Question

Иследовать функцию y=x-2/x+1

Answer 1

Ответ:

Пошаговое объяснение:

y"=(x+1-(x-2))/(x+1)^2=3/(x+1)^2,   производная>0, значит ф-я возрастает на (-~;-1) и (-1;+~)  , обл. опр. все знач-я кроме х=1  ,мн-во значений (-~;1) и (1; +~) , точек max и min нет. у=(х+1-3)/(х+1)=1 -   3/х+1-преобразование ф-ции,гипербола

Answer 2

To explore the function $y = \frac{x - 2}{x + 1}$, we can analyze its properties, graph it, and examine some key points.

1. Domain and Vertical Asymptote: The function is defined for all real values of $x$ except when the denominator $x + 1$ is equal to zero, which leads to division by zero. So, the domain of the function is all real numbers except $x = -1$. At $x = -1$, there is a vertical asymptote, which means the function approaches infinity or negative infinity as $x$ approaches -1.

2. Horizontal Asymptote: To find the horizontal asymptote, we can analyze the behavior of the function as $x$ approaches positive or negative infinity. The function can be rewritten as $y = \frac{x}{x+1} - \frac{2}{x+1}$. As $x$ becomes very large (positive or negative), the $x$ terms dominate, and the function approaches the value $y = 1$ since the $-\frac{2}{x+1}$ term becomes negligible compared to the $\frac{x}{x+1}$ term. Therefore, the function has a horizontal asymptote at $y = 1$.

3. x-intercept: To find the x-intercept, we set $y$ equal to zero and solve for $x$: $0 = \frac{x - 2}{x + 1}$. The numerator $x - 2$ can only be equal to zero when $x = 2$. However, $x = 2$ is not a valid x-value for the function since it would lead to division by zero in the denominator. Therefore, the function has no x-intercept.

4. y-intercept: To find the y-intercept, we set $x$ equal to zero and calculate the value of $y$: $y = \frac{0 - 2}{0 + 1} = -2$. So, the y-intercept is at the point (0, -2).

5. Critical Points: Critical points occur when the derivative of the function is equal to zero or undefined. Let's find the derivative of the function and then solve for $x$ when the derivative is equal to zero.

$y = \frac{x - 2}{x + 1}$

$y' = \frac{(x + 1)(1) - (x - 2)(1)}{(x + 1)^2} = \frac{3}{(x + 1)^2}$

The derivative is never equal to zero, so there are no critical points.

Now, let's graph the function to visually understand its behavior:

As you can see from the graph, the function has a vertical asymptote at $x = -1$ and a horizontal asymptote at $y = 1$. The function approaches the horizontal asymptote as $x$ becomes large in both the positive and negative directions.

Overall, the function $y = \frac{x - 2}{x + 1}$ has no x-intercept, a y-intercept at (0, -2), a vertical asymptote at $x = -1$, and a horizontal asymptote at $y = 1$.