Tangent Line Equations

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Lesson Priority: High

 
Objectives
  • Know how to find the tangent lines of a function at a point
  • Use the same technique to find the equation of a normal line at a point
  • Learn a guaranteed-to-work approach for finding equations of tangent lines that go through a specified point
Lesson Description

On quizzes and exams, we are commonly asked to find not only the instantaneous slope at a point on a function, but also the linear equation of the straight line at that point. More advanced questions ask for equations of lines that are tangent to the function but pass through a point that is not on the function. Each question has its own short solution process, and this lesson will cover both cases.

 
Practice Problems

Practice problems and worksheet coming soon!

 

Tangency

In a geometry sense, the word tangency refers to touching at exactly one point without cutting through. For example, the line in the following figure is tangent to the circle:As we've been seeing, derivatives are all about finding slope, and the derivative of a function at a point will give us the instantaneous slope of a tangent line at that point. The most common question we're asked when it comes to function tangent lines is "find the equation of a line that is tangent to the function at the specified point."

Equations of Tangent Lines

With an understanding of how to find a specified derivative, and some of our old-school Algebra knowledge, we can handle the oft-asked task of finding the equation of the tangent line of a function at a given $x$ value. For the most part, finding the equation of a line is just a matter of remembering some $y=mx+b$ knowledge from yesteryear.Your job is simply this:
  • Find the coordinates of the point of tangency
  • Find the instantaneous slope of the function at that point
  • Write a line equation using either point-slope or slope-intercept form, whichever is easiest (almost always the former)
Let's jump right in with an example.
 
Example 1 Find the line tangent to the curve $f(x) = -3(x-2)^2 + 1$ at $x = 0$. $\blacktriangleright$ At the point of tangency, when $x=0$, the function value is $f(0)$, which we can compute right away: $$f(0)= -3(0-2)^2 + 1 = -11$$ Now we know that the coordinates of the point of tangency is the ordered pair $(0,-11)$. Next we need the slope of tangency, also known as the instantaneous slope. This is obtained by finding the derivative of $f(x)$ at $x = 0$. Using what we've learned about finding derivatives, let's compute $f'(x)$ and then plug in $x=0$.$$f(x) = -3(x-2)^2 + 1$$$$\Rightarrow f'(x) = -6(x-2)$$Therefore, $f'(0)$ is$$f'(0) = -6((0)-2) = 12$$ The slope of $f(x)$ at $x=0$ is $12$. Knowing the slope and a point on the line, we can quickly conjure up a point-slope form linear equation:$$y - y_1 = m(x - x_1)$$$$y - (-11) = 12(x - 0)$$$$y + 11 = 12x$$
Pro Tip
The process for finding tangent line equations is always the same: find the actual location of the place of tangency, find the slope at that point, and then plug the point and slope into the point-slope equation. Virtually every teacher and professor will be ok with point-slope equations rather than $y=mx+b$ slope-intercept equations, but as always, pay attention to your instructions when taking a quiz or test.

Equations of Normal Lines

Beside being asked for the equation of the tangent line at a point, teachers love to also ask for the equation of the normal line at a point. This will require virtually the same process as the one needed for finding tangent line equations, but teachers love to ask for normal lines to try and catch students off guard.Remember from geometry that we say a normal line is one that is instantaneously perpendicular to a curve. For example, the blue line in the following figure is the normal line to the shown function at the point of intersection.The process for finding equations of normal lines is:
  • Find the coordinates of the point of interest
  • Find the perpendicular slope to the instantaneous slope of the function at that point
  • Write a line equation using either point-slope or slope-intercept form, whichever is easiest
Hopefully that looks refreshingly familiar. The only real difference between equations of normal and tangent lines is the extra step of identifying the slope that is perpendicular to the instantaneous slope.Recall from Algebra that Parallel and Perpendicular Lines » have specific numeric relationships - particularly, perpendicular slopes are negative reciprocals of one another.
 
Example 2Find the equation of the normal line to the function $g(x) = \sqrt{x}$ at $x = 4$.$\blacktriangleright$ First we'll locate the place that the normal line crosses the function. If $x=4$, then the point $(4,g(4))$ is on the function:$$g(4) = \sqrt{4} = 2$$As you may have suspected via mental math, we're working at the point $(4,2)$.Now, let's identify the instantaneous slope of $g(x)$ at $x=4$. Using the power rule for derivatives:$$g(x) = x^{1/2}$$$$g'(x) = \frac{1}{2\sqrt{x}}$$$$g'(4) = \frac{1}{4}$$If this question wanted us to find a tangent line equation, we would want a linear equation with slope $1/4$. However, as this is a normal line, we want the slope that is perpendicular to $1/4$, which is $-4$ (the negative reciprocal).Since we want a line that passes through the point $(4,2)$, and has slope $-4$, the equation of that line is$$y - 2 = -4(x - 4)$$
You Should Know
Your instructions are king. In addition to the twist you're about to see under the next heading, teachers can also ask for things like "using the limit definition of the derivative to find the tangent slope". This is a topic where it's particularly important to read the instructions on a quiz or test.

Mr. Math Makes It Mean

External PointsTangent and normal line equations aren't always difficult on their own, since the only thing that makes them different from straight derivative practice problems is the addition of some linear equation knowledge that you've already had for a few years.Tests often spice this topic up by asking us to find the equation of lines that are tangent to a function but that also pass through a specified point that is not on the function itself.We'll use the same derivative and linear ideas to solve this type of problem, and hopefully you can appreciate why this catches people off guard when we get so used to the relatively straight-forward tangent and normal line questions. We'll learn by doing with an example.
 
Example 3Find the equation of a line that is tangent to $f(x) = -x^2 + 4$ and passes through the point $(-4,-3)$.$\blacktriangleright$ While we're going to use similar ideas to solve this problem as the ones we used for the previous problems, it is in fact asking us a different question than before. The reason for this is that $(-4,-3)$ is not on the function:
Warning!
If you weren't paying attention or didn't consider this issue, a common mistake is to find the instantaneous slope of the function at $x=-4$ and assume that's the slope we want, and plug into point-slope form. This is not the case!
What we need to do is set up an equation with variable expressions, because the fact is while we could probably eyeball the answer, we don't actually know where this tangent line will touch the function.The tangent line we seek has slope $f'(x)$ and goes through the point $(-4,-3)$, so we can't start by saying that the equation of this tangent line is$$y - y_1 = m(x - x_1)$$$$y - (-3) = f'(x) \cdot (x - (-4))$$$$\longrightarrow y + 3 = (-2x) \cdot(x+4)$$where $-2x$ is the derivative of $f(x) = -x^2 + 4$, obtained by the power rule for derivatives.We can't say this is the final answer, however, because we don't know the slope. The trick is that we can replace y with $-x^2 + 4$, since $y=f(x)$, and solve this equation for $x$, thus identifying the $x$ value at which the tangent line touches the function.$$y + 3 = (-2x) \cdot(x+4)$$becomes$$\left(-x^2 + 4\right) + 3 = -2x \cdot (x + 4)$$$$-x^2 + 7 = -2x^2 - 8x$$$$x^2 + 8x + 7 = 0$$$$(x + 1)(x + 7) = 0$$$$\longrightarrow x = -7, \; -1$$As it turns out, there are two candidates, and without any further specification from your instructions (in fact, the instructions said to find a line, not the line) either choice is correct. We'll pick $x = -1$ simply because it's visible on the graph shown above for this function.If $x=-1$, then $f(-1)=3$, and $f'(-1) = -2(-1) = 2$. Finally, we have enough information to write our point-slope tangent line equation.Using the point $(-4,-3)$ in the point-slope equation:$$y + 3 = 2(x+4)$$It would also be fine to use the point $(-1,3)$ in the point-slope equation:$$y - 3 = 2(x+1)$$Both of these equations represent the same line.

Put It To The Test

This topic is a frequent flier on quizzes due to the detail involved and the fact that you need to do a little critical thinking beyond a usual "find the derivative" type question. Make sure you're ready to read the directions and execute!
 
Example 4Find the equation of a tangent line of the function $g(x) = -2x^2 + 3x - 1$ at $x = 2$.
Show solution
$\blacktriangleright$ First, find the coordinate of the ordered pair on the function when $x=2$.$$g(2) = -2(2)^2 + 3(2) - 1 = -3$$Therefore the point of tangency is $(2,-3)$. The next step is to find the instantaneous slope at this point.$$g'(x) = -4x + 3$$$$g'(2) = -4(2) + 3 = -5$$Finally, let's get this information into a linear equation. As usual we'll choose to use point-slope form, as it's the fastest equation to use.$$y - y_1 = m(x - x_1)$$$$y - (-3) = -5(x - (2))$$$$y + 3 = -5(x-2)$$
 
Example 5Find the equation of a normal line of the function $f(x) = \frac{1}{x}$ at $x=-3$.
Show solution
$\blacktriangleright$ We want the equation of a line that is on the function, so this is nearly identical to the prior problem except that we have to remember to take the negative reciprocal of the tangent slope.First, identify the location of this normal line in terms of where on the function it intersects. When $x=-3$, $f(-3) = -1/3$.Next, let's get the instantaneous slope at that point, so that we can find the slope that is perpendicular to that instantaneous slope.$$f(x) = x^{-1}$$$$f'(x) = -\frac{1}{x^2}$$At $x=-3$,$$f'(-3)= -\frac{1}{9}$$If the instantaneous slope of the function at $x=-3$ is $-1/9$, then the slope perpendicular to it is $9$ (negative reciprocal rule).Having what we need to fill in the point-slope formula, the solution follows.$$y - (-1/3) = 9 (x - (-3))$$$$y + \frac{1}{3} = 9 ( x + 3)$$A picture of the situation confirms that this solution is reasonable.
 
Example 6Find a normal line to the function$$f(x) = -\frac{x^3}{3} + \frac{x^2}{4} + 5x - \frac{9}{4}$$that passes through the point $(-2,4)$ and intersects the function in quadrant 1.
Show solution
$\blacktriangleright$ Let's start with the derivative of the function:$$f'(x) = -x^2 + \frac{x}{2} + 5$$We want a linear equation with a slope that is perpendicular to this slope, that passes through the point $(-2,4)$, so let's set up a point-slope equation.The slope we need is the negative reciprocal of this slope:$$\frac{1}{f'(x)} = \frac{-1}{-x^2 + \frac{x}{2} + 5}$$Plugging in to point-slope form:$$y - y_1 = m \cdot (x - x_1)$$$$y - 4 = \frac{-1}{\left( -x^2 + \frac{x}{2} + 5 \right)} \cdot (x + 2)$$Now replace $y$ with $f(x)$, even though this will be labor intensive.$$\left[-\frac{x^3}{3} + \frac{x^2}{4} + 5x - \frac{9}{4}\right] - 4 = -\frac{x+2}{\left( -x^2 + \frac{x}{2} + 5 \right)}$$Using a little algebra, distributing, and moving everything to the left side of the equation will yield$$(2x-5)\left( -4x^3 +3x^2 + 60x - 75 \right) = 24$$$$\longrightarrow -8x^4 +26x^3 +105x^2 - 450x + 351 = 0$$Now, this isn't a very nice thing to be asked to solve, but when you are asked to solve a third or fourth order equation, you should expect one of two things: the solution will be a relatively small integer, or you'll be allowed a calculator.Either way, hopefully $1$, $-1$, $3$, and $-3$ came to mind to try as solutions, based on the rational roots theorem », and with a few trial and error attempts, you should find via long division » that $x-3$ is a factor of this polynomial.$$-(x-3)\left( 8x^3 -2x^2 - 111x + 117 \right) = 0$$This means that $x=3$ is a solution to the equation, and $x=3$ is the location on the function that will have a normal line that passes through $(-2,4)$.This works out well because $f(3)$ is $6$, so this is a quadrant 1 location, just as we were asked for.Additionally, $f'(3)$ is $-5/2$, so the normal slope at that point is $2/5$. Finally, we have all the information we need to answer the question.The normal line of $f(x)$ that passes through $(-2,4)$ is$$y - 4 = \frac{2}{5} \cdot (x + 2)$$
Pro Tip
Normal line questions can be cumbersome. However, if you run into a normal line equation question that has a lot of messy algebra, chances are the problem was designed so that the solution has relatively simple integer or fraction locations and slopes. Professors are much more likely to ask about tangent line equations anyway, which are slightly simpler.
 
Lesson Takeaways
  • Know what information you need to write a tangent or normal line equation
  • Be ready to pay attention to your instructions, since a few words can change what you need to do entirely
  • Practice the slightly convoluted (but commonly tested) process of finding tangent or normal lines through points that are not on the function

Lesson Metrics

At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).

Key Lesson Sections

Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.

Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).

Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.

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