How to Find the Equation of a Tangent Line
Unlike a straight line, a curve's slope constantly changes as you move along the graph. Calculus introduces students to the idea that each point on this graph could be described with a slope, or an "instantaneous rate of change." The tangent line is a straight line with that slope, passing through that exact point on the graph. To find the equation for the tangent, you'll need to know how to take the derivative of the original equation.
Finding the Equation of a Tangent Line

Sketch the function and tangent line (recommended). A graph makes it easier to follow the problem and check whether the answer makes sense. Sketch the function on a piece of graph paper, using a graphing calculator as a reference if necessary. Sketch the tangent line going through the given point. (Remember, the tangent line runs 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}. Draw the tangent going through point (6, 1).You don't know the tangent's equation yet, but you can already tell that its slope is negative, and that its yintercept is negative (well below the parabola vertex with y value 5.5). If your final answer doesn't match these details, you'll know to check your work for mistakes.

Take the first derivative to find the equation for the slope of the tangent line. For function f(x), the first derivative f'(x) represents the equation for the slope of the tangent line at any point on f(x). There are many ways to take derivatives. Here's 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}.Recall the power rule when taking derivatives: ddxxn=nxn−1{\displaystyle {\frac {d}{dx}}x^{n}=nx^{n1}}.The function's first derivative = f'(x) = (2)(0.5)x + 3  0.f'(x) = x + 3. Plug any value a for x into this equation, and the result will be the slope of the line tangent to f(x) at the point were x = a.

Enter the x value of the point you're investigating. Read the problem to discover the coordinates of the point for which you're finding the tangent line. Enter the xcoordinate of this point into f'(x). The output is the slope of the tangent line at this point.
 Example 1 (cont.): The point mentioned in the problem is (6, 1). Use the xcoordinate 6 as the input for f'(x):f'(6) = 6 + 3 = 3The slope of the tangent line is 3.

Write the tangent line equation in pointslope form. The pointslope form of a linear equation is y−y1=m(x−x1){\displaystyle yy_{1}=m(xx_{1})}, where m is the slope and (x1,y1){\displaystyle (x_{1},y_{1})} is a point on the line. You now have all the information you need to write the tangent line's equation in this form.
 Example 1 (cont.): y−y1=m(x−x1){\displaystyle yy_{1}=m(xx_{1})}The slope of the line is 3, so y−y1=−3(x−x1){\displaystyle yy_{1}=3(xx_{1})}The tangent line passes through (6, 1), so the final equation is y−(−1)=−3(x−(−6)){\displaystyle y(1)=3(x(6))}Simplify to y+1=−3x−18{\displaystyle y+1=3x18}y=−3x−19{\displaystyle y=3x19}

Confirm the equation on your graph. If you have a graphing calculator, graph the original function and the tangent line to check that you have the correct answer. If working on paper, refer to your earlier graph to make sure there are no glaring mistakes in your answer.
 Example 1 (cont.): The initial sketch showed that the slope of the tangent line was negative, and the yintercept was well below 5.5. The tangent line equation we found is y = 3x  19 in slopeintercept form, meaning 3 is the slope and 19 is the yintercept. Both of these attributes match the initial predictions.

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} at x = 2:
 Using the power rule, the first derivative f′(x)=3x2+4x+5{\displaystyle f'(x)=3x^{2}+4x+5}. This function will tell us the slope of the tangent.
 Since x = 2, find f′(2)=3(2)2+4(2)+5=25{\displaystyle f'(2)=3(2)^{2}+4(2)+5=25}. This is the slope at x = 2.
 Notice we do not have a point this time, only an xcoordinate. To find the ycoordinate, plug x = 2 into the initial function: f(2)=23+2(2)2+5(2)+1=27{\displaystyle f(2)=2^{3}+2(2)^{2}+5(2)+1=27}. The point is (2,27).
 Write the tangent line equation in pointslope form: y−y1=m(x−x1){\displaystyle yy_{1}=m(xx_{1})}y−27=25(x−2){\displaystyle y27=25(x2)}If required, simplify to y = 25x  23.
Solving Related Problems

Find the extreme points on a graph. These are points where the graph reaches a local maximum (a point higher than the points on either side), or local minimum (lower than the points on either side). The tangent line always has a slope of 0 at these points (a horizontal line), but a zero slope alone does not guarantee an extreme point. Here's how to find them:
 Take the first derivative of the function to get f'(x), the equation for the tangent's slope.
 Solve for f'(x) = 0 to find possible extreme points.
 Take the second derivative to get f''(x), the equation that tells you how quickly the tangent's slope is changing.
 For each possible extreme point, plug the xcoordinate 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, find f(a) to get the ycoordinate.

Find the equation of the normal. The "normal" to a curve at a particular point passes through that point, but has a slope perpendicular to a tangent. To find the equation for the normal, take advantage of the fact that (slope of tangent)(slope of normal) = 1, when they both pass through the same point on the graph. In other words:
 Find f'(x), the slope of the tangent line.
 If the point is at x = a, find f'(a) to find the slope of the tangent at that point.
 Calculate −1f′(a){\displaystyle {\frac {1}{f'(a)}}} to find the slope of the normal.
 Write the normal equation in slopepoint form.
Tips
 If necessary, start by rewriting the initial equation in standard form: f(x) = ... or y = ...
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