangent is $y = f(a)$ and the normal is the vertical line $x = a$. You can’t divide by $0$, so the formula $-\dfrac{1}{m}$ doesn’t apply.

Finding only one tangent. Equations like $f'(x) = 9$ or $a^2 - 2a - 3 = 0$ often have two solutions, and each gives its own tangent. Solve fully before you write anything down.

Rounding the gradient too early. In technology questions, store the GDC’s gradient and use the stored value. Rounding $-1.39221\ldots$ to $-1.4$ before writing the equation changes the answer.

Practice

1. (Warm-up) Find the gradient of the tangent to $y = x^3$ at $x = 2$, and hence the equation of the tangent in the form $y = mx + c$.

Solution

$\dfrac{dy}{dx} = 3x^2$, so the gradient at $x = 2$ is $12$. The point is $(2, 8)$:

$$y - 8 = 12(x - 2) \quad\Rightarrow\quad y = 12x - 16$$

2. (Warm-up) The tangent to a curve at a point has the given gradient. Write down the gradient of the normal at that point.

• (a) $3$
• (b) $-\dfrac{3}{4}$
• (c) $0$

Solution

(a) $-\dfrac{1}{3}$

(b) $-\dfrac{1}{-3/4} = \dfrac{4}{3}$

(c) The tangent is horizontal, so the normal is vertical. Its gradient is undefined, and its equation has the form $x = \text{constant}$.

3. (Core) Find the equation of the normal to $y = 2x^2 - 3x + 1$ at the point where $x = 1$. Give your answer in the form $ax + by + d = 0$.

Solution

At $x = 1$: $y = 2 - 3 + 1 = 0$, so the point is $(1, 0)$.

$\dfrac{dy}{dx} = 4x - 3$, which is $1$ at $x = 1$. The normal gradient is $-1$:

$$y - 0 = -1(x - 1) \quad\Rightarrow\quad y = -x + 1 \quad\Rightarrow\quad x + y - 1 = 0$$

4. (Core) Let $f(x) = x + \dfrac{4}{x}$, for $x \ne 0$. Find the equations of the tangent and the normal at the point where $x = 2$.

Solution

$f(2) = 2 + 2 = 4$, so the point is $(2, 4)$.

Write $f(x) = x + 4x^{-1}$, so $f'(x) = 1 - 4x^{-2} = 1 - \dfrac{4}{x^2}$, and $f'(2) = 1 - 1 = 0$.

The tangent is horizontal: $y = 4$. The normal is vertical: $x = 2$.

5. (Core) Find the point on the curve $y = x^2 - 5x + 8$ where the tangent is parallel to the line $y = 3x - 2$, and find the equation of that tangent.

Solution

$\dfrac{dy}{dx} = 2x - 5 = 3$, so $x = 4$. Then $y = 16 - 20 + 8 = 4$, so the point is $(4, 4)$.

$$y - 4 = 3(x - 4) \quad\Rightarrow\quad y = 3x - 8$$

6. (Core) Let $f(x) = e^x - 2x$.

• (a) Use technology to find $f'(1)$, to 3 s.f.
• (b) Find the equation of the tangent at $x = 1$, and show that it passes through the origin.
• (c) Find the equation of the normal at $x = 1$, with coefficients to 3 s.f.

Solution

(a) The GDC’s $\dfrac{dy}{dx}$ feature at $x = 1$ gives $f'(1) \approx 0.718282$, so $f'(1) \approx 0.718$. (Once you’ve met the derivative of $e^x$, in AA SL or AI HL, you can also do this by hand: $f'(x) = e^x - 2$, so $f'(1) = e - 2$ exactly.)

(b) $f(1) = e - 2 \approx 0.718282$ as well. The tangent is

$$y - (e - 2) = (e - 2)(x - 1) \quad\Rightarrow\quad y = (e - 2)x \approx 0.718x$$

There’s no constant term, so $x = 0$ gives $y = 0$: the tangent passes through the origin.

(c) The normal gradient is $-\dfrac{1}{e - 2} \approx -1.39221$:

$$y - 0.718282 = -1.39221(x - 1) \quad\Rightarrow\quad y \approx -1.39x + 2.11 \text{ (3 s.f.)}$$

7. (Core) The line $y = kx - 7$ is a tangent to the curve $y = x^2 + 2$. Find the possible values of $k$.

Solution

Let the line touch the curve at $x = a$. Two conditions must hold:

• equal gradients: $2a = k$
• the point is on both: $a^2 + 2 = ka - 7$

Substitute $k = 2a$ into the second equation:

$$a^2 + 2 = 2a^2 - 7 \quad\Rightarrow\quad a^2 = 9 \quad\Rightarrow\quad a = \pm 3$$

So $k = 6$ (touching at $(3, 11)$) or $k = -6$ (touching at $(-3, 11)$).

Check: $6(3) - 7 = 11$ ✓ and $-6(-3) - 7 = 11$ ✓.

8. (Challenge) The normal to the curve $y = x^2$ at the point $A(1, 1)$ meets the curve again at $B$. Find the coordinates of $B$.

Solution

At $x = 1$ the tangent gradient is $2$, so the normal gradient is $-\dfrac{1}{2}$:

$$y - 1 = -\tfrac{1}{2}(x - 1) \quad\Rightarrow\quad y = -\tfrac{1}{2}x + \tfrac{3}{2}$$

Set this equal to $x^2$ and multiply by $2$:

$$\begin{aligned} 2x^2 &= -x + 3 \\ 2x^2 + x - 3 &= 0 \\ (2x + 3)(x - 1) &= 0 \end{aligned}$$

$x = 1$ is the point $A$ we started from, so $B$ has $x = -\dfrac{3}{2}$ and $y = \left(-\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$. So $B = \left(-\dfrac{3}{2}, \dfrac{9}{4}\right)$.

Check on the normal: $-\tfrac{1}{2}\left(-\tfrac{3}{2}\right) + \tfrac{3}{2} = \tfrac{3}{4} + \tfrac{6}{4} = \tfrac{9}{4}$. ✓

9. (Challenge, AA) Find the equations of the tangents to the curve $y = x^2 - 2x + 4$ that pass through the origin, and the points where they touch the curve.

Solution

The origin isn’t on the curve ($y = 4$ when $x = 0$). Let the tangent touch at $x = a$. The point is $(a, a^2 - 2a + 4)$ and the gradient is $2a - 2$, so the tangent is

$$y - (a^2 - 2a + 4) = (2a - 2)(x - a)$$

Substitute $(0, 0)$:

$$\begin{aligned} -(a^2 - 2a + 4) &= (2a - 2)(-a) \\ -a^2 + 2a - 4 &= -2a^2 + 2a \\ a^2 &= 4 \end{aligned}$$

So $a = 2$ or $a = -2$.

• $a = 2$: the point is $(2, 4)$ and the gradient is $2$, so the tangent is $y = 2x$.
• $a = -2$: the point is $(-2, 12)$ and the gradient is $-6$, so the tangent is $y = -6x$.

Both pass through the origin, as required. Check the points: $2(2) = 4$ ✓ and $-6(-2) = 12$ ✓.