How Do You Find the Equation of a Tangent Line at a Point?

Differentiate the function, evaluate the derivative at the given point to obtain the gradient of the tangent, and substitute this gradient and the point into the point-gradient equation of a line.

The Method

For a curve $y=f(x)$, the derivative (f'(x)) gives the gradient at any value of $x$. At the point where $x=a$, the tangent therefore has gradient

$$m=f'(a).$$

Geometrically, the derivative is the limiting gradient of secant lines through nearby points on the curve. This is why evaluating (f'(x)) at $x=a$ produces the gradient of the unique tangent at that point, provided the function is differentiable there.

If the point is ((a,f(a))), use the point-gradient form

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

You can then rearrange the equation into $y=mx+c$ if required.

If only the $x$-coordinate is given, first calculate the corresponding $y$-coordinate using $y=f(a)$. The complete process is:

1. Find (f'(x)).
2. Calculate (f'(a)) to obtain the tangent's gradient.
3. Calculate $f(a)$ if the full point is not given.
4. Substitute into the equation of a straight line.

Worked Example

Find the equation of the tangent to $f(x)=x^2+3x$ at $x=2$.

First find the point:

$$f(2)=2^2+3(2)=10.$$

Therefore, the tangent passes through $(2,10)$. Differentiate:

$$f'(x)=2x+3.$$

Evaluate the derivative at $x=2$:

$$m=f'(2)=2(2)+3=7.$$

Using point-gradient form:

$$y-10=7(x-2).$$

Hence,

$$y=7x-4.$$

A quick check is to substitute $x=2$ into $y=7x-4$: this gives $y=10$, confirming that the line passes through the required point.

A common misconception is that the gradient is $f(a)$. This is incorrect: $f(a)$ is the point's $y$-coordinate, while (f'(a)) is the tangent's gradient.

Do not use the normal gradient by mistake. The normal would have gradient $-1/7$, whereas this question asks for the tangent, whose gradient is $7$.

IB Exam Technique

This is IB Mathematics AA SL 5.4: Tangents and normals and may appear on Paper 1 or Paper 2. For the command term Find, show the derivative, its evaluation at the point, and substitution into a line equation; a final equation without working may not earn full marks.