Lesson Explainer: Equations of Tangent Lines and Normal Lines Mathematics • Higher Education

In this explainer, we will learn how to find the slope and equation of the tangent and normal to a curve at a given point using derivatives.

The derivative of a curve at a point tells us the slope of the tangent line to the curve at that point and there are many different techniques for finding the derivatives of different functions. We can utilize these differentiation techniques to help us find the equation of tangent lines to various differentiable functions.

First, let us recall exactly what we mean by the tangent to a curve a point.

Definition: The Tangent Line to a Curve at a Point

For a curve 𝑦=𝑓(π‘₯) and point (π‘₯,𝑦) on the curve, we say that the line π‘Žπ‘₯+𝑏𝑦+𝑐=0 is the tangent line to the curve at the point (π‘₯,𝑦) if

  • the tangent line passes through the point (π‘₯,𝑦);
  • the curve and tangent line have the same slope at the point (π‘₯,𝑦).

In the definition above, we state that our tangent line and curve will have the same slope at the point (π‘₯,𝑦). This means that, around the point (π‘₯,𝑦), the line will only touch the curve.

The equation of a straight line can be found using the value of its slope and the coordinates of a point which lies on the line. From the definition above, we know that the tangent line and the curve both pass through the point (π‘₯,𝑦). Hence, the only missing piece of information is the slope.

We can then use differentiation to find the slope of the curve, 𝑦=𝑓(π‘₯), at this point. For a function 𝑓, which is differentiable at π‘₯, this slope is given by 𝑓′(π‘₯).

Let us see an example of how we can use this to find the equation of the tangent line to a curve at a point.

Example 1: Finding the Equation of the Tangent to the Curve of a Polynomial Function at a Given Value for π‘₯

Find the equation of the tangent to the curve 𝑦=βˆ’2π‘₯+8π‘₯βˆ’19 at π‘₯=2.

Answer

To find the equation of the tangent to a curve at a point, we need two pieces of information: the coordinates of the point and the slope of the curve at this point.

The question wants us to find the tangent when π‘₯=2, so the π‘₯-coordinate is 2. We can find the 𝑦-coordinate for this value of π‘₯ by substituting π‘₯=2 into the equation of our curve: 𝑦=βˆ’2(2)+8(2)βˆ’19=βˆ’16+32βˆ’19=βˆ’3.

This gives us the point (2,βˆ’3) on our curve that our tangent line must pass through.

Next, we need the slope of the curve when π‘₯=2; to find this we need to differentiate: dddd𝑦π‘₯=π‘₯ο€Ήβˆ’2π‘₯+8π‘₯βˆ’19=βˆ’6π‘₯+16π‘₯.

We then substitute in π‘₯=2 to find the slope of the tangent line at this point: dd𝑦π‘₯|||=βˆ’6(2)+16(2)=βˆ’24+32=8.ο—οŠ²οŠ¨οŠ¨

Now, to find the equation of our tangent line, we recall that the equation of a line of slope π‘š passing through (π‘₯,𝑦) is given by π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

In our case, we have π‘¦βˆ’(βˆ’3)=8(π‘₯βˆ’2)𝑦+3=8π‘₯βˆ’16π‘¦βˆ’8π‘₯+19=0.

Therefore, the equation of the tangent line to our curve at π‘₯=2 is given by the equation π‘¦βˆ’8π‘₯+19=0.

This method gives us a really useful result for finding the equation of a tangent line to a curve at a point (provided the derivative of the curve exists at this point). If we have a curve 𝑦=𝑓(π‘₯) and a point (π‘₯,𝑦) on our curve, then the tangent line to our curve at this point must have slope 𝑓′(π‘₯). Recall that the point-slope form of a straight line tells us the equation of a line passing through the point (π‘₯,𝑦) with slope π‘š, given by π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

So, if we know the slope of this tangent line and a point it passes through, this is all the information we need to find its equation.

Definition: The Equation of a Tangent Line

The equation of the tangent line to the curve 𝑦=𝑓(π‘₯) at the point (π‘₯,𝑦) is given by π‘¦βˆ’π‘¦=𝑓′(π‘₯)β‹…(π‘₯βˆ’π‘₯).

This formula assumes 𝑓 is differentiable at π‘₯. If our function 𝑓 is not differentiable at π‘₯, then we cannot use this formula to find the equation of the tangent line at π‘₯. Instead, we will need to consider the problem graphically. Consider trying to find the tangent lines to the following two curves at π‘₯=0.

In our first graph, we can see that the curve is not defined at π‘₯=0, which also means it is not differentiable at π‘₯=0. If our curve is not defined at π‘₯=0, then it cannot have a tangent line at this value of π‘₯.

In our second graph, we can see that the tangent line at π‘₯=0 should be vertical. If we were to differentiate our function 𝑓(π‘₯)=√π‘₯,𝑓′(π‘₯)=13π‘₯.

We could then attempt to find the slope of our curve at π‘₯=0: 𝑓′(0)=13ο€½0=10.

We then see this is undefined. When 𝑓 is not differentiable at a point, sketching the graph can often help us determine whether this point has a vertical tangent line.

So far, we have been focused on tangent lines. However, there is another important type of line we need to consider called a normal line. A normal line to a curve at a point is very similar to the tangent line; the only difference is that the normal line will be perpendicular to the tangent line.

Definition: The Normal Line to a Curve at a Point

For a curve 𝑦=𝑓(π‘₯) and point (π‘₯,𝑦) on the curve, we say that the line π‘Žπ‘₯+𝑏𝑦+𝑐=0 is the normal line to the curve at the point (π‘₯,𝑦) if

  • the point (π‘₯,𝑦) lies on our line;
  • this line is perpendicular to the tangent line of our curve at this point.

It is worth pointing out that we can define the normal line based on the information that it is perpendicular to the curve at this point; however, it can be easier to think about the line being perpendicular to the tangent line.

Finding the equation of the normal line will take a little bit more work since the derivative of the function only gives us the slope of the tangent line. To find the equation of the normal to a curve at a point, we need a point on the line and its slope to find the point-slope equation.

Since we already know the normal passes through the point (π‘₯,𝑦), we only need to find the slope of the normal line to find its point-slope equation.

We want to find an expression for the slope of our normal line in terms of the slope of the tangent line. To do this, we first note that if the tangent line is horizontal, then the normal line needs to be perpendicular, so it must be a vertical line, and the same will be true in reverse.

This means we can now assume we are dealing with a tangent line which is not horizontal or vertical. So, we can write the equation of the tangent line at the point (π‘₯,𝑦) in the form 𝑦=π‘šπ‘₯+𝑐, where π‘š is not zero. We will also say that the equation of our normal line is 𝑦=𝑛π‘₯+𝑑.

To find an expression for the slope of our normal line, 𝑛, we will start with a sketch.

To find an expression for the slope, 𝑛, we will add in the line π‘₯=π‘₯+1.

We can now see we have a triangle with a right angle at (π‘₯,𝑦). We can find the coordinates of the vertices. We already know the coordinates of the vertex at (π‘₯,𝑦); we will compare the other two vertices to this point to find their coordinates.

Since we chose our vertical line to be one unit to the right of (π‘₯,𝑦), our other two vertices will be one unit to the right. We can find the 𝑦-coordinates of these two vertices by recalling that the slope of a line tells us the change in 𝑦 for every 1 unit change in π‘₯. Since the change in π‘₯ is 1 unit for both vertices, the change in 𝑦 for the tangent line will be π‘š and the change in 𝑦 for the normal line will be 𝑛.

The coordinates of the vertices are therefore (π‘₯,𝑦), (π‘₯+1,𝑦+π‘š), and (π‘₯+1,𝑦+𝑛).

We can then find the lengths of each side in this triangle, by using the distance between two points formula. We will not go through the individual calculation steps, but the results are shown below.

Finally, since this is a right triangle, we can apply the Pythagorean theorem to this triangle, where the hypotenuse of this triangle is the side opposite the right angle, in this case the vertical line. In fact, because this is a vertical line, the length of this line is the difference between its 𝑦-coordinates. In this case, we will use |π‘šβˆ’π‘›|, since we do not know the sign of π‘šβˆ’π‘›: ο€»βˆš1+π‘šο‡+ο€»βˆš1+𝑛=|π‘šβˆ’π‘›|1+π‘š+1+𝑛=π‘šβˆ’2π‘šπ‘›+𝑛2=βˆ’2π‘šπ‘›π‘›=βˆ’1π‘š.

This gives us an equation to find the slope of our normal line; it is the negative of the reciprocal of the slope of the tangent line. We also know how to find the slope of the tangent by using the derivative.

This means we can use the fact that π‘š=𝑓′(π‘₯) to find a formula for the equation of the normal line.

Definition: The Equation of a Normal Line to a Curve

If 𝑓′(π‘₯)β‰ 0, then the equation of the normal line to 𝑦=𝑓(π‘₯) at the point (π‘₯,𝑦) is given by π‘¦βˆ’π‘¦=βˆ’1𝑓′(π‘₯)(π‘₯βˆ’π‘₯).

If the slope of our curve at π‘₯ is zero, then the normal line at this point will be vertical, and its equation will be π‘₯=π‘₯. If the slope of our curve is undefined at a point, there are two possibilities.

  1. The tangent line to the curve at this point is vertical; in this case, the normal line will be horizontal.
  2. The tangent line to the curve at this point does not exist; in this case, the normal line does not exist.

Let us see some examples of applying these formulae to some curves.

Example 2: Finding the Equation of the Normal to the Curve of a Polynomial Function at a Point with the Given π‘₯-Coordinate

Find the equation of the normal to the curve 𝑦=βˆ’2π‘₯βˆ’7π‘₯+2 at π‘₯=βˆ’2.

Answer

We want to find the equation of the normal line to a curve at a point. To do this, we need to find a point on the line and its slope. We can find a point on the line by substituting π‘₯=βˆ’2 into the equation of our curve: 𝑦=βˆ’2(βˆ’2)βˆ’7(βˆ’2)+2=βˆ’10.

So, the normal line passes through the point (βˆ’2,βˆ’10).

Next, we recall that we can find the slope of the normal line by using the slope of the tangent line.

If we let 𝑓(π‘₯)=βˆ’2π‘₯βˆ’7π‘₯βˆ’2, then the tangent line will have slope 𝑓′(βˆ’2): 𝑓′(π‘₯)=βˆ’6π‘₯βˆ’14π‘₯,𝑓′(βˆ’2)=βˆ’6(βˆ’2)βˆ’14(βˆ’2)=4.

We have shown that the tangent will have slope 4, but the slope of the normal line is the negative of the reciprocal of this value: normallineslope=βˆ’1𝑓′(βˆ’2)=βˆ’14.

We know the slope of our line is βˆ’14 and we also know it passes through (βˆ’2,βˆ’10). This gives us the equation π‘¦βˆ’(βˆ’10)=βˆ’14(π‘₯βˆ’(βˆ’2))𝑦+10=βˆ’14(π‘₯+2)βˆ’4π‘¦βˆ’40=π‘₯+24𝑦+π‘₯+42=0.

Therefore, the equation of the normal line to the curve when π‘₯=βˆ’2 is given by 4𝑦+π‘₯+42=0.

In our next example, we will consider how to find the points on a curve where its tangent line at that point will be parallel to a given line.

Example 3: Finding the π‘₯-Coordinate of the Point on the Curve of a Quadratic Function Where the Tangent Line Is Parallel to the π‘₯-Axis

What is the π‘₯-coordinate of the point where the tangent line to 𝑦=π‘₯+12π‘₯+11 is parallel to the π‘₯-axis?

Answer

We want to find the π‘₯-coordinate where the tangent line to this curve will be parallel to the π‘₯-axis. We know that the π‘₯-axis is horizontal, so any line parallel to this must also be horizontal; in other words, the slope of this tangent line must be equal to zero.

We also know that, for a curve 𝑦=𝑓(π‘₯), the slope of the tangent line to this curve at the point (π‘₯,𝑦) will be its derivative at that point, in this case 𝑓′(π‘₯).

Therefore, in order to solve this question, we must find the values of π‘₯ for which the derivative is equal to zero.

Since our function is a polynomial, we can differentiate it by using the power rule for differentiation: ddπ‘₯ο€Ήπ‘₯+12π‘₯+11=2π‘₯+12.

By setting our derivative equal to zero, we can find the values of π‘₯ where the tangent line is parallel to the π‘₯-axis: 2π‘₯+12=0π‘₯=βˆ’6.

Therefore, the tangent line to this curve is parallel to the π‘₯-axis when π‘₯=βˆ’6.

In our next example, we will find the equation of a tangent line to a curve that makes a specific angle with the positive π‘₯-axis.

Example 4: Finding the Equation of the Tangent to the Curve of a Cubic Function given the Angle the Tangent Makes with the π‘₯-Axis

Find the equation of the tangent to the curve 𝑦=π‘₯+9π‘₯+26π‘₯ that makes an angle of 135∘ with the positive π‘₯-axis.

Answer

In this question, we want to find the tangent to a curve which makes an angle of 135∘ with the positive π‘₯-axis. This means that to answer this question we are going to need to work out the corresponding slope for the line that makes this angle with the positive π‘₯-axis.

First, if we translate a line, it will not change the angle it makes with the positive π‘₯-axis. So, we can start by sketching our line passing through the origin (since this is the simplest case), making an angle of 135∘ with the positive π‘₯-axis. This will have the same slope as our tangent line.

We can then see that 135=90+45∘∘∘, giving us the following diagram:

There are then two ways to find the slope of this line; we can use the fact that the slope of a line is the tangent of the angle it makes with the positive π‘₯-axis; in this case, tan135=βˆ’1∘; or we could use trigonometry to find the slope of this line. In either case, we see the question wants us to find the tangent line with slope βˆ’1.

The slope of the tangent line at a point is the same as the derivative of the curve at that point, so we want to set the derivative equal to βˆ’1 and solve for π‘₯: dddd𝑦π‘₯=π‘₯ο€Ήπ‘₯+9π‘₯+26π‘₯=3π‘₯+18π‘₯+26.

This means we want to solve 3π‘₯+18π‘₯+26=βˆ’13π‘₯+18π‘₯+27=0π‘₯+6π‘₯+9=0(π‘₯+3)=0.

The solution is π‘₯=βˆ’3.

To find the equation of the tangent line to the curve when π‘₯=βˆ’3, we need to find the coordinates of a point on the line. We can find this by substituting π‘₯=βˆ’3 into the equation for our curve: 𝑦=(βˆ’3)+9(βˆ’3)+26(βˆ’3)=βˆ’24.

The tangent we are looking for passes through (βˆ’3,βˆ’24) and has a slope of βˆ’1.

We can use this to find the equation of the line π‘¦βˆ’(βˆ’24)=βˆ’1(π‘₯βˆ’(βˆ’3))𝑦+24=βˆ’(π‘₯+3)𝑦+24=βˆ’π‘₯βˆ’3𝑦+π‘₯+27=0.

Therefore, the tangent to the curve which makes an angle of 135∘ with the positive π‘₯-axis has the equation 𝑦+π‘₯+27=0.

It is not always as simple as differentiating a polynomial to find the slope of the tangent or normal line to our curve. We will sometimes need to apply other derivative rules to help us find this value. Let us see an example of this.

Example 5: Finding the Equation of the Normal to the Curve of a Function Involving Trigonometric Functions at a Given π‘₯-Coordinate

Find all points with π‘₯-coordinates in [0,πœ‹[ where the curve 𝑦=2π‘₯sin has a tangent that is parallel to the line 𝑦=βˆ’π‘₯βˆ’18.

Answer

First, for a line to be parallel to the line 𝑦=βˆ’π‘₯βˆ’18, it must have the same slope. Therefore, our tangent line must have a slope of βˆ’1. Recall that the slope of the tangent line to a curve 𝑦=𝑓(π‘₯) at the point π‘₯ is 𝑓′(π‘₯). In our case, 𝑓(π‘₯)=2π‘₯sin. We can differentiate this using the fact that, for any constant 𝑛 where π‘₯ is measured in radians, ddsincosπ‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯.

Hence, 𝑓′(π‘₯)=22π‘₯.cos

Setting the slope of our tangent line to be equal to βˆ’1 gives us 22π‘₯=βˆ’1.cos

We can then solve this for π‘₯ in the interval [0,πœ‹[: cos2π‘₯=βˆ’12.

We can sketch this as follows.

This has solutions π‘₯=πœ‹3 and π‘₯=2πœ‹3. Finally, we need to find the coordinates of these points by substituting these π‘₯-values into the function sin2π‘₯: sinandsinο€»2ο€»πœ‹3=√32ο€Ό2ο€Ό2πœ‹3=βˆ’βˆš32, giving us the coordinates ο€Ώπœ‹3,√32 and ο€Ώ2πœ‹3,βˆ’βˆš32.

Hence, the coordinates of the points with π‘₯-coordinates in [0,πœ‹[, where the curve 𝑦=2π‘₯sin has a tangent that is parallel to the line 𝑦=βˆ’π‘₯βˆ’18, are ο€Ώπœ‹3,√32 and ο€Ώ2πœ‹3,βˆ’βˆš32.

In our final example, we will determine the point of intersection between two curves where the intersection is orthogonal.

Example 6: Finding the Point Where Two Quadratic Curves Intersect Orthogonally

The curves 𝑦=2π‘₯βˆ’3π‘₯βˆ’2 and 𝑦=βˆ’3π‘₯+5π‘₯βˆ’5 intersect orthogonally at a point. What is this point?

Answer

We say that two curves intersect orthogonally if they intersect at right angles. Equivalently, the tangent lines to both curves at the point of intersection are orthogonal (meet at right angles).

We recall that the slope of a curve at a point is given by the value of its derivative at this point. We start by finding all of the points of intersection between these two curves by setting the functions to be equal to one another and solving for π‘₯: 2π‘₯βˆ’3π‘₯βˆ’2=βˆ’3π‘₯+5π‘₯βˆ’52π‘₯βˆ’3π‘₯βˆ’2+3π‘₯βˆ’5π‘₯+5=05π‘₯βˆ’8π‘₯+3=0(5π‘₯βˆ’3)(π‘₯βˆ’1)=0.

Hence, the curves intersect when π‘₯=35 and when π‘₯=1. We need to find the slopes of both curves at each of these π‘₯-values to determine if they are orthogonal. We do this by differentiating each curve by using the power rule for differentiation. For the first curve: ddπ‘₯ο€Ή2π‘₯βˆ’3π‘₯βˆ’2=4π‘₯βˆ’3.

We can use this to find the slope at both π‘₯-values.

At π‘₯=35, ddπ‘₯ο€Ή2π‘₯βˆ’3π‘₯βˆ’2||=4ο€Ό35οˆβˆ’3=βˆ’35.οŠ¨ο—οŠ²οŽ’οŽ€

At π‘₯=1, ddπ‘₯ο€Ή2π‘₯βˆ’3π‘₯βˆ’2||=4(1)βˆ’3=1.οŠ¨ο—οŠ²οŠ§

We can do the same with the second curve: ddπ‘₯ο€Ήβˆ’3π‘₯+5π‘₯βˆ’5=βˆ’6π‘₯+5.

At π‘₯=35, ddπ‘₯ο€Ήβˆ’3π‘₯+5π‘₯βˆ’5||=βˆ’6ο€Ό35+5=725.οŠ¨ο—οŠ²οŽ’οŽ€

At π‘₯=1, ddπ‘₯ο€Ήβˆ’3π‘₯+5π‘₯βˆ’5||=βˆ’6(1)+5=βˆ’1.οŠ¨ο—οŠ²οŠ§

Since the slopes of all four lines are nonzero, for the lines to be orthogonal, their slopes must be negative reciprocals of each other. Taking the negative reciprocal of βˆ’35 gives us βˆ’ο€Όβˆ’35=βˆ’ο€Όβˆ’53=53, which is not equal to the slope of the second curve at this point, so the tangent lines are not orthogonal.

Taking the negative reciprocal of 1 gives us βˆ’(1)=βˆ’1, which is equal to the slope of the second curve at this point. Hence, the tangent lines are orthogonal.

We can find the coordinates of this point by substituting π‘₯=1 into the equation of either curve: 𝑦=2(1)βˆ’3(1)βˆ’2=2βˆ’3βˆ’2=βˆ’3.

Hence, the curves intersect orthogonally at the point (1,βˆ’3).

Let us finish by recapping some of the things we covered when finding the equations of tangent lines and normal lines to curves.

Key Points

  • The equation of the tangent line to the curve 𝑦=𝑓(π‘₯) at the point (π‘₯,𝑦) is given by π‘¦βˆ’π‘¦=𝑓′(π‘₯)(π‘₯βˆ’π‘₯).
  • If 𝑓′(π‘₯)β‰ 0, then the equation of the normal line to 𝑦=𝑓(π‘₯) at the point (π‘₯,𝑦) is given by π‘¦βˆ’π‘¦=βˆ’1𝑓′(π‘₯)(π‘₯βˆ’π‘₯).
  • If 𝑓′(π‘₯)=0, then the tangent line to 𝑦=𝑓(π‘₯) at the point (π‘₯,𝑦) is horizontal and has the equation 𝑦=𝑦.
  • If 𝑓′(π‘₯)=0, the normal line to 𝑦=𝑓(π‘₯) at the point (π‘₯,𝑦) is vertical and has the equation π‘₯=π‘₯.
  • Two curves intersect orthogonally at the point (π‘₯,𝑦) if both curves intersect this point and the slopes of their tangents at this point are orthogonal.
  • If 𝑓′(π‘₯) is undefined, we may still be able to find the tangent and normal lines at π‘₯. However, it is not always possible.

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