What if the derivative is undefined at x=a?

If the derivative approaches positive or negative infinity at $x=a$, the tangent line is vertical, and its equation is simply $x=a$. If it is undefined due to a cusp, corner, or discontinuity, no tangent line exists at that point.

How do I find the tangent line if I am given a y-coordinate instead of an x-coordinate?

Set the original function $f(x)$ equal to the given y-coordinate and solve for $x$. There may be multiple solutions. Use each valid x-coordinate as your $a$ and perform the standard procedure for each point.

What if the curve is an implicit equation instead of a function y=f(x)?

Use implicit differentiation to find the derivative $dy/dx$. Then, substitute both the x-coordinate and the y-coordinate of your specific point into the derivative expression to calculate the numerical slope.

How to find the equation of a tangent line

The equation of a tangent line to a function $f(x)$ at $x=a$ is found by evaluating the function and its derivative at that point. This method applies whenever the function is differentiable at $x=a$ . The derivative $f'(a)$ provides the slope of the tangent line, while $f(a)$ provides the y-coordinate. These values are substituted into the point-slope form of a linear equation.

The setup

You require a function $f(x)$ and an x-coordinate $x=a$ . The function must be continuous and differentiable at the given point. Recall the point-slope formula for a line: $y - y_1 = m(x - x_1)$ .

The steps

Evaluate $f(a)$ to find the y-coordinate. The point of tangency is $(a, f(a))$ . 2. Compute the derivative $f'(x)$ using standard differentiation rules. 3. Evaluate $f'(a)$ to find the slope $m$ of the tangent line. 4. Substitute $m$ , $a$ , and $f(a)$ into the point-slope formula $y - f(a) = m(x - a)$ . 5. Isolate $y$ to rewrite the equation in slope-intercept form $y = mx + b$ if requested.

Checking the result

Substitute $x=a$ into your final tangent line equation. It must output the exact same y-coordinate as $f(a)$ . Next, take the derivative of your tangent line; it must equal the constant $f'(a)$ .

Common errors

The most frequent error is leaving the slope as $f'(x)$ instead of evaluating it at the constant $x=a$ . A tangent line is a linear equation, so its slope must be a scalar, not a function of $x$ . Another common error is sign mistakes when isolating $y$ .

Worked example

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

Step 1: Find the y-coordinate. $f(3) = 3^2 - 3(3) + 2 = 9 - 9 + 2 = 2$ . The point of tangency is $(3, 2)$ . Step 2: Compute the derivative. $f'(x) = 2x - 3$ . Step 3: Find the slope at $x = 3$ . $m = f'(3) = 2(3) - 3 = 6 - 3 = 3$ . Step 4: Substitute into the point-slope formula. $y - 2 = 3(x - 3)$ . Step 5: Convert to slope-intercept form. $y - 2 = 3x - 9$ simplifies to $y = 3x - 7$ . The final equation is $y = 3x - 7$ .

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References: Stewart Calculus, Chapter 2: Derivatives · OpenStax Calculus Volume 1, Chapter 3: Derivatives · Khan Academy: Derivatives - Tangent lines