Mathematical Methods · Units 3 & 4

Tangents and Normals

Learn how to find the equation of the tangent and the normal to a curve at a point, with plain English intuition, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Balance a ruler against a curved bowl and it touches at exactly one spot, resting flush against the curve like a perfect kiss. That ruler is the tangent. Now stand a second ruler straight up from that same spot, dead square to the first, and you have the normal. These two lines are how calculus pins down the direction a curve is heading at any single point, and once you can find them you can solve a huge slice of the exam.

The whole topic in three chunks

1 · Tangent gradient

The point comes from substituting $x_1$ into the curve to get $y_1$
The tangent gradient is the derivative at that point, $m_{tangent} = \left.\frac{dy}{dx}\right|_{x = x_1}$
Differentiate first, then substitute the number in

2 · Normal gradient

The normal meets the tangent at a right angle
Its gradient is the negative reciprocal, $m_{normal} = -\dfrac{1}{m_{tangent}}$
Flip the fraction over and change the sign, doing both

3 · Build the line

Both lines share the same point $(x_1, y_1)$
Use point-gradient form $y - y_1 = m(x - x_1)$
Tidy into $y = mx + c$ , feeding in the right gradient

What a tangent and a normal really are

Think of a tiny ant walking along the curve. At any instant the ant is facing one exact direction. The tangent is the straight line that points the same way the ant is facing right at that point. It just grazes the curve and carries on in a straight line.

The normal is the line that crosses the tangent at a perfect right angle at the same point. If the tangent is the direction of travel, the normal is “straight out to the side”. Picture the surface of a hill: the tangent runs along the slope, and the normal points directly away from the slope, the way a flagpole would stand.

Both are just straight lines, so each one needs only two things: a point to pass through and a gradient.

Step one: get the gradient from the derivative

The point is the easy part. You read off or work out the $x$ value, then substitute it into the original curve to get the matching $y$ value. That gives you the point $(x_1, y_1)$ .

For the gradient you use the derivative. The derivative $\dfrac{dy}{dx}$ is the gradient function, which gives the steepness at any $x$ . To get the gradient at your specific point, substitute the $x$ value of the point into the derivative.

m_{tangent} = \left. \frac{dy}{dx} \right|_{x = x_1}

The single biggest error here is stopping too early. The derivative on its own is still a function full of $x$ . You must put your number in to turn it into an actual gradient. Differentiate, then substitute the number in. Leaving the answer as the gradient function is one of the most common slips assessors report.

The tangent y = x - 4 (dashed) touches the curve y = x² - 3x at the point (2, -2), matching its gradient of 1 there.

Step two: flip and change the sign for the normal

The tangent and the normal meet at a right angle. There is a neat rule for the gradients of two perpendicular lines: multiply them together and you always get $-1$ .

m_{tangent} \times m_{normal} = -1

Rearranged, this says the gradient of the normal is the negative reciprocal of the tangent gradient. In plain words: take the tangent gradient, flip it upside down, and change its sign.

m_{normal} = -\frac{1}{m_{tangent}}

So if the tangent gradient is $2$ , the normal gradient is $-\dfrac{1}{2}$ . If the tangent gradient is $-\dfrac{3}{4}$ , the normal gradient is $\dfrac{4}{3}$ . Notice both the flip and the sign change happen every single time. Forgetting one of them is the classic mistake: changing only the sign gives the wrong line, and so does only flipping.

Step three: build the equation of the line

Once you have a point $(x_1, y_1)$ and a gradient $m$ , both the tangent and the normal are found the same way, using the point gradient form of a straight line.

y - y_1 = m(x - x_1)

Substitute your point and your gradient, then tidy it into the form the question asks for, usually $y = mx + c$ . The only difference between finding the tangent and finding the normal is which gradient you feed into this formula.

Here is the full recipe.

Find the point. Substitute the $x$ value into the original curve to get $y_1$ .
Differentiate the curve to get $\dfrac{dy}{dx}$ .
Substitute the $x$ value into the derivative to get the tangent gradient $m$ .
For a normal, replace $m$ with its negative reciprocal $-\dfrac{1}{m}$ .
Put the point and the gradient into $y - y_1 = m(x - x_1)$ and simplify.

A few things the examiners watch for. Write coordinates with round brackets, as in $(2, -2)$ , and never lose the closing bracket. Watch your signs carefully when you expand a bracket like $-\dfrac{1}{2}(x - 3)$ , since a dropped negative is one of the most common ways marks slip away. And read the question: a request for the gradient wants a single number, while a request for the equation wants a full line written as $y = mx + c$ .

Learn it, then fade the help

1 · I do — fully worked

Tangent to $y = x^2 - 3x$ at $x = 2$ . First the point: $y = (2)^2 - 3(2) = -2$ , so the point is $(2, -2)$ . Differentiate for the gradient function, $\dfrac{dy}{dx} = 2x - 3$ , then substitute $x = 2$ to get $m = 2(2) - 3 = 1$ . Put the point and gradient into $y - y_1 = m(x - x_1)$ : $y - (-2) = 1(x - 2)$ , which tidies to $y = x - 4$ .

2 · We do — your turn, guided

Same steps, your turn. Find the tangent to $y = x^2 + x$ at $x = 1$ . Start with the point: $y = (1)^2 + 1 = \;?$ , then $\dfrac{dy}{dx} = 2x + 1$ so $m = \;?$ , and finish with $y - y_1 = m(x - x_1)$ .

Check your step

The point is $(1, 2)$ and $m = 2(1) + 1 = 3$ . Then $y - 2 = 3(x - 1)$ , which tidies to $y = 3x - 1$ .

3 · You do — solo

No scaffolding now: find the equation of the normal to $y = x^2 - 3x$ at $x = 2$ .

Reveal solution

The point is $(2, -2)$ and the tangent gradient is $m = 1$ (as above). The normal gradient is the negative reciprocal, $-\dfrac{1}{1} = -1$ . Then $y - (-2) = -1(x - 2)$ , which tidies to $y = -x$ .

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

How do you find the gradient of the tangent at a point?

How do you get the gradient of the normal from the tangent gradient?

Write the point-gradient form of a straight line.

If the tangent gradient is

-\dfrac{3}{4}

, what is the normal gradient?

What is the single most common slip when finding a tangent gradient?

Recall · Stationary Points

What is special about the tangent at a stationary point?

Recall · Functions and Graphs — straight lines

What is the product of the gradients of two perpendicular lines?

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked Example 1Tangent through the origin, from a real exam

For $g(x) = 2^x + 5$ , find the equation of the tangent to $g$ that passes through the origin, correct to three decimal places.

1
The tangent at $x = a$ has gradient $g'(a) = \log_e 2 \cdot 2^a$ at the point $(a, 2^a + 5)$ , giving $y = \log_e 2 \cdot 2^a (x - a) + 2^a + 5$ .
$y = \log_e 2 \cdot 2^a (x - a) + 2^a + 5$
2
Require the tangent to pass through $(0, 0)$ : substitute the origin and solve numerically for $a$ .
$0 = \log_e 2 \cdot 2^a(0 - a) + 2^a + 5 \;\Rightarrow\; a \approx 2.61784$
3
Compute the gradient at this $a$ to write the tangent through the origin. An equation, not just $a$ , was required.
$m = \log_e 2 \cdot 2^a \approx 4.255, \quad y \approx 4.255x$

Answer

y \approx 4.255x

VCAA 2023 Mathematical Methods Exam 2, Section B Q3cii

Worked Example 2Tangent to a sine curve, from a real exam

For $f(x) = \sin(x) + 1$ , find the equation of the tangent to the graph of $y = f(x)$ at the point where $x = \dfrac{2\pi}{3}$ .

1
The gradient is $f'\left(\tfrac{2\pi}{3}\right) = \cos\tfrac{2\pi}{3}$ .
$f'(x) = \cos x, \quad f'\!\left(\tfrac{2\pi}{3}\right) = -\tfrac12$
2
Use the point $\left(\tfrac{2\pi}{3},\, \tfrac{\sqrt3}{2} + 1\right)$ in point-gradient form. An equation (not just a gradient) was required.
$y = -\tfrac12\!\left(x - \tfrac{2\pi}{3}\right) + \tfrac{\sqrt3}{2} + 1 = -\tfrac12 x + \tfrac{\pi}{3} + \tfrac{\sqrt3}{2} + 1$

Answer

y = -\dfrac{1}{2}x + \dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2} + 1

VCAA 2025 Mathematical Methods Exam 2, Section B Q4d

Worked Example 3Tangent to a parabola

Find the equation of the tangent to $y = x^2 - 3x$ at the point where $x = 2$ .

1
Find the $y$ value of the point by substituting $x = 2$ into the curve.
$y = (2)^2 - 3(2) = -2$
2
Differentiate to get the gradient function.
$\frac{dy}{dx} = 2x - 3$
3
Substitute $x = 2$ to get the gradient of the tangent at the point.
$m = 2(2) - 3 = 1$
4
Use $y - y_1 = m(x - x_1)$ with the point $(2, -2)$ and $m = 1$ .
$y - (-2) = 1(x - 2)$

Answer

y = x - 4

Worked Example 4Normal to the same parabola

Find the equation of the normal to $y = x^2 - 3x$ at the point where $x = 2$ .

1
The point is $(2, -2)$ and the tangent gradient is $m = 1$ , as found above.
$m_{tangent} = 1$
2
The normal is perpendicular, so flip and change the sign to get the perpendicular gradient.
$m_{normal} = -\frac{1}{1} = -1$
3
Use $y - y_1 = m(x - x_1)$ with the point $(2, -2)$ and $m = -1$ .
$y - (-2) = -1(x - 2)$

Answer

y = -x

Worked Example 5Tangent involving a fraction gradient

Find the equation of the tangent to $y = \dfrac{4}{x}$ at the point where $x = 1$ .

1
Find the point. When $x = 1$ , $y = \frac{4}{1} = 4$ , so the point is $(1, 4)$ .
$y = \frac{4}{1} = 4$
2
Write the curve as a power, then differentiate.
$y = 4x^{-1}, \quad \frac{dy}{dx} = -4x^{-2} = -\frac{4}{x^2}$
3
Substitute $x = 1$ for the gradient of the tangent.
$m = -\frac{4}{(1)^2} = -4$
4
Use $y - y_1 = m(x - x_1)$ with $(1, 4)$ and $m = -4$ .
$y - 4 = -4(x - 1)$

Answer

y = -4x + 8

Practice test

Try it yourself

9 questions, 11 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.A polynomial has the equation $y = x(3x - 1)(x + 3)(x + 1)$ . The number of tangents to this curve that pass through the positive $x$ -intercept is:

1mark

VCAA 2023 Mathematical Methods Exam 2, Section A Q14

Q2.Let $A$ be a point on the line $y = x + c$ and $B$ be a point on the curve $y = \log_e(x - 1)$ . If $A$ and $B$ are placed such that the line segment $AB$ has the minimum possible length, and this length is $\sqrt{2}$ , the value of $c$ must be:

1mark

VCAA 2025 Mathematical Methods Exam 2, Section A Q19

Q3.The following algorithm applies Newton's method using a For loop with 3 iterations. Define $\text{newton}(f(x), df(x), x0)$ : For $i$ from 1 to 3, if $df(x0) = 0$ then Return "Error: Division by zero" else $x0 \leftarrow x0 - f(x0) \div df(x0)$ , EndFor, Return $x0$ . The Return value of the function $\text{newton}(x^3 + 3x - 3,\ 3x^2 + 3,\ 1)$ is closest to:

1mark

VCAA 2023 Mathematical Methods Exam 2, Section A Q13

Q4.The gradient of the tangent to $y = x^3 - x$ at the point where $x = 2$ is:

1mark

Need a hint?

Differentiate first to get the gradient function, then substitute

x = 2

to turn it into a single number.

Q5.The tangent to a curve at a point has gradient $m = \tfrac{2}{3}$ . The gradient of the normal at that point is:

1mark

Need a hint?

The normal gradient is the negative reciprocal: flip the fraction over and change the sign, doing both.

Q6.The equation of the tangent to $y = x^2$ at the point $(3, 9)$ is:

1mark

Need a hint?

Find the gradient at

x = 3

, then substitute the point and gradient into

y - y_1 = m(x - x_1)

Q7.The normal to $y = x^2 - 4x + 1$ at the point where $x = 3$ has equation:

1mark

Need a hint?

Find the point and the tangent gradient first, then take the negative reciprocal before building the line.

Q8.The tangent to $y = e^x$ at the point where $x = 0$ has equation:

1mark

Need a hint?

Recall that the derivative of

e^x

e^x

, and that

e^0 = 1

for both the point and the gradient.

Q9.Find the equation of the normal to $y = x^2 + 2x$ at the point where $x = 1$ . Show every step.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

How do you find the equation of a tangent to a curve at a point?

Work out the point by substituting the x value into the curve to get y. Differentiate the curve and substitute the same x value to get the gradient. Then put the point and gradient into y minus y one equals m times x minus x one and simplify.

What is the difference between a tangent and a normal to a curve?

The tangent is the straight line that just grazes the curve and points the same way the curve is heading at that point. The normal is the line through the same point at a right angle to the tangent. They share the point but their gradients are negative reciprocals of each other.

How do you find the gradient of the normal?

Take the gradient of the tangent, flip it upside down and change its sign. This is called the negative reciprocal. For example, if the tangent gradient is 2 the normal gradient is minus one half. You must do both the flip and the sign change every time.

Tangents and Normals

Learn

What a tangent and a normal really are

Step one: get the gradient from the derivative

Step two: flip and change the sign for the normal

Step three: build the equation of the line

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions