How To Draw A Tangent Line On A Graph

Quick Overview

To observe 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.

Examples

Example i

Suppose $$f(ten) = x^iii$$. Find the equation of the tangent line at the point where $$10 = 2$$.

Step one

Find the signal of tangency.

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

$$ f(two) = ii^3 = eight $$

The point is $$(ii,viii)$$.

Footstep 2

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

$$ f'(10) = 3x^two\longrightarrow f'(2) = 3(two^2) = 12 $$

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

Pace three

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

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

Answer

$$y - 8 = 12(10-2)$$

For reference, here is the graph of the role and the tangent line nosotros just found.

Example two

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

Step 1

Find the derivative.

$$ f'(10) = 2x -1 $$

Step 2

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

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

Pace iii

Find the point on the part where $$10 = -one$$.

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

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

Step 4

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

$$ \begin{marshal*} y -y_1 & = g(x-x_1)\\[6pt] y - 2 & = -3(x - (-one))\\[6pt] y - 2 & = -three(10+1) \finish{align*} $$

Answer

$$ y - 2 = -three(x+ane) $$

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

Tangent Lines to Implicit Curves

The procedure doesn't alter when working with implicitly defined curves.

Example 3

Suppose $$x^2 + y^2 = xvi$$. Detect the equation of the tangent line at $$ten = 2$$ for $$y>0$$.

Footstep 1

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

$$ \brainstorm{align*} \blue{x^2} + y^2 & = 16\\[6pt] \bluish{2^2} + y^2 & = 16\\[6pt] \blueish{4} + y^two & = 16\\[6pt] y^2 & = 12\\[6pt] y & = \pm\sqrt{12}\\[6pt] y & = \pm\sqrt{4\cdot 3}\\[6pt] y & = \pm2\sqrt iii \finish{align*} $$

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

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

Step 2

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

Since the equation is implicitly defined, nosotros utilise implicit differentiation.

$$ \begin{align*} 2x + 2y\,\frac{dy}{dx} & = 0\\[6pt] 2y\,\frac{dy}{dx} & = -2x\\[6pt] \frac{dy}{dx} & = -\frac{2x}{2y}\\[6pt] \frac{dy}{dx} & = -\frac 10 y \end{align*} $$

Footstep three

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

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

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

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

Step 4

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

At the signal $$(two,2\sqrt iii)$$, the slope of the tangent line is

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

Answer

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

For reference, the graph of the curve and the tangent line nosotros plant is shown below.

Normal Lines

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 aforementioned 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 gradient of the tangent line is $$m = f'(x)$$, the slope of the normal line is $$chiliad = -\frac ane {f'(x)}$$.

Example four

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

Footstep ane

Find the point on the function.

$$ f\left(\frac \pi 6\right) = \cos \frac \pi 6 = \frac{\sqrt iii} two $$

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

Step two

Find the value of the derivative at $$ten = \frac \pi half-dozen$$.

$$ f'(10) = -\sin x\longrightarrow f'\left(\frac \pi vi\right) = -\sin\frac\pi 6 = -\frac 1 2 $$

The slope of the tangent line is $$1000 = -\frac 1 two$$. Since nosotros are looking for the line that is perpendicular to the tangent line, we want to utilise $$m = 2$$.

Step 3

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

$$ \begin{align*} y -y_1 & = thou(10-x_1)\\[6pt] y - \frac{\sqrt 3} ii & = 2\left(x - \frac \pi 6\right) \cease{align*} $$

Respond

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

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