Question

Y=x+1/x-3 ИССЛЕДОВАТЬ ФУНКЦИЮ

Answer 1

Найдем область определения выражения, отыскав промежутки, на которых оно определено.

(

−

∞

;

3

)

∪

(

3

;

∞

)

(-∞;3)∪(3;∞)

{

x

|

x

≠

3

}

Answer 2

To investigate the function $y = \frac{x+1}{x-3}$, we'll examine its properties, such as domain, asymptotes, intercepts, symmetry, and behavior as $x$ approaches certain values.

1. Domain: The function is defined for all real numbers except when the denominator is zero ($x - 3 = 0$). So, the domain is all real numbers except $x \neq 3$.

2. Vertical Asymptote: The vertical asymptote occurs where the denominator becomes zero. Thus, there is a vertical asymptote at $x = 3$.

3. Horizontal Asymptote: To determine the horizontal asymptote, we'll check the behavior of the function as $x$ approaches positive and negative infinity.

   As $x \to \infty$, both the numerator and denominator have the highest power of $x$ as $x$, so we can apply the rule of "Divide the coefficients of the highest power of $x$". Thus, the horizontal asymptote is $y = \frac{1}{1} = 1$ as $x \to \infty$.

   As $x \to -\infty$, again both the numerator and denominator have the highest power of $x$ as $x$, so the horizontal asymptote remains $y = 1$ as $x \to -\infty$.

4. Intercepts:

   • x-intercept: To find the x-intercept, set $y$ to zero and solve for $x$: $0 = \frac{x + 1}{x - 3}$. Since the fraction is zero only when the numerator is zero, we have $x + 1 = 0 \Rightarrow x = -1$. So the x-intercept is $(-1, 0)$.
   • y-intercept: To find the y-intercept, set $x$ to zero: $y = \frac{0 + 1}{0 - 3} = \frac{1}{-3}$. So the y-intercept is $(0, -1/3)$.

5. Symmetry: To check for symmetry, let's evaluate the function at $-x$: $y(-x) = \frac{-x + 1}{-x - 3}$. It doesn't match the original function $y = \frac{x+1}{x-3}$. Hence, the function is not symmetric about the origin.

6. Vertical Shift: The function has a vertical shift of 1 unit up (compared to $y = \frac{x}{x-3}$) due to the constant term of $1$ in the numerator.

7. Critical Points: To find the critical points, we'll look for values of $x$ where the derivative of the function is zero or undefined. The derivative of $y$ with respect to $x$ is:

   $\frac{dy}{dx} = \frac{(x-3) - (x+1)}{(x-3)^2} = \frac{-4}{(x-3)^2}$.

   The derivative is never zero since the numerator is always -4. However, it is undefined when the denominator is zero (i.e., when $x = 3$). But we know that the function has a vertical asymptote at $x = 3$, so it never crosses the vertical asymptote, and hence, there are no critical points.

8. Increasing/Decreasing Intervals: Since the derivative is always negative (since the numerator is negative and the denominator is squared), the function is decreasing for all $x$ except where it is undefined (at $x = 3$).

Now, armed with this information, you can sketch the graph of the function. The graph will approach the vertical asymptote $x = 3$ without crossing it. As $x$ approaches infinity, the graph will approach the horizontal asymptote $y = 1$. As $x$ approaches negative infinity, the graph will also approach $y = 1$. The graph will have an x-intercept at $(-1, 0)$ and a y-intercept at $(0, -1/3)$. It will be a decreasing function with a vertical shift of 1 unit up compared to the function $y = \frac{x}{x-3}$.