how to find a tangent line

Here's an interesting fact: The derivative at a point is the slope of the tangent line at that point on the graph.

Jenn (B.S., M.Ed.) of Calcworkshop® teaching how to find the tangent line of a derivative

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

Why is this fascinating?

Because if we are ever asked to solve problems involving the slope of a tangent line, all we need are the same skills we learned back in algebra for writing equations of lines.

Equations Of Lines

So, what do we remember about equations for lines?

Well, they require just two elements:

Point
Slope

Example

For instance, if we desired to write the equation of a line given the point (6,1) and slope m = 3. All we will do is substitute the given information into the point-slope formula and simplify, as indicated below.

\begin{equation}
\begin{array}{l}
y-y_{1}=m\left(x-x_{1}\right) \text { if }(6,1) \text { and } m=3 \\
y-1=3(x-6) \\
y=3 x-17
\end{array}
\end{equation}

Equation Of Tangent Line

This means that to find the equation of a tangent line to a curve, f(x), we simply need two elements: point and slope. The only difference is that to find our slope (i.e., rate of change), we will use derivatives!

Is your mind blown yet?

Example

Alright, suppose we are asked to write the equation of the line tangent to the curve \(y=x^{2} \text { at } x=3\).

First, we will find our point by substituting x = 3 into our function to identify the corresponding y-value.

\begin{equation}
\begin{array}{l}
f(x)=x^{2} \quad x=3 \\
f(3)=(3)^{2}=9 \\
(3,9)
\end{array}
\end{equation}

Next, we take the derivative of our curve to find the rate of change.

\begin{equation}
f^{\prime}(x)=2 x
\end{equation}

We will then swap our given x-value into our derivative to find the slope at x = 3.

\begin{equation}
f^{\prime}(3)=2(3)=6
\end{equation}

Lastly, we will substitute our point (3,9) and slope m = 6 into the formula for point-slope form and write the equation of the tangent line.

\begin{equation}
\begin{array}{l}
y-y_{1}=m\left(x-x_{1}\right) \text { if }(3,9) \text { and } m=6 \\
y-9=6(x-3) \\
y=6 x-9
\end{array}
\end{equation}

See, finding the equation of the tangent line is easy!

We were able to use our algebra skills to find the equation of the line tangent to a curve.

find the equation of the tangent line

Find The Equation Of The Tangent Line

Therefore, let's formally lay out the steps for writing the tangent line equation to a curve, as this particular skill is pivotal for future lessons dealing with linearization and differentials.

Substitute the given x-value into the function to find the y-value or point.
Calculate the first derivative of f(x).
Plug the ordered pair into the derivative to find the slope at that point.
Substitute both the point and the slope from steps 1 and 3 into point-slope form to find the equation for the tangent line.

Normal Line Equation

Likewise, we can even extend this concept to writing equations of normal lines, which are also called perpendicular lines. The only difference will be that we will simply use the negative reciprocal slope of the line tangent.

Example

For this problem, consider the curve \(f(x)=2^{3 x}\). Find the tangent line equation and normal line to f(x) at x = 1.

First, we will find our point by substituting x = 1 into our function to identify the corresponding y-value.

\begin{equation}
\begin{aligned}
&f(x)=2^{3 x} \quad x=1\\
&f(1)=2^{3(1)}=8\\
&(1,8)
\end{aligned}
\end{equation}

Next, we take the derivative of f(x) to find the rate of change.

\begin{equation}
f^{\prime}(x)=2^{3 x} \cdot \ln (2) \cdot 3
\end{equation}

Next, we will swap our given x-value into our derivative to find the slope at x = 1.

\begin{equation}
f^{\prime}(1)=2^{3(1)} \cdot \ln (2) \cdot 3=24 \ln 2 \approx 16.64
\end{equation}

This means that the slope of the tangent line is 16.64, and the slope of the normal line is -1/16.64 or -0.06, which is the negative reciprocal slope!

Lastly, we will write the equation of the tangent line and normal lines using the point (1,8) and slope tangent slope of m = 16.64 and normal slope of -0.06, respectively.

find equations tangent normal lines curve given point

Find The Equations Of The Tangent And Normal Lines Of The Curve For A Given Point

Simple!

Together we will walk through three examples and learn how to use the point-slope form to write the equation of tangent lines and normal lines.

Let's get to it!

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

calcworkshop jenn explaining equation of tangent line

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