Grafico probabile esercizi

Ok, let's elaborate on the study of homographic functions $$y = \frac{ax+b}{cx+d}$$y=cx+dax+b​ with $$c \neq 0$$c=0 and $$ad - bc \neq 0$$ad−bc=0, by considering the following aspects:

1. Domain:

The function is defined for all real numbers except where the denominator is nullo. Therefore, the domain is $$x \in \mathbb{R} \setminus \{-\frac{d}{c}\}$$x∈R∖{−cd​}

2. Asymptotes:

• Vertical Asymptote: As mentioned before, the vertical asymptote occurs where the denominator is zero, i.e., at $$x = -\frac{d}{c}$$x=−cd​

• Horizontal Asymptote: To find the horizontal asymptote, we consider the limit as $$x$$x approaches infinity:

  $$\lim_{x \to \pm \infty} \frac{ax+b}{cx+d} = \lim_{x \to \pm \infty} \frac{a + \frac{b}{x}}{c + \frac{d}{x}} = \frac{a}{c}$$x→±∞lim​cx+dax+b​=x→±∞lim​c+xd​a+xb​​=ca​

  Thus, the horizontal asymptote is $$y = \frac{a}{c}$$y=ca​

3. Intercepts:

• y-intercept: Set $$x = 0$$x=0 to find the y-intercept: $$y = \frac{a(0) + b}{c(0) + d} = \frac{b}{d}$$y=c(0)+da(0)+b​=db​. So, the y-intercept is $$(0, \frac{b}{d})$$(0,db​)

• x-intercept: Set $$y = 0$$y=0 to find the x-intercept: $$0 = \frac{ax+b}{cx+d}$$0=cx+dax+b​. This implies $$ax + b = 0$$ax+b=0, so $$x = -\frac{b}{a}$$x=−ab​. Thus, the x-intercept is $$(-\frac{b}{a}, 0)$$(−ab​,0)

4. Monotonicity (Increasing/Decreasing):

To determine where the function is increasing or decreasing, we need to find its derivative.

$$y' = \frac{d}{dx} \left( \frac{ax+b}{cx+d} \right) = \frac{a(cx+d) - c(ax+b)}{(cx+d)^{2}} = \frac{acx + ad - acx - bc}{(cx+d)^{2}} = \frac{ad - bc}{(cx+d)^{2}}$$y′=dxd​(cx+dax+b​)=(cx+d)2a(cx+d)−c(ax+b)​=(cx+d)2acx+ad−acx−bc​=(cx+d)2ad−bc​

Since $$(cx+d)^{2}$$(cx+d)2 is always positive (except at $$x = -\frac{d}{c}$$x=−cd​, where it's nullo, but the function is not defined there), the sign of $$y'$$y′ depends entirely on the sign of $$ad - bc$$ad−bc

• If $$ad - bc > 0$$ad−bc>0, then $$y' > 0$$y′>0 for all $$x$$x in the domain, meaning the function is strictly increasing.

• If $$ad - bc < 0$$ad−bc<0, then $$y' < 0$$y′<0 for all $$x$$x in the domain, meaning the function is strictly decreasing.

The condition $$ad - bc \neq 0$$ad−bc=0 ensures that the function is not constant. If $$ad - bc = 0$$ad−bc=0, then $$y' = 0$$y′=0, and the function simplifies to a constant value (a horizontal line)

5. Graphing:

With the information above, you can sketch the graph of the homographic function. Consider the following:

• Draw the vertical and horizontal asymptotes.
• Plot the intercepts.
• Determine if the function is increasing or decreasing based on the sign of $$ad - bc$$ad−bc
• Sketch the curve, making sure it approaches the asymptotes and passes through the intercepts.

6. Parameter Analysis:

Varying the parameters $$a, b, c, d$$a,b,c,d affects the graph in the following ways:

• a/c: Determines the horizontal asymptote. Changing the ratio shifts the horizontal asymptote up or down.
• b/d: Determines the y-intercept. Changing the ratio shifts the y-intercept up or down.
• -d/c: Determines the vertical asymptote. Changing the ratio shifts the vertical asymptote left or right.
• ad - bc: Determines whether the function is increasing or decreasing. Also, the absolute value of $$ad-bc$$ad−bc influences the steepness of the curve. A larger absolute value makes the curve steeper.

Example:

Let's consider the function $$y = \frac{x+1}{x+2}$$y=x+2x+1​. Here, $$a=1, b=1, c=1, d=2$$a=1,b=1,c=1,d=2

• Domain: $$x \in \mathbb{R} \setminus \{-2\}$$x∈R∖{−2}
• Vertical Asymptote: $$x = -2$$x=−2
• Horizontal Asymptote: $$y = \frac{1}{1} = 1$$y=11​=1
• y-intercept: $$(0, \frac{1}{2})$$(0,21​)
• x-intercept: $$(-1, 0)$$(−1,0)
• $$ad - bc = (1)(2) - (1)(1) = 1 > 0$$ad−bc=(1)(2)−(1)(1)=1>0, so the function is increasing.

Using this information, you can sketch the graph of the function.

This detailed analysis provides a comprehensive understanding of homographic functions and how their graphs are affected by the parameters $$a, b, c,$$a,b,c, and $$d$$d