Unlike a straight line, the slope of a curve constantly changes as you move along the graph. The calculus introduces students to the idea that each point on this graph can be described with a slope or "instantaneous rate of change." The tangent line is a straight line with that slope, which passes through that exact point on the graph. To find the equation of the tangent, you will need to know how to calculate the derivative of the original equation.
Steps
Method 1 of 2: Finding the Equation of the Tangent Line
Step 1. Draw the function and the tangent line (recommended)
A graph makes it easier to follow up on the problem and see if the answer makes sense. Draw the function on a piece of graph paper. If necessary, refer to a graphing calculator. Draw the tangent line through the given point. (Remember, the tangent line passes through that point and has the same slope as the graph at that point.)

Example 1:
Sketch the graph of the parabola f (x) = 0.5x2 + 3x − 1 { displaystyle f (x) = 0.5x ^ {2} + 3x1}
. Dibuja la tangente que pasa a través de ese punto (6, 1).
Aun no conoces la ecuación de la tangente, pero ya puedes saber que la pendiente es negativa y que su intersección en y es negativa (muy por debajo del vértice de la parábola con valor en y de 5, 5). Si la respuesta final no coincide con estos detalles, sabrás que tienes que revisarla y buscar errores.
Step 2. Take the first derivative to find the equation for 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 any point on f (x). There are many ways to drift. 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}
Al derivar, recuerda la regla de la potencia: ddxxn=nxn−1{displaystyle {frac {d}{dx}}x^{n}=nx^{n1}}
La primera derivada de la función es = f'(x) = (2)(0, 5)x + 3  0.
f'(x) = x + 3. Sustituye cualquier valor de x en esta ecuación y el resultado será la pendiente de la línea tangente a f(x) en el punto donde x = a.
Step 3. Substitute in the x value of the point you want to investigate
Read the problem to find the coordinates of the point for which you must find the tangent line. Substitute the x coordinate of this point into f '(x). The result will be the slope of the tangent line at this point.

Example 1 (cont.):
The point mentioned in the problem is (6, 1). Use the coordinate x 6 and substitute it into f '(x):
f '( 6) = 6 + 3 = 3
The slope of the tangent line is 3.
Step 4. Record the equation of the tangent line in pointslope form
The pointslope form of a linear equation is y − y1 = m (x − x1) { displaystyle yy_ {1} = m (xx_ {1})}
, donde m es la pendiente y (x1, y1){displaystyle (x_{1}, y_{1})}
es un punto sobre la línea. Ahora tienes toda la información que necesitas para escribir la ecuación de la línea tangente de esta forma.

Ejemplo 1 (cont.):
y−y1=m(x−x1){displaystyle yy_{1}=m(xx_{1})}
La pendiente de la línea es 3, por lo que y−y1=−3(x−x1){displaystyle yy_{1}=3(xx_{1})}
La línea tangente pasa a través de (6, 1), por lo que la ecuación final es y−(−1)=−3(x−(−6)){displaystyle y(1)=3(x(6))}
Simplifícalo a y+1=−3x−18{displaystyle y+1=3x18}
y=−3x−19{displaystyle y=3x19}
Step 5. Confirm the equation on the graph
If you have a graphing calculator, graph the original function and the tangent line to check that you have the correct answer. If you are doing it on paper, check the chart above to make sure there are no obvious errors in the answer.

Example 1 (cont.):
The initial drawing shows that the slope of the tangent line is negative and that the yintercept is less than 5.5. The equation of the tangent line that we have found is y = 3x  19 in slopeintercept form, which means that 3 is the slope and 19 is the yintercept. Both attributes match the initial predictions.
Step 6. Try a more difficult problem
Here's a runthrough of the whole process again. This time, the goal is to find the line tangent to f (x) = x3 + 2x2 + 5x + 1 { displaystyle f (x) = x ^ {3} + 2x ^ {2} + 5x + 1}
en x = 2:
 Usando la regla de la potencia, la primera derivadaf′(x)=3x2+4x+5{displaystyle f'(x)=3x^{2}+4x+5}
 Dado que x = 2, calcula f′(2)=3(2)2+4(2)+5=25{displaystyle f'(2)=3(2)^{2}+4(2)+5=25}
 Toma en cuenta que esta vez no tenemos un punto, solo una coordenada en x. Para encontrar la coordenada en y, sustituye x = 2 en la función inicial: f(2)=23+2(2)2+5(2)+1=27{displaystyle f(2)=2^{3}+2(2)^{2}+5(2)+1=27}
 Anota la ecuación de la línea tangente en la forma puntopendiente: y−y1=m(x−x1){displaystyle yy_{1}=m(xx_{1})}
. Esta función nos dirá la pendiente de la tangente.
. Esta es la pendiente en x = 2.
. El punto es (2, 27).
y−27=25(x−2){displaystyle y27=25(x2)}
Si se requiere, simplifícalo a y = 25x  23.
Método 2 de 2: Resolver problemas relacionados
Step 1. Determine the end points of a graph
These are the points where the graph reaches a local maximum (one point higher than the points on each side) or a local minimum (lower than the points on each side). The tangent line always has a slope of 0 at these points (a horizontal line), but a slope of zero does not guarantee that a point is extreme. Here's how to determine them:
 Determine the first derivative of the function to obtain f '(x), the equation for the slope of the tangent.
 Solve for f '(x) = 0 to find "possible" end points.
 Determine the second derivative to get f '' (x), the equation that tells you how fast the slope of the tangent changes.
 For each possible extreme point, substitute the coordinate x a into f '' (x). If f '' (a) is positive, there is a local minimum at a. If f '' (a) is negative, there is a local maximum. If f '' (a) is 0, there is an inflection point, not an extreme point.
 If there is a maximum or minimum at a, determine f (a) to get the ycoordinate.
Step 2. Find the equation of the normal
The "normal" of a curve 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 (slope of the tangent) (slope of the normal) = 1, when both pass through the same point on the graph. In other words:
 Determine f '(x), the slope of the tangent line.
 If the point is at x = a, determine f '(a) to find the slope of the tangent at that point.

Find −1f ′ (a) { displaystyle { frac {1} {f '(a)}}}
para encontrar la pendiente de la normal.
 anota la ecuación de la normal en la forma pendientepunto.