Question
Given the function defined by f(x)=6/x-3 2.1 Write down the equations of the asymptotes of f. 2.2 Determine the x intercept of f. 2.3 Sketch the graph of f. Show clearly the intercepts with the axis and the asymptote. 2.4 Write down the equation of the line of symmetry with negative gradient.
Ask by Adkins Cook.
Jan 28,2025 18:26
UpStudy AI Solution
Tutor-Verified Answer
Answer
**Asymptotes:** Vertical: x = 0, Horizontal: y = -3
**x Intercept:** (2, 0)
**Line of Symmetry with Negative Gradient:** y = -x - 3
Solution
Certainly! Let's address each part of the problem step by step for the function \( f(x) = \frac{6}{x} - 3 \).
---
### **2.1. Equations of the Asymptotes of \( f \)**
**a. Vertical Asymptote:**
A vertical asymptote occurs where the function is undefined, which is when the denominator is zero.
\[
\frac{6}{x} - 3 \text{ is undefined when } x = 0.
\]
**Vertical Asymptote:**
\[
x = 0
\]
**b. Horizontal Asymptote:**
A horizontal asymptote describes the behavior of the function as \( x \) approaches infinity (\( \infty \)) or negative infinity (\( -\infty \)).
\[
\lim_{x \to \pm\infty} f(x) = \lim_{x \to \pm\infty} \left( \frac{6}{x} - 3 \right) = -3
\]
**Horizontal Asymptote:**
\[
y = -3
\]
---
### **2.2. Determining the \( x \)-Intercept of \( f \)**
An \( x \)-intercept occurs where the function crosses the \( x \)-axis, i.e., where \( f(x) = 0 \).
\[
0 = \frac{6}{x} - 3
\]
Solve for \( x \):
\[
\frac{6}{x} = 3 \\
6 = 3x \\
x = 2
\]
**\( x \)-Intercept:**
\[
(2, \, 0)
\]
---
### **2.3. Sketching the Graph of \( f \)**
While a visual sketch isn't possible here, we can describe the key features of the graph based on the asymptotes and intercepts:
1. **Asymptotes:**
- **Vertical Asymptote:** \( x = 0 \)
- **Horizontal Asymptote:** \( y = -3 \)
2. **Intercepts:**
- **\( x \)-Intercept:** \( (2, \, 0) \)
- **\( y \)-Intercept:** None (the function is undefined at \( x = 0 \))
3. **Behavior:**
- As \( x \) approaches 0 from the positive side (\( x \to 0^+ \)), \( f(x) \to +\infty \).
- As \( x \) approaches 0 from the negative side (\( x \to 0^- \)), \( f(x) \to -\infty \).
- As \( x \to \pm\infty \), \( f(x) \) approaches the horizontal asymptote \( y = -3 \).
4. **Symmetry:**
- The function exhibits central symmetry about the point where the asymptotes intersect, which is \( (0, -3) \).
**Summary of Key Points:**
- **Vertical Asymptote:** \( x = 0 \)
- **Horizontal Asymptote:** \( y = -3 \)
- **\( x \)-Intercept:** \( (2, \, 0) \)
- **No \( y \)-Intercept**
---
### **2.4. Equation of the Line of Symmetry with Negative Gradient**
The line of symmetry with a negative gradient for the function \( f(x) = \frac{6}{x} - 3 \) can be determined as follows:
1. **Understanding Symmetry:**
- The function has a central point of symmetry at the intersection of its asymptotes, which is \( (0, -3) \).
- A line of symmetry with a negative slope (gradient) passes through this central point.
2. **Determining the Equation:**
- Let’s assume the line has the form \( y = -x + b \).
- Since it passes through \( (0, -3) \), substitute \( x = 0 \) and \( y = -3 \) to find \( b \):
\[
-3 = -0 + b \\
b = -3
\]
- Therefore, the equation of the line is:
\[
y = -x - 3
\]
**Equation of the Line of Symmetry:**
\[
y = -x - 3
\]
---
### **Final Summary:**
1. **Asymptotes:**
- Vertical: \( x = 0 \)
- Horizontal: \( y = -3 \)
2. **\( x \)-Intercept:** \( (2, \, 0) \)
3. **Graph Features:**
- Approaches \( x = 0 \) and \( y = -3 \).
- Passes through \( (2, 0) \).
- No \( y \)-intercept.
4. **Line of Symmetry with Negative Gradient:**
\[
y = -x - 3
\]
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Extra Insights
The function \( f(x) = \frac{6}{x} - 3 \) has a vertical asymptote at \( x = 0 \) (where the denominator is zero) and a horizontal asymptote at \( y = -3 \) (as \( x \) approaches \( \pm \infty \)). These two asymptotes indicate that the function does not cross these lines, providing boundaries for the graph.
To find the \( x \)-intercept, we set \( f(x) = 0 \): \( \frac{6}{x} - 3 = 0 \). Solving this yields \( \frac{6}{x} = 3 \) or \( x = 2 \). Therefore, the \( x \) intercept is \( (2, 0) \). The function can be sketched by plotting the asymptotes and the intercept, demonstrating its behavior in the quadrants around the asymptotes.
The line of symmetry for this function, given its nature, would not typically accommodate a line of symmetry since it's a hyperbolic function; however, we can note the general trend that functions of this type do not reflect symmetrically across a line but can possess other symmetries.
Related Questions
Déterminer le domaine de définition de la fonction f(x) = ln((x - 9)²). Ecrire la réponse sous forme D_{f}=... Réponse :
Pre Calculus
Dec 18, 2024
Convert 3 from radians to decimal degrees. Round to 1 decimal place if necessary. 3 radians
Pre Calculus
Mar 23, 2025
4) Data la funzione definita mediante la legge
la curva di livello \( k=3 \) è l'insieme
A) \( \left\{(x, y) \in \mathbb{R}^{2}: x^{2}+5-x=3\right\} \)
B) \( \left\{(x, y) \in \mathbb{R}^{2}: x^{2}+5 y^{2}-x<3\right\} \)
C) \( \left\{(x, y) \in \mathbb{R}^{2}: x^{2}+5 y^{2}-x>3\right\} \)
Pre Calculus
Feb 02, 2025
On étudie un groupe de 3000 sportifs qui pratiquent soit l'athlétisme dans le club A, soit le basketball dans le club B.
En 2023, le club A compte 1700 membres et le club B en compte 1300.
On décide de modéliser le nombre de membres du club A et du club B respectivement par deux suites a_{n} et b_{n}, où n désigne le rang de l'année à partir de 2023.
L'année 2023 correspond au rang 0 . On a alors a_{0}=1700 et b_{0}=1300.
Pour notre étude, on fait les hypothèses suivantes :
- durant l'étude, aucun sportif ne quitte le groupe;
- chaque année, 15% des sportifs du club A quittent ce club et adhèrent au club B;
- chaqueannée, 10% des sportifs du club B quittent ce club et adhèrent au club A.
1) Calculer les nombres de membres de chaque club en 2024 .
2) Pour tout entier naturel n, déterminer une relation liant a_{n} et b_{n}.
3) Montrer que la suite a_{n} vérifie la relation suivante pour tout entier naturel n :
a_{n+1}=0,75 a_{n}+300
4)
a) Démontrer par récurrence que pour tout entier naturel n, on a: 1200 leq a_{n+1} leq a_{n} leq 1700.
b) En déduire que la suite a_{n} converge.
5) Soit v_{n} la suite définie pour tout entier naturel n par v_{n}=a_{n}-1200.
a) Démontrer que la suite v_{n} est géométrique.
b) Exprimer v_{n} en fonction de n.
c) En déduire que pour tout entier naturel n, a_{n}=500 * 0,75^{n}+1200 .
6)
a) Déterminer la limite de la suite a_{n}.
b) Interpréter le résultat de la question précédente dans le contexte de l'exercice.
Pre Calculus
Feb 02, 2025
