The problem of drawing up the equation of a tangent to a function graph comes down to need of commission of selection from a set of direct subjects which can meet the set requirements. Everything it straight lines can be set by either points, or slope. To solve the graph of function and a tangent, it is necessary to perform certain operations.
Instruction
1. Read attentively a task of drawing up the equation of a tangent. As a rule, there is a certain equation of a function graph expressed through x and y and also coordinates of one of tangent points.
2. Construct a function graph in coordinates of axes x and y. For this purpose it is necessary to make the table of a ratio of equality of y at a preset value x. If the function graph nonlinear, then for its construction is required, at least, five values of coordinates. Draw axes of coordinates and a function graph. Put also the end which is specified in a statement of the problem.
3. Find value of an abscissa of a point of contact which are designated by the letter "and". If it coincides with the tangent set by a point, then "and" will be equal it x - coordinate. Define value of the f(a) function, having substituted abscissa size in the function equation.
4. Define the first derivative of the equation of function f’ (x) and substitute in it value of a point "and".
5. Take the general equation of a tangent which is defined as y = by f(a) = f (a) (x – a), also substitute in it the found values a, f(a), f' (a). A solution of a function graph and tangent will be as a result found.
6. Solve a problem otherwise if the set point of a tangent did not coincide with a contact point. In this case it is necessary in the tangent equation instead of figures to substitute the letter "and". After that instead of the letters "x" and "y" substitute value of coordinates of the set point. Solve the turned-out equation in which the letter "and" is an unknown. Put the received value in the tangent equation.
7. Work out the tangent equation with the letter "and" if in a statement of the problem the equation of function and the equation of the parallel line of rather required tangent is set. After that it is necessary to find a derivative of function of a parallel straight line to determine coordinate at a point "and". Substitute the corresponding value in the equation of a tangent and solve function.