The slope of the tangent to the line. Equation of the tangent to the graph of a function

In mathematics, one of the parameters that describes the position of a line on the Cartesian coordinate plane is the angular coefficient of this line. This parameter characterizes the slope of the straight line to the abscissa axis. To understand how to find the slope, first recall the general form of the equation of a straight line in the XY coordinate system.

In general, any line can be represented by the expression ax+by=c, where a, b and c are arbitrary real numbers, but a 2 + b 2 ≠ 0.

Using simple transformations, such an equation can be brought to the form y=kx+d, in which k and d are real numbers. The number k is the slope, and the equation of a line of this type is called an equation with a slope. It turns out that to find the slope, you simply need to reduce the original equation to the form indicated above. For a more complete understanding, consider a specific example:

Problem: Find the slope of the line given by the equation 36x - 18y = 108

Solution: Let's transform the original equation.

Answer: The required slope of this line is 2.

If, during the transformation of the equation, we received an expression like x = const and as a result we cannot represent y as a function of x, then we are dealing with a straight line parallel to the X axis. The angular coefficient of such a straight line is equal to infinity.

For lines expressed by an equation like y = const, the slope is zero. This is typical for straight lines parallel to the abscissa axis. For example:

Problem: Find the slope of the line given by the equation 24x + 12y - 4(3y + 7) = 4

Solution: Let's bring the original equation to its general form

24x + 12y - 12y + 28 = 4

It is impossible to express y from the resulting expression, therefore the angular coefficient of this line is equal to infinity, and the line itself will be parallel to the Y axis.

Geometric meaning

For a better understanding, let's look at the picture:

In the figure we see a graph of a function like y = kx. To simplify, let’s take the coefficient c = 0. In the triangle OAB, the ratio of side BA to AO will be equal to the angular coefficient k. At the same time, the ratio BA/AO is the tangent of the acute angle α in the right triangle OAB. It turns out that the angular coefficient of the straight line is equal to the tangent of the angle that this straight line makes with the abscissa axis of the coordinate grid.

Solving the problem of how to find the angular coefficient of a straight line, we find the tangent of the angle between it and the X axis of the coordinate grid. Boundary cases, when the line in question is parallel to the coordinate axes, confirm the above. Indeed, for a straight line described by the equation y=const, the angle between it and the abscissa axis is zero. The tangent of the zero angle is also zero and the slope is also zero.

For straight lines perpendicular to the x-axis and described by the equation x=const, the angle between them and the X-axis is 90 degrees. The tangent of a right angle is equal to infinity, and the angular coefficient of similar straight lines is also equal to infinity, which confirms what was written above.

Tangent slope

A common task often encountered in practice is also to find the slope of a tangent to the graph of a function at a certain point. A tangent is a straight line, therefore the concept of slope is also applicable to it.

To figure out how to find the slope of a tangent, we will need to recall the concept of derivative. The derivative of any function at a certain point is a constant numerically equal to the tangent of the angle that is formed between the tangent at the specified point to the graph of this function and the abscissa axis. It turns out that to determine the angular coefficient of the tangent at the point x 0, we need to calculate the value of the derivative of the original function at this point k = f"(x 0). Let's look at the example:

Problem: Find the slope of the line tangent to the function y = 12x 2 + 2xe x at x = 0.1.

Solution: Find the derivative of the original function in general form

y"(0.1) = 24. 0.1 + 2. 0.1. e 0.1 + 2. e 0.1

Answer: The required slope at point x = 0.1 is 4.831

Tangent is a straight line passing through a point on the curve and coinciding with it at this point up to first order (Fig. 1).

Another definition: this is the limiting position of the secant at Δ x→0.

Explanation: Take a straight line intersecting the curve at two points: A And b(see picture). This is a secant. We will rotate it clockwise until it finds only one common point with the curve. This will give us a tangent.

Strict definition of tangent:

Tangent to the graph of a function f, differentiable at the point xO, is a straight line passing through the point ( xO; f(xO)) and having a slope f′( xO).

The slope has a straight line of the form y =kx +b. Coefficient k and is slope this straight line.

The angular coefficient is equal to the tangent of the acute angle formed by this straight line with the abscissa axis:

k = tan α

Here angle α is the angle between the straight line y =kx +b and positive (that is, counterclockwise) direction of the x-axis. It is called angle of inclination of a straight line(Fig. 1 and 2).

If the angle of inclination is straight y =kx +b acute, then the slope is a positive number. The graph is increasing (Fig. 1).

If the angle of inclination is straight y =kx +b is obtuse, then the slope is a negative number. The graph is decreasing (Fig. 2).

If the straight line is parallel to the x-axis, then the angle of inclination of the straight line is zero. In this case, the slope of the line is also zero (since the tangent of zero is zero). The equation of the straight line will look like y = b (Fig. 3).

If the angle of inclination of a straight line is 90º (π/2), that is, it is perpendicular to the abscissa axis, then the straight line is given by the equality x =c, Where c– some real number (Fig. 4).

Equation of the tangent to the graph of a functiony = f(x) at point xO:

Example: Find the equation of the tangent to the graph of the function f(x) = x 3 – 2x 2 + 1 at the point with abscissa 2.

Solution .

We follow the algorithm.

1) Touch point xO is equal to 2. Calculate f(xO):

f(xO) = f(2) = 2 3 – 2 ∙ 2 2 + 1 = 8 – 8 + 1 = 1

2) Find f′( x). To do this, we apply the differentiation formulas outlined in the previous section. According to these formulas, X 2 = 2X, A X 3 = 3X 2. Means:

f′( x) = 3X 2 – 2 ∙ 2X = 3X 2 – 4X.

Now, using the resulting value f′( x), calculate f′( xO):

f′( xO) = f′(2) = 3 ∙ 2 2 – 4 ∙ 2 = 12 – 8 = 4.

3) So, we have all the necessary data: xO = 2, f(xO) = 1, f ′( xO) = 4. Substitute these numbers into the tangent equation and find the final solution:

y = f(xO) + f′( xO) (x – x o) = 1 + 4 ∙ (x – 2) = 1 + 4x – 8 = –7 + 4x = 4x – 7.

Answer: y = 4x – 7.

You are already familiar with the concept of a tangent to the graph of a function. The graph of the function f differentiable at the point x 0 near x 0 practically does not differ from the tangent segment, which means it is close to the secant segment l passing through the points (x 0 ; f (x 0)) and (x 0 +Δx; f ( x 0 + Δx)). Any of these secants passes through point A (x 0 ; f (x 0)) of the graph (Fig. 1). In order to uniquely define a line passing through a given point A, it is enough to indicate its slope. The angular coefficient Δy/Δx of the secant as Δх→0 tends to the number f ‘(x 0) (we will take it as the angular coefficient of the tangent) They say that the tangent is the limiting position of the secant at Δх→0.

If f'(x 0) does not exist, then the tangent either does not exist (like the function y = |x| at the point (0; 0), see figure) or is vertical (like the graph of the function at the point (0 ; 0), Fig. 2).

So, the existence of a derivative of the function f at the point xo is equivalent to the existence of a (non-vertical) tangent at the point (x 0, f (x 0)) of the graph, while tangent slope is equal to f" (x 0). This is geometric meaning of derivative

The tangent to the graph of a function f differentiable at the point xo is a straight line passing through the point (x 0 ; f (x 0)) and having an angular coefficient f ‘(x 0).

Let's draw tangents to the graph of the function f at points x 1, x 2, x 3 (Fig. 3) and note the angles they form with the abscissa axis. (This is the angle measured in the positive direction from the positive direction of the axis to the straight line.) We see that angle α 1 is acute, angle α 3 is obtuse, and angle α 2 is zero, since straight line l is parallel to the Ox axis. The tangent of an acute angle is positive, the tangent of an obtuse angle is negative, tan 0 = 0. Therefore

F"(x 1)>0, f’(x 2)=0, f’(x 3)
Constructing tangents at individual points allows you to more accurately sketch graphs. So, for example, to construct a sketch of a graph of the sine function, we first find that at points 0; π/2 and π derivative of sine is equal to 1; 0 and -1 respectively. Let's construct straight lines passing through the points (0; 0), (π/2,1) and (π, 0) with angular coefficients of 1, 0 and -1, respectively (Fig. 4). It remains to fit into the resulting trapezoid formed by these straight lines and straight line Ox, graph of the sine so that for x equal to 0, π/2 and π, it touches the corresponding straight lines.

Note that the graph of the sine in the vicinity of zero is practically indistinguishable from the straight line y = x. Let, for example, the scales along the axes be chosen so that a unit corresponds to a segment of 1 cm. We have sin 0.5 ≈ 0.479425, i.e. |sin 0.5 - 0.5| ≈ 0.02, and on the chosen scale this corresponds to a segment 0.2 mm long. Therefore, the graph of the function y = sin x in the interval (-0.5; 0.5) will deviate (in the vertical direction) from the straight line y = x by no more than 0.2 mm, which approximately corresponds to the thickness of the drawn line.

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You will need

• - mathematical reference book;
• - notebook;
• - a simple pencil;
• - pen;
• - protractor;
• - compass.

Instructions

Please note that the graph of the differentiable function f(x) at the point x0 is no different from the tangent segment. Therefore, it is quite close to the segment l, to the one passing through the points (x0; f(x0)) and (x0+Δx; f(x0 + Δx)). To specify a straight line passing through point A with coefficients (x0; f(x0)), specify its slope. Moreover, it is equal to Δy/Δx secant tangent (Δх→0), and also tends to the number f‘(x0).

If there are no values ​​for f‘(x0), then there is no tangent, or it runs vertically. Based on this, the derivative of the function at the point x0 is explained by the existence of a non-vertical tangent, which is in contact with the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent is equal to f "(x0). The geometric derivative, that is, the angular coefficient of the tangent, becomes clear.

That is, in order to find the slope of the tangent, you need to find the value of the derivative of the function at the point of tangency. Example: find the angular coefficient of the tangent to the function y = x³ at the point with abscissa X0 = 1. Solution: Find the derivative of this function y΄(x) = 3x²; find the value of the derivative at the point X0 = 1. у΄(1) = 3 × 1² = 3. The angle coefficient of the tangent at the point X0 = 3.

Draw additional tangents in the figure so that they touch the graph of the function at the points: x1, x2 and x3. Mark the angles formed by these tangents with the abscissa axis (the angle is counted in the positive direction - from the axis to the tangent line). For example, angle α1 will be acute, angle (α2) will be obtuse, and the third (α3) will be equal to zero, since the tangent line drawn is parallel to the OX axis. In this case, the tangent of an obtuse angle is a negative value, and the tangent of an acute angle is positive, with tg0 and the result is zero.

A tangent to a given circle is a straight line that has only one common point with this circle. A tangent to a circle is always perpendicular to its radius drawn to the point of tangency. If two tangents are drawn from one point that does not belong to the circle, then the distances from this point to the points of tangency will always be the same. Tangents to circles are built in different ways, depending on their location relative to each other.

Instructions

Constructing a tangent to one circle.
1. Construct a circle of radius R and take A, which the tangent will pass through.
2. A circle is constructed with a center in the middle of the segment OA and radii equal to this segment.
3. The intersection of two tangent points drawn through point A to a given circle.

External tangent to two circles.

2. Draw a circle of radius R – r with center at point O.
3. A tangent from O1 is drawn to the resulting circle, the point of tangency is designated M.
4. Radius R passing through point M to point T – the tangent point of the circle.
5. Through the center O1 of the small circle, a radius r is drawn parallel to R of the large circle. The radius r points to point T1 – the point of tangency of the small circle.
circles.

Internal tangent to two circles.
1. Two circles of radius R and r are constructed.
2. Draw a circle of radius R + r with center at point O.
3. A tangent is drawn to the resulting circle from point O1, the point of tangency is designated by the letter M.
4. Ray OM intersects the first circle at point T - at the point of tangency of the great circle.
5. Through the center O1 of the small circle, a radius r is drawn parallel to the ray OM. The radius r points to point T1 – the point of tangency of the small circle.
6. Straight line TT1 – tangent to the given circles.

Sources:

• internal tangent

Angular closet– ideal for empty corners in the apartment. In addition, the corner configuration closet ov gives the interior a classic atmosphere. As a finishing for corners closet ov any material that is suitable for this purpose can be used.

You will need

• Fiberboard, MDF, screws, nails, saw blade, frieze.

Instructions

Cut a template 125 mm wide and 1065 mm long from plywood or fiberboard. The edges must be filed at an angle of 45 degrees. Using the finished template, determine the dimensions of the side walls, as well as the place where it will be located closet.

Connect the lid to the side walls and triangular shelves. The cover must be secured to the upper edges of the side walls using screws. For structural strength, additional glue is used. Attach the shelves to the slats.

Angle the saw blade at a 45-degree angle and bevel the leading edge of the side walls along the guide bar. Attach fixed shelves to MDF strips. Connect the side walls with screws. Make sure there are no gaps.

Make marks in the wall, between which place the frame of the corner closet A. Attach using screws closet to Wall. The length of the dowel should be 75 mm.

Cut out the front frame from a solid MDF board. Using a circular saw, cut the openings in it using a ruler. Finish the corners.

Find the abscissa value of the tangent point, which is denoted by the letter “a”. If it coincides with a given tangent point, then "a" will be its x-coordinate. Determine the value functions f(a) by substituting into the equation functions abscissa value.

Determine the first derivative of the equation functions f’(x) and substitute the value of point “a” into it.

Take the general tangent equation, which is defined as y = f(a) = f (a)(x – a), and substitute the found values ​​of a, f(a), f "(a) into it. As a result, the solution to the graph will be found and tangent.

Solve the problem in a different way if the given tangent point does not coincide with the tangent point. In this case, it is necessary to substitute “a” instead of numbers in the tangent equation. After this, instead of the letters “x” and “y”, substitute the value of the coordinates of the given point. Solve the resulting equation in which “a” is the unknown. Plug the resulting value into the tangent equation.

Write an equation for a tangent with the letter “a” if the problem statement specifies the equation functions and the equation of a parallel line relative to the desired tangent. After this we need the derivative functions, to the coordinate at point “a”. Substitute the appropriate value into the tangent equation and solve the function.

When composing the equation of a tangent to the graph of a function, the concept of “abscissa of the point of tangency” is used. This value can be specified initially in the conditions of the problem or it must be determined independently.

Instructions

Draw the x and y coordinate axes on a piece of paper. Study the given equation for the graph of a function. If it is , then it is enough to have two values ​​for the parameter y for any x, then plot the found points on the coordinate axis and connect them with a line. If the graph is nonlinear, then make a table of the dependence of y on x and select at least five points to construct the graph.

Determine the value of the abscissa of the tangent point for the case when the given tangent point does not coincide with the graph of the function. We set the third parameter with the letter “a”.

Write down the equation of the function f(a). To do this, substitute a instead of x in the original equation. Find the derivative of the function f(x) and f(a). Substitute the required data into the general tangent equation, which has the form: y = f(a) + f "(a)(x – a). As a result, obtain an equation that consists of three unknown parameters.

Substitute into it, instead of x and y, the coordinates of the given point through which the tangent passes. After this, find the solution to the resulting equation for all a. If it is square, then there will be two values ​​for the abscissa of the tangent point. This is that the tangent passes twice near the graph of the function.

Draw a graph of the given function and , which are specified according to the conditions of the problem. In this case, it is also necessary to specify the unknown parameter a and substitute it into the equation f(a). Equate the derivative f(a) to the derivative of the equation of a parallel line. This comes from the condition of parallelism of the two. Find the roots of the resulting equation, which will be the abscissa of the point of tangency.

The straight line y=f(x) will be tangent to the graph shown in the figure at point x0 if it passes through the point with coordinates (x0; f(x0)) and has an angular coefficient f"(x0). Find such a coefficient, Knowing the features of a tangent, it’s not difficult.

You will need

• - mathematical reference book;
• - a simple pencil;
• - notebook;
• - protractor;
• - compass;
• - pen.

Instructions

If the value f‘(x0) does not exist, then either there is no tangent, or it runs vertically. In view of this, the presence of a derivative of the function at the point x0 is due to the existence of a non-vertical tangent tangent to the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent will be equal to f "(x0). Thus, the geometric meaning of the derivative becomes clear - the calculation of the angular coefficient of the tangent.

Determine the general one. This kind of information can be obtained by referring to census data. To determine the overall fertility, mortality, marriage and divorce rates, you will need to find the product of the total population and the calculation period. Write the resulting number into the denominator.

Put on the numerator the indicator corresponding to the desired relative. For example, if you are faced with determining the total fertility rate, then in place of the numerator there should be a number that reflects the total number of births for the period of interest to you. If your goal is the mortality rate or marriage rate, then in place of the numerator put the number of deaths in the calculation period or the number of marriages, respectively.

Multiply the resulting number by 1000. This will be the overall coefficient you are looking for. If you are faced with the task of finding the overall growth rate, then subtract the mortality rate from the birth rate.

Video on the topic

Sources:

• General vital rates

The main indicator of extraction efficiency is coefficient distribution. It is calculated by the formula: Co/Sw, where Co is the concentration of the extracted substance in the organic solvent (extractor), and St is the concentration of the same substance in water, after equilibrium has reached. How can you experimentally find the distribution coefficient?