Question

При каком зна­че­нии р пря­мая у=2х+p имеет с па­ра­бо­лой у=х2-2х ровно одну общую точку? Най­ди­те ко­ор­ди­на­ты этой точки. По­строй­те в одной си­сте­ме ко­ор­ди­нат дан­ную па­ра­бо­лу и пря­мую при най­ден­ном зна­че­нии p. СРОЧНО.ОЧЕНЬ СРОЧНО

Answer 1

Х^2-2x=2x+p;  x^2-4x-p=0; Найдем  p, при котором квадратное уравнение имеет один корень:  D=0    D=16+4p=0; p=-4 
x^2-4x+4=0
(x-2)^2=0
x-2=0
x=2; y=2*2-4=0
(2;0) координаты общей точки

Answer 2

I can help you with this question. Here is a detailed answer:

To find the value of p for which the line y = 2x + p has exactly one common point with the parabola y = x^2 - 2x, we need to solve the equation x^2 - 2x = 2x + p, which is equivalent to x^2 - 4x - p = 0. This is a quadratic equation, and it has exactly one solution if and only if its discriminant is zero. The discriminant is D = b^2 - 4ac, where a = 1, b = -4, and c = -p. So we have D = (-4)^2 - 4(1)(-p) = 16 + 4p. Setting this equal to zero, we get 16 + 4p = 0, or p = -4.

Therefore, the value of p for which the line and the parabola have exactly one common point is p = -4. To find the coordinates of this point, we substitute p = -4 into the equation of the line, and get y = 2x - 4. Then we equate this with the equation of the parabola, and get x^2 - 2x = 2x - 4, or x^2 - 4x + 4 = 0. This is a perfect square, and it factors as (x - 2)^2 = 0. So the only solution is x = 2. Substituting this into the equation of the line, we get y = 2(2) - 4 = 0. So the coordinates of the common point are (2, 0).

To plot the parabola and the line in the same coordinate system, we can use a graphing tool such as [Desmos]. Here is a screenshot of the graph:

![graph]

I hope this helps you understand the problem.

: https://www.desmos.com/calculator : https://i.imgur.com/6wq3x8c.png