Unlike a straight line, the slope of a curve is constantly changing as it moves along the graph. Calculus introduces students to the concept that each point on this graph can be described with a slope, or an "instant rate of change." The tangent line is a straight line relative to that slope that passes through the same point on the graph. To find out what the tangent equation is, you will need to know how to extract the derivative of the original equation.
Steps
Method 1 of 2: Finding the equation of a tangent
Step 1. Sketch the function and tangent (recommended)
The chart helps you track the problem and see if the answer makes sense. Sketch the function on a piece of graph paper, using a graphing calculator if necessary. Draw the tangent that passes through the given point (remember that it passes through that point and has the same slope as the graph there).

Example 1:
Sketch the graph of the parabola f(x)=0, 5x2+3x−1{displaystyle f(x)=0, 5x^{2}+3x1}
. Desenhe a tangente que passa pelo ponto (6, 1).
Você ainda não conhece a equação da tangente, mas pode observar que o declive é negativo e que sua intercepção y é também negativa (bem abaixo do vértice da parábola, com valor y = 5, 5). Se a sua resposta final não for igual a esses detalhes, você poderá conferir os cálculos em busca de erros.
Step 2. Obtain the firstorder derivative to find the equation for the slope of the tangent
For the function f(x), the first derivative f'(x) represents the equation of the slope of the tangent at any point of f(x). There are many ways to derive. Here is a simple example using the power rule:

Example 1 (cont.):
the graph is described by the function f(x)=0, 5x2+3x−1{displaystyle f(x)=0, 5x^{2}+3x1}
Lembrese da regra das potências ao fazer derivadas: ddxxn=nxn−1{displaystyle {frac {d}{dx}}x^{n}=nx^{n1}}
A primeira derivada da função será igual a f'(x) = (2)(0, 5)x + 3  0.
f'(x) = x + 3. Insira qualquer valor “a” para o x dessa equação e o resultado será igual ao declive da tangente de f(x) no ponto em que x = a.
Step 3. Enter the x value of the point to be investigated
Read the problem to find the coordinates of the point whose tangent you want to find. Enter the x coordinate of this point in f'(x). The result will be the slope of the tangent at that point.

Example 1 (cont.):
the point mentioned in the problem is (6, 1). Use the coordinate x = 6 as the value of the independent variable in f'(x):
f'(6) = 6 + 3 = 3
The slope of the tangent is equal to 3.
Step 4. Write the tangent equation in fundamental form
The fundamental form of a linear equation is represented by y−y1=m(x−x1){displaystyle yy_{1}=m(xx_{1})}
, onde m representa o declive (coeficiente angular da reta) e (x1, y1){displaystyle (x_{1}, y_{1})}
representa um ponto da reta.Agora, você tem toda a informação necessária para escrever a equação da tangente nessa forma.

Exemplo 1 (cont.):
y−y1=m(x−x1){displaystyle yy_{1}=m(xx_{1})}
O coeficiente angular da reta é igual a 3 e, por isso, y−y1=−3(x−x1){displaystyle yy_{1}=3(xx_{1})}
A tangente passa pelo ponto (6, 1), de modo que a equação final pode ser representada por y−(−1)=−3(x−(−6)){displaystyle y(1)=3(x(6))}
Simplifiquea para y+1=−3x−18{displaystyle y+1=3x18}
y=−3x−19{displaystyle y=3x19}
Step 5. Confirm the equation on your graph
If you have a graphing calculator, assemble the original function and the tangent to check that the result is correct. If you are working on paper, go back to the previous chart to ensure there are no errors in the answer.

Example 1 (cont.):
the initial sketch revealed that the slope of the tangent was negative, and the yintercept was well below 5. 5. The tangent equation we found is represented by y = 3x  19 in fundamental form, indicating that 3 represents the slope and 19, the yintercept. Both attributes are the same as the initial predictions.
Step 6. Try to solve a more difficult problem
Here's a followup to the entire process once again. Now, the goal is to find the tangent of f(x)=x3+2x2+5x+1{displaystyle f(x)=x^{3}+2x^{2}+5x+1}
em x = 2:
 Com a regra das potências, a derivada primeira será igual a f′(x)=3x2+4x+5{displaystyle f'(x)=3x^{2}+4x+5}
 Uma vez que x = 2, encontre f′(2)=3(2)2+4(2)+5=25{displaystyle f'(2)=3(2)^{2}+4(2)+5=25}
 Observe que não temos o valor do ponto nesse momento, mas apenas uma coordenada x. Para descobrir qual é a coordenada y, insira x = 2 na função inicial: f(2)=23+2(2)2+5(2)+1=27{displaystyle f(2)=2^{3}+2(2)^{2}+5(2)+1=27}
 Escreva a equação da tangente na forma fundamental: y−y1=m(x−x1){displaystyle yy_{1}=m(xx_{1})}
. Essa função nos mostrará qual é o declive da tangente.
. Esse é o declive da função quando x = 2.
. O ponto será (2, 27).
y−27=25(x−2){displaystyle y27=25(x2)}
Se necessário, simplifiquea para y = 25x  23.
Método 2 de 2: Solucionando problemas relacionados
Step 1. Find the extreme points of a graph
These are the points at which the graph reaches a local maximum (higher point than the points on any side) or a local minimum (lower than all points on any side). The tangent will always have a slope equal to 0 at these points (horizontal line), which does not necessarily indicate an extreme point. Learn how to find them here:
 Find the first derivative of the function to get f'(x), the equation for the slope of the tangent.
 Solve f'(x) = 0 to find possible extreme points.
 Take the second derivative to get f''(x), the equation that tells you how quickly the slope of the tangent changes.
 For each possible extreme point, enter the coordinate x = a into f''(a). If the value of f''(a) is positive, there is a local minimum in a. If the value of f''(a) is negative, it is a local maximum. If the value of f''(a) is equal to 0, there is an inflection point, not an extreme point.
 If there is a maximum or a minimum in a, find the value of f''(a) to know what the ycoordinate is.
Step 2. Find the normal equation
The "normal" of a slope at a particular point passes through that point, but has a slope perpendicular to a tangent. To find the equation of the normal, take advantage of the fact that the product (slope of tangent).(slope of normal) = 1, when both pass through the same point on the graph. In other words:
 Find f'(x), the slope of the tangent.
 If the point is at x = a, find f'(a) to find the slope of the tangent at that location.
 Calculate −1f′(a){displaystyle {frac {1}{f'(a)}}}
para encontrar o declive da normal.
 escreva a equação normal na forma fundamental.
dicas

se necessário, comece a reescrever a equação inicial na forma geral:
f(x) = … ou y = …