This instruction contains the answer to a question how to find the tangent equation to a function graph. Exhaustive reference information is given. Application of theoretical calculations is investigated on a concrete example.
Instruction
1. Reference material. For a start we will give definition of a tangent. A tangent to a curve in this point of M is called the limit provision of a secant of NM when the point of N approaches along a curve a point M. Naydem the tangent equation to y function graph = f (x).
2. We define slope of a tangent to a curve in the M point. The curve representing y function graph = f(x) is continuous in some vicinity of a point of M (including a point of M). Let's carry out the secant of MN1 forming with the positive direction of an axis Ox a corner α.Координаты M points (x; y), N1 point coordinates (x+ ∆ x; y+ ∆ y). From the received triangle of MN1N it is possible to find slope of this secant: tg α = Δy/ΔxMN = ∆xNN1 = upri aspiration of a point of N1 on a curve to a point the M secant of MN1 turns around a point of M, and the corner α aspires to a corner ϕ between tangent MT and the positive direction of an axis Ox. k = tg ϕ = 〖 lim 〗┬ (∆x→0) 〖 〗 Δy/Δx = f' (x) Thus, slope of a tangent to a function graph is equal to value of derivative this function in a contact point. The geometrical meaning of a derivative consists in it.
3. The tangent equation to the set curve in the set point of M has an appearance: y-y0 = f' (x0) (x - x0), where (x0; y0) – contact point coordinates, (x; y) – the current coordinates, i.e. coordinates of any point belonging to a tangent, f' (x0) = k = tg α – slope of a tangent.
4. Let's find the tangent equation on an example. y=x2 function graph – 2x is given. It is necessary to find the tangent equation in a point with x0 abscissa = 3. From the equation of this curve we find ordinate of a point of contact of y0 = 32 - 2∙3 = 3. We find a derivative, and then we calculate its value in x0 point = 3. We have: y '=2x – 2f' (3) = 2∙3 – 2 = 4. Now, knowing a point (3; 3) on a curve and slope of f' (3) = 4 tangents in this point, we receive the required equation: y – 3 = 4 (x – 3) or y – 4x + 9 = 0