How to Find the Equation of a Tangent

Unlike a straight line, the slope (slope) is constantly changing as it moves along the curve. Calculus gives the idea that each point on a graph can be represented by a slope, or “instantaneous rate of change”. A tangent at a point is a line with the same slope and passing through the same point. To find the equation of the tangent line, you need to know how to derive the original equation.

Table of Contents

Find the equation of the tangent line

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Graph the function and the tangent line (this step is optional but recommended). The chart will make it easier for you to understand the question and check if the answer makes sense. Graph the function on grid paper, use a scientific calculator with graphing function for reference if needed. Draw a tangent line through the given point (Remember that the tangent line passes through that point and has the same slope as the graph there).

Example 1: Draw parabp $f$ $($ $x$ $)$ $=$ $0.5$ $x$ $2$ $+$ $3$ $x$ $-$ $first$ ${displaystyle f(x)=0.5x^{2}+3x-1}$ $f(x)=0.5x^{2}+3x-1$ . Draw a tangent line through the point (-6, -1).
Even if you don’t know the tangent equation, you can still see that its slope is negative and the coordinate is negative (it’s far below the parabp vertex with the coordinate equal to -5.5). If the final answer found does not match these details, there must be an error in the calculation and you need to check again.

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Take the first derivative to find the equation of the slope of the tangent line. For the function f(x), the first derivative f'(x) represents the equation of the slope of the tangent line at every point on f(x). There are many ways to take the derivative. Here is a simple example using the power rule: ^{[1] X Research Source}

Example 1 (cont): The graph is given by the function $f$ $($ $x$ $)$ $=$ $0.5$ $x$ $2$ $+$ $3$ $x$ $-$ $first$ ${displaystyle f(x)=0.5x^{2}+3x-1}$ $f(x)=0.5x^{2}+3x-1$ .
Remember the rule of exponentiation when taking derivatives: $d$ $d$ $x$ $x$ $n$ $=$ $n$ $x$ $n$ $-$ $first$ ${displaystyle {frac {d}{dx}}x^{n}=nx^{n-1}}$ ${frac {d}{dx}}x^{n}=nx^{{n-1}}$ .
First derivative of the function = f'(x) = (2)(0.5)x + 3 – 0.
f'(x) = x + 3. Replace x with any value of a, the equation will give us the slope of the tangent to the function f(x) at the point x = a.

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Enter the x value of the point under consideration. Read the problem to find the coordinates of the point to find the tangent line. Enter the coordinates of this point in f'(x). The result obtained is the slope of the tangent line at the upper point.

Example 1 (cont’d): The score mentioned in the question is (-6, -1). Use the -6 coordinate to replace f'(x):
f'(-6) = -6 + 3 = -3
The slope of the tangent line is -3.

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Write an equation for a tangent that has the form of a straight line when the slope is known and a point lies on it. This linear equation is written in the form

y

-

y

first

=

m

(

x

-

x

first

)

{displaystyle y-y_{1}=m(x-x_{1})}

y-y_{1}=m(x-x_{1})

. where m is the slope and

(

x

first

,

y

first

)

{displaystyle (x_{1},y_{1})}

(x_{1},y_{1})

is a point on the tangent line. ^{[2] X Research Source} You now have all the information you need to write the equation of the tangent line in this form.

Example 1 (cont’d): $y$ $-$ $y$ $first$ $=$ $square meter$ $($ $x$ $-$ $x$ $first$ $)$ ${displaystyle y-y_{1}=m(x-x_{1})}$ $y-y_{1}=m(x-x_{1})$
The slope of the tangent line is -3, so: $y$ $-$ $y$ $first$ $=$ $-$ $3$ $($ $x$ $-$ $x$ $first$ $)$ ${displaystyle y-y_{1}=-3(x-x_{1})}$ $y-y_{1}=-3(x-x_{1})$
The tangent line passes through the point (-6, -1), so the final equation is: $y$ $-$ $($ $-$ $first$ $)$ $=$ $-$ $3$ $($ $x$ $-$ $($ $-$ $6$ $)$ $)$ ${displaystyle y-(-1)=-3(x-(-6))}$ $y-(-1)=-3(x-(-6))$
Simplify we get: $y$ $+$ $first$ $=$ $-$ $3$ $x$ $-$ $18$ ${displaystyle y+1=-3x-18}$ $y+1=-3x-18$
$y$ $=$ $-$ $3$ $x$ $-$ $19$ ${displaystyle y=-3x-19}$ $y=-3x-19$

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Graphical confirmation. If you have a graphing calculator, plot the original function and the tangent line to check if the answer is correct. If calculating on paper, use the graph you drew earlier to make sure there are no obvious errors in your answer.

Example 1 (cont): The initial drawing shows that the tangent line has negative slope and the origin is far below -5.5. The tangent equation just found is y = -3x -19, which means -3 is the slope and -19 is the origin.

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Try solving a more difficult problem. We go through all the steps above again. Now, the goal is to find the tangent to

f

(

x

)

=

x

3

+

2

x

2

+

5

x

+

first

{displaystyle f(x)=x^{3}+2x^{2}+5x+1}

f(x)=x^{3}+2x^{2}+5x+1

at x = 2:

Find the first derivative using the power rule: $f$ $s$ $($ $x$ $)$ $=$ $3$ $x$ $2$ $+$ $4$ $x$ $+$ $5$ ${displaystyle f'(x)=3x^{2}+4x+5}$ $f'(x)=3x^{2}+4x+5$ . This function will give us the slope of the tangent.
For x = 2, find $f$ $s$ $($ $2$ $)$ $=$ $3$ $($ $2$ $)$ $2$ $+$ $4$ $($ $2$ $)$ $+$ $5$ $=$ $25$ ${displaystyle f'(2)=3(2)^{2}+4(2)+5=25}$ $f'(2)=3(2)^{2}+4(2)+5=25$ . This is the slope at x = 2.
Notice that this time, we don’t have a point and only the x coordinates. To find the y coordinate, replace x = 2 in the original function: $f$ $($ $2$ $)$ $=$ $2$ $3$ $+$ $2$ $($ $2$ $)$ $2$ $+$ $5$ $($ $2$ $)$ $+$ $first$ $=$ $27$ ${displaystyle f(2)=2^{3}+2(2)^{2}+5(2)+1=27}$ $f(2)=2^{3}+2(2)^{2}+5(2)+1=27$ . The score obtained is (2.27).
Write an equation of the tangent line that passes through a point and has a given slope: $y$ $-$ $y$ $first$ $=$ $m$ $($ $x$ $-$ $x$ $first$ $)$ ${displaystyle y-y_{1}=m(x-x_{1})}$ $y-y_{1}=m(x-x_{1})$
$y$ $-$ $27$ $=$ $25$ $($ $x$ $-$ $2$ $)$ ${displaystyle y-27=25(x-2)}$ $y-27=25(x-2)$
If necessary, reduce to y = 25x – 23.

Solve related problems

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Find the extreme point on the graph. They are the points at which the graph approaches a local maximum (points higher than neighbors on both sides) or local minimums (lower than neighbors on both sides). The tangent always has zero slope at these points (a horizontal line). However, zero slope is not enough to conclude that it is the extreme point. Here’s how to find them: ^{[3] X Research Sources}

Take the first derivative of the function to get f'(x), the tangent slope equation.
Solve the equation f'(x) = 0 to find the potential extreme point .
Taking the second derivative to get f”(x), the equation tells us the rate of change of the slope of the tangent.
At each potential extreme point, substitute the coordinate a into f”(x). If f”(a) is positive, we have a local minimum at a . If f”(a) is negative, we have a local maximum. If f”(a) is 0, that’s not an extreme but an inflection point.
If the maximum or minimum is reached at a , find f(a) to determine the coordinates.

Advice

If necessary, rewrite the original equation in standard form: f(x) = … or y = …

Steps

Find the equation of the tangent line

Solve related problems

Advice

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