Calculation of the line tangent to a curve at a point. Exercises resolved.

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.

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:

line tangent to a curve

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:

equation of the line tangent exercises solved

exercises straight tangent

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:

recta tangente

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:

straight tangent exercises

at point x=-1.

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

ejercicios de recta tangente

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:

equation straight tangent solved exercises

straight line tangent to a curve at a point resolved exercises

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:

equation of the tangent line

straight tangent and normal resolved exercises

The y coordinate is y=3.

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

ejercicios de tangente

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:

straight line tangent to a curve at a point

Therefore, we calculate the derivative of the function:

exercises of the tangent line

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

equation of the line tangent to the curve

equation of the tangent line at a given point

The slope of the line is equal to 2.

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

slope of the line tangent examples

equuacion recta tangente

Therefore, in the point-slope equation:

tangente ejercicios resueltos

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

line tangent and normal line to a curve at an example point

And we operate:

tangent exercises resolved

tangent to a curve

straight tangent resolved exercises

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:

examples of tangent resolved

so that the tangent is parallel to the equation line:

derived tangent line exercises solved

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:

calculate tangent line

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

tangent problems solved

calculadora de recta tangente

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:

line tangent and line normal to a curve at a point

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:

ejercicio recta tangente

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:

exercises solved from tangent line

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:

straight line tangent to curve

The derivative of the function at any point shall be:

calculate tangent line at one point

We match it to the value of the slope:

straight line tangent at one point

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:

tangent of a curve

exercices recta tangente 2 bachillerato

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:

straight line tangent to an exercise curve

normal and tangent line exercises

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:

straight line exercises tangent to a curve

equation exercises of the tangent line

In the slope point equation:

straight line tangent to a circumference resolved exercises

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

line tangent problems

We operate:

exercises of tangent lines

line tangent passing through a point

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:

exercises resolved straight line tangent

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

the equation of the line tangent to the curve with equation x2-3x at the point (-2, 1), 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:

slope of the line tangent to a curve

slope of a straight line tangent resolved exercises

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

equuación recta tangente

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

equation of the derived tangent line

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:

recta tangente formula

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:

equation of the tangent line exercises

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

equation of the tangent line

We apply the formula:

equation of the tangent line at one point

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

straight equation tangent to a curve

exercises equation straight tangent

The derivative of the function at any point will be:

tangent exercises

We match this derivative to the previously calculated slope:

calculo de la recta tangente

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:

straight tangent examples

We pass all the terms on to the same member:

slope of a curve examples

We operate:

tangent exercises solved easily

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

resolved tangent exercises

calculate the tangent line

tangent exercise

straight tangent exercises resolved

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:

find the tangent line to a curve

First we replace the x with 5 and operate:

find the equation of the tangent line

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

calculadora recta tangente

We do the same with x=1

exercices recta tangente 1 bachillerato

So the coordinates of the other point are:

equation of the line tangent to the curve

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:

line tangent to a function

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

examples of tangent line

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

exercises tangent

Where we get the value of c:

the equation of the line tangent to the curve with equation x2-3x at the point (0, 1), is:

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:

equation of the tangent line at a given point

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

exercises straight tangent and normal

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

calculate the tangent line at one point

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

exercises with tangent

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):

equation straight tangent at one point

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

equation of the tangent and normal line derived from resolved exercises

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

formula recta tangente

We operate and we have it:

exercises solved straight tangent 2 high school

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:

tangentes a una curva

From the second equation we clear b:

the equation of the line tangent to the graph f(x)=x2 at the point (1,2) is:

We substitute the expression of b in the first equation:

line tangent to a function at a point

We operate to remove the parenthesis:

equation of the line tangent to the curve at one point

We operate on the first member:

how to calculate the tangent line

We clear “a” and solve:

equation of the tangent line to a curve

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

equation of the tangent line

We replace and operate, reaching the value of b:

tangente ejerciciosTherefore the values of a, b and c are:

pendiente de la recta tangente