How to Find Equations of Tangent Lines and Normal Lines

• To find the equation of a line you need a point and a slope.
• The slope of the tangent line is the value of the derivative at the point of tangency.
• The normal line is a line that is perpendicular to the tangent line and passes through the point of tangency.

Example 1

Suppose $$f(x) = x^3$$. Find the equation of the tangent line at the point where $$x = 2$$.

Find the point of tangency.

Since $$x=2$$, we evaluate $$f(2)$$.

$$ f(2) = 2^3 = 8 $$

The point is $$(2,8)$$.

Find the value of the derivative at $$x = 2$$.

$$ f'(x) = 3x^2\longrightarrow f'(2) = 3(2^2) = 12 $$

The the slope of the tangent line is $$m = 12$$.

Find the point-slope form of the line with slope $$m = 12$$ through the point $$(2,8)$$.

$$ \begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - 8 & = 12(x-2) \end{align*} $$

$$y - 8 = 12(x-2)$$

For reference, here is the graph of the function and the tangent line we just found.

Example 2

Suppose $$f(x) = x^2 - x$$. Find the equation of the tangent line with slope $$m = -3$$.

Find the derivative.

$$ f'(x) = 2x -1 $$

Find the $$x$$-value where $$f'(x)$$ equals the slope.

$$ \begin{align*} f'(x) & = 2x -1\\[6pt] -3 & = 2x -1\\[6pt] -2 & = 2x\\[6pt] x & = -1 \end{align*} $$

Find the point on the function where $$x = -1$$.

$$ f(-1) = (-1)^2 - (-1) = 1 + 1 = 2 $$

The point is $$(-1, 2)$$.

Find the equation of the line through the point $$(-1,2)$$ with slope $$m=-3$$.

$$ \begin{align*} y -y_1 & = m(x-x_1)\\[6pt] y - 2 & = -3(x - (-1))\\[6pt] y - 2 & = -3(x+1) \end{align*} $$

$$ y - 2 = -3(x+1) $$

For reference, here's the graph of the function and the tangent line we just found.

Example 3

Suppose $$x^2 + y^2 = 16$$. Find the equation of the tangent line at $$x = 2$$ for $$y>0$$.

Find the $$y$$-value of the point of tangency.

$$ \begin{align*} \blue{x^2} + y^2 & = 16\\[6pt] \blue{2^2} + y^2 & = 16\\[6pt] \blue{4} + y^2 & = 16\\[6pt] y^2 & = 12\\[6pt] y & = \pm\sqrt{12}\\[6pt] y & = \pm\sqrt{4\cdot 3}\\[6pt] y & = \pm2\sqrt 3 \end{align*} $$

Since the problem states we are interested in $$y>0$$, we use $$y = 2\sqrt 3$$.

The point of tangency is $$(2, 2\sqrt 3)$$.

Find the equation for $$\frac{dy}{dx}$$.

Since the equation is implicitly defined, we use implicit differentiation.

$$ \begin{align*} 2x + 2y\,\frac{dy}{dx} & = 0\\[6pt] 2y\,\frac{dy}{dx} & = -2x\\[6pt] \frac{dy}{dx} & = -\frac{2x}{2y}\\[6pt] \frac{dy}{dx} & = -\frac x y \end{align*} $$

Find the slope of the tangent line at the point of tangency.

At the point $$(2,2\sqrt 3)$$, the slope of the tangent line is

$$ \begin{align*} \frac{dy}{dx}\bigg|_{(\blue{2},\red{2\sqrt 3})} & = -\frac {\blue 2} {\red{2\sqrt 3}}\\[6pt] & = -\frac 1 {\sqrt 3}\\[6pt] & = -\frac 1 {\sqrt 3}\cdot \blue{\frac{\sqrt 3}{\sqrt 3}}\\[6pt] & = -\frac{\sqrt 3} 3 \end{align*} $$

The slope of the tangent line is $$m = -\frac{\sqrt 3} 3$$.

Find the equation of the tangent line through $$(2,2\sqrt 3)$$ with a slope of $$m=-\frac{\sqrt 3} 3$$.

At the point $$(2,2\sqrt 3)$$, the slope of the tangent line is

$$ \begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - 2\sqrt 3 & = -\frac{\sqrt 3} 3(x-2) \end{align*} $$

The equation of the tangent line is $$y - 2\sqrt 3 = -\frac{\sqrt 3} 3(x-2)$$

For reference, the graph of the curve and the tangent line we found is shown below.

Suppose we have a a tangent line to a function. The function and the tangent line intersect at the point of tangency. The line through that same point that is perpendicular to the tangent line is called a normal line.

Recall that when two lines are perpendicular, their slopes are negative reciprocals. Since the slope of the tangent line is $$m = f'(x)$$, the slope of the normal line is $$m = -\frac 1 {f'(x)}$$.

Example 4

Suppose $$f(x) = \cos x$$. Find the equation of the line that is normal to the function at $$x = \frac \pi 6$$.

Find the point on the function.

$$ f\left(\frac \pi 6\right) = \cos \frac \pi 6 = \frac{\sqrt 3} 2 $$

The point is $$\left(\frac \pi 6, \frac{\sqrt 3} 2\right)$$.

Find the value of the derivative at $$x = \frac \pi 6$$.

$$ f'(x) = -\sin x\longrightarrow f'\left(\frac \pi 6\right) = -\sin\frac\pi 6 = -\frac 1 2 $$

The slope of the tangent line is $$m = -\frac 1 2$$. Since we are looking for the line that is perpendicular to the tangent line, we want to use $$m = 2$$.

Find the equation of the line through the point $$\left(\frac \pi 6, \frac{\sqrt 3} 2\right)$$ with a slope of $$m =2$$.

$$ \begin{align*} y -y_1 & = m(x-x_1)\\[6pt] y - \frac{\sqrt 3} 2 & = 2\left(x - \frac \pi 6\right) \end{align*} $$

The line normal to the function at $$x = \frac \pi 6$$ is $$y - \frac{\sqrt 3} 2 = 2\left(x - \frac \pi 6\right)$$.

For reference, here is the graph of the function and the normal line we found.