How do you find the equation of a line that is tangent to a curve at a given point?

Category: Derivatives »

Course: Calculus »

Relevant MM Lessons:This question involves a few major concepts across several lessons - for a deeper dive and for a full gambit of practice and common tricks, make sure to check out each Relevant Mister Math Lesson in the list above.Finding the line that is tangent to a function at a specified point is a task that every Calculus instructor asks of his or her students. To accomplish this task, we can always follow the same major steps.

Step 1 - The Tangent Slope

Recall that one way we can think of the derivative of a function is to consider it to be the "tangent slope function" of the original function. Therefore we will somehow use the function's derivative to get the tangent slope that we want.This goes down in one of two ways: either they tell us the value of $f'(x)$ at the point of interest, or, more commonly, we find the function's derivative and plug in the $x$ value to get the tangent slope.Example of Way #1: Find the line tangent to the unspecified function $f(x)$ at $x=3$ given that $f(3) = 5$ and $f'(3) = 2/3$.$\longrightarrow$ In this case, the slope of the tangent line at $x=3$ is $2/3$ because that's what it means to say $f'(3) = 2/3$.Example of Way #2: Find the line tangent to $f(x) = x^3 - 2x + 3$ at $x=-2$.$\longrightarrow$ Here, we need to obtain the tangent slope at $x=-2$ by finding $f'(x)$ and plugging in $-2$. Using our knowledge of The Power Rule » we can quickly see that $f'(x) = 3x^2 - 2$, and so $f'(-2) = 10$, meaning the tangent line slope of $f(x)$ at $x=-2$ is $10$.We will continue to work both of these examples as we go through the rest of the steps.

Step 2 - A Point

In order to write the equation of a line, we need a point on the line in addition to needing the slope of the line. We have the slope from Step 1, so to get a point on the tangent line, we look no further than the point of tangency itself. Since we want the tangent line at a specified $x$ value, we just need to get the actual function $y$ value, because if $(x,y)$ is on the function at the point of tangency, it is also on the tangent line.Way #1 Example Continued: At $x=3$, we were explicitly told that $f(3) = 5$. Therefore, the point $(3,5)$ is on the function, and, more importantly, on the tangent line.Way #2 Example Continued: At $x=-2$, we know the function definition, so it's up to us to plug in $x=-2$ and obtain the $y$ value. Since $f(x) = x^3 - 2x + 3$, it follows that $f(-2) = (-2)^3 - 2(-2) + 3 = -1$. Therefore the point $(-2,-1)$ is on the function and also on the tangent line.

Step 3 - The Equation

Most teachers accept answers in the "point-slope" linear equation form:$$y - y_1 = m(x - x_1)$$This requires very little work once you've performed Steps 1 and 2. Here are the answers to our two examples:Way #1 Example Continued: The equation of a line with slope $2/3$ that goes through the point $(3,5)$ is$$y - 5 = \frac{2}{3}\,(x - 3)$$Way #2 Example Continued: The equation of a line with slope $10$ that goes through the point $(-2,-1)$ is$$y + 1 = 10(x+2)$$You should be able to leave your answer this way if no specific form was requested, because point-slope linear equations are a standardized form. If your teacher specifies orally or in the exam instructions that you must turn in your answer in "slope-intercept" form (a.k.a. $y=mx+b$ form), my advice would be to start with the point-slope form above, isolate $y$, and clean up.Way #1 Example Continued:$$\begin{align} y & = \frac{2}{3}\,(x - 3) + 5 \\ & = \frac{2x}{3} - \frac{2}{3} \cdot 3 + 5 \\ & = \frac{2x}{3} + 3 \end{align} $$Way #2 Example Continued:$$\begin{align} y & = 10(x + 2) - 1 \\ & = 10x + 20 - 1 \\ & = 10x + 19 \end{align} $$