In this lesson I will explain everything you need to know to calculate the **equation of the tangent line to a curve defined by a function**.

Índice de Contenidos

## How to calculate the tangent line to a curve at a point

To calculate the equation of the tangent line, we will use the point-slope equation:

So we need to know the coordinates of a point P that passes through the line, which will be the point where the line is tangent to the curve and also the slope of that line:

To calculate the coordinates of the point where the line is tangent, if you give us the x-coordinate of the point, we only have to replace the x-coordinate with the y-coordinate in the function and we get the y-coordinate, since the y-coordinate coincides with the value of the function for that x-coordinate.

On the other hand, the slope of the line tangent to a point of a function coincides with the value of the value derived from the function at that point:

So by deriving the function of the curve and replacing it with the value of x of the point where the curve is tangent, we will obtain the value of the slope m.

If we know the slope m, but we do not know the coordinates of the point where the line is tangent to the curve, we can clear the x from the previous formula.

Other times we are asked that the tangent line should be parallel to another given line. Two parallel lines have the same slope, so from the given line, we can obtain the slope.

To be clear, let’s look at each of these cases with several exercises solved.

## Determined straight line calculation exercises tangent to a curve

### Exercise 1

Find the equation of the line tangent to the cure:

at point x=-1.

As I mentioned before, for the calculation of the tangent line equation, we will use the point-slope equation:

We need to obtain the coordinates of point P, which will be the point where the line is tangent to the curve and the slope of the tangent line:

We are going to calculate the coordinates of point P. We already have the x-coordinate, since it is given to us by the statement. To calculate the “y” coordinate we only have to replace the x by -1 in the y function and operate:

The y coordinate is y=3.

Therefore, the coordinates of point P, where the line is tangent is:

Now we are going to calculate the slope of the tangent line, which will be equal to the one derived from the function at point P, that is, when x=-1:

Therefore, we calculate the derivative of the function:

We obtain the value of the derivative of the function for x=-1:

The slope of the line is equal to 2.

We already have what we need to calculate the equation of the line:

Therefore, in the point-slope equation:

We replace the coordinates of the point and the slope by its values:

And we operate:

Coming up to the equation of the tangent line we were looking for.

Let’s go with a slightly different exercise.

### Exercise 2

Find the equation of the line tangent to the curve:

so that the tangent is parallel to the equation line:

In this case we don’t know any coordinate of the point where the line is tangent to the curve, but they give us the data that it must be parallel to another line, so they are indirectly giving us the slope.

As before, the equation of the tangent line will be obtained from the point-slope equation:

So we need to know the coordinates of the point and the slope:

The slope is obtained from the line given to us by the statement. Let’s calculate the slope of that line. To do this, clear the “y” and the number in front of the x will be the slope:

The slope of the line given by the statement is -3, and since it is parallel to the tangent line, the slope of the tangent line is also -3:

Now let’s calculate the coordinates of point P.

We know that the slope at any point of the curve is equal to the value of the derivative at that point:

As we already know the slope, we only have to calculate the slope and we will obtain the x-coordinate that complies with the slope of the line tangent to that point being -3

We calculate the derivative of the function:

The derivative of the function at any point shall be:

We match it to the value of the slope:

And we have a first-degree equation left, from which we have to clear X0, which is the x-coordinate of the point we are looking for:

To calculate the “y” coordinate, we obtain the value of the function for x=-1, so we replace the x with -1 in the function y we operate:

The value of the function in x=-1 is 0, so the y=0 coordinate.

We already have the coordinates of the point and the value of the slope:

In the slope point equation:

We replace the slope and coordinates of the point by its values:

We operate:

And finally we come to the equation of the tangent line that meets the conditions required by the result.

We’re going to solve another exercise like this, a little more complicated.

### Exercise 3

Find, if they exist, the coordinates “x” and “y” of the points on the curve defined by the formula:

Where the tangent line is parallel to the line’r’, the equation of which is:

This exercise is very similar to the previous one, only the calculations are a little more complex. Besides, they are not asking for the equation of the line, but the coordinates of the points where the line is tangent.

We begin by calculating the slope of the line given to us by the statement. To do this we must clear the y:

Separate the fraction in two terms and the value of the slope is the fraction in front of the x:

As it is parallel to the tangent line we are looking for, the slope of the tangent line is the same:

As in the previous exercise, the slope of the line tangent to a point of the cure will be equal to the value of the derivative at that point:

So from this equality, we will obtain the values of x for which the slope has that value.

Let’s calculate the derivative of the function. In this case we have a ratio of two functions:

The derivative of a quotient is equal to the following formula:

We apply the formula:

And we operate (very carefully the minus sign that separates both terms in the denominator):

The derivative of the function at any point will be:

We match this derivative to the previously calculated slope:

We have one equation left from which we have to clear X0.

First of all, we multiply them in a cross to eliminate the denominators:

We pass all the terms on to the same member:

We operate:

We’ve come to a complete second degree equation, which we’ve come to solve:

We obtain two solutions of x: x1=5 and x2=1

To obtain the y-coordinate for each x-value, we calculate the value of the function:

First we replace the x with 5 and operate:

The value of “y” is -5/2, so the first point where the line is tangent is:

We do the same with x=1

So the coordinates of the other point are:

### Exercise 4

It calculates the values of a, b and c in the function f(x)=ax²+bx+c, knowing that it passes through the point (0.1) and that the slope of the tangent line at the point (2.-1) is equal to 0.

We have the following function:

If it passes through point (0.1) it means that the value of the function in x=0 is 1:

To obtain the value of the function in x=o, replace the x with 0 and equal it to 1:

Where we get the value of c:

On the other hand, the statement also tells us about the slope at the point (2.-1), which means that the function also passes through that point, so that the function in x=2 is equal to -1:

To obtain the value of the function in x=2, replace the x with 2 and equal it to -1:

We operate and replace c with its value. We’ve got some left:

We leave the terms with a and b in the first member and pass the numbers to the second member:

We have an equation that depends on a and b. We need another equation to form a system of two equations with two unknowns and be able to find their values.

The slope of the line tangent to the function at a point is equal to the value of the derivative at that point. First, we get the derivative of f(x):

The statement tells us that the slope at x=2 is equal to 0, or the same thing:

To obtain the value of the derivative in x=2, replace x with 2 and equal it to 0:

We operate and we have it:

We already have the second equation that we were missing, which together with the previous one, forms the following system, which we will now solve:

From the second equation we clear b:

We substitute the expression of b in the first equation:

We operate to remove the parenthesis:

We operate on the first member:

We clear “a” and solve:

This value of “a” is replaced by the expression where we clear b:

We replace and operate, reaching the value of b:

Therefore the values of a, b and c are: