Answer :
Certainly! Let's break down the given function \( y = 3x^4 - 2x^2 + 8 \).
### 1. Understanding the Components
The function is a polynomial:
- \( 3x^4 \): This term is the highest degree term (degree 4). It dominates the behavior of the function for large values of \( x \).
- \( -2x^2 \): This is a middle degree term (degree 2), which affects the curvature of the polynomial.
- \( +8 \): This is the constant term, which shifts the entire graph of the polynomial upwards by 8 units.
### 2. Analyzing the Graph
1. Degree and Leading Coefficient:
The leading term \( 3x^4 \) has a positive coefficient, indicating that as \( x \) approaches \( \pm \infty \), \( y \) will also approach \( +\infty \).
2. Intercepts:
- y-intercept: Set \( x = 0 \):
[tex]\[ y = 3(0)^4 - 2(0)^2 + 8 = 8 \][/tex]
So, the y-intercept is at \( (0, 8) \).
3. Behavior Near the Origin:
- For small values of \( x \), the terms \( 3x^4 \) and \( -2x^2 \) are relatively small, and the function is dominated by the constant term \( +8 \).
### 3. Critical Points and Concavity
To find critical points and analyze concavity, we take the first and second derivatives.
1. First derivative \( y' \) to find critical points:
[tex]\[ y' = \frac{d(3x^4 - 2x^2 + 8)}{dx} = 12x^3 - 4x \][/tex]
Set \( y' = 0 \) to find critical points:
[tex]\[ 12x^3 - 4x = 0 \][/tex]
[tex]\[ 4x(3x^2 - 1) = 0 \][/tex]
This gives solutions:
[tex]\[ x = 0 \quad \text{or} \quad 3x^2 - 1 = 0 \][/tex]
[tex]\[ 3x^2 - 1 = 0 \implies x^2 = \frac{1}{3} \implies x = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3} \][/tex]
So, the critical points are \( x = 0 \) and \( x = \pm\frac{\sqrt{3}}{3} \).
2. Second derivative \( y'' \) to determine concavity:
[tex]\[ y'' = \frac{d(12x^3 - 4x)}{dx} = 36x^2 - 4 \][/tex]
Evaluate \( y'' \) at the critical points:
- At \( x = 0 \):
[tex]\[ y''(0) = 36(0)^2 - 4 = -4 \quad (\text{concave down}) \][/tex]
- At \( x = \pm\frac{\sqrt{3}}{3} \):
[tex]\[ y''\left(\pm\frac{\sqrt{3}}{3}\right) = 36\left(\frac{(\sqrt{3}/3)^2}{\}} - 4 = 36 \left( \frac{1}{3} \right) - 4 = 12 - 4 = 8 \quad (\text{concave up}) \][/tex]
### 4. Summarizing Findings
- The function \( y = 3x^4 - 2x^2 + 8 \) has its y-intercept at \( (0, 8) \).
- The critical points are at \( x = 0 \), where it has a local maximum due to \( y''(0) < 0 \); and \( x = \pm\frac{\sqrt{3}}{3} \), where it has local minima since \( y''\left(\pm\frac{\sqrt{3}}{3}\right) > 0 \).
- As \( x \to \pm \infty \), \( y \to \infty \) due to the dominance of the \( 3x^4 \) term.
This provides a thorough understanding of the function's behavior.
### 1. Understanding the Components
The function is a polynomial:
- \( 3x^4 \): This term is the highest degree term (degree 4). It dominates the behavior of the function for large values of \( x \).
- \( -2x^2 \): This is a middle degree term (degree 2), which affects the curvature of the polynomial.
- \( +8 \): This is the constant term, which shifts the entire graph of the polynomial upwards by 8 units.
### 2. Analyzing the Graph
1. Degree and Leading Coefficient:
The leading term \( 3x^4 \) has a positive coefficient, indicating that as \( x \) approaches \( \pm \infty \), \( y \) will also approach \( +\infty \).
2. Intercepts:
- y-intercept: Set \( x = 0 \):
[tex]\[ y = 3(0)^4 - 2(0)^2 + 8 = 8 \][/tex]
So, the y-intercept is at \( (0, 8) \).
3. Behavior Near the Origin:
- For small values of \( x \), the terms \( 3x^4 \) and \( -2x^2 \) are relatively small, and the function is dominated by the constant term \( +8 \).
### 3. Critical Points and Concavity
To find critical points and analyze concavity, we take the first and second derivatives.
1. First derivative \( y' \) to find critical points:
[tex]\[ y' = \frac{d(3x^4 - 2x^2 + 8)}{dx} = 12x^3 - 4x \][/tex]
Set \( y' = 0 \) to find critical points:
[tex]\[ 12x^3 - 4x = 0 \][/tex]
[tex]\[ 4x(3x^2 - 1) = 0 \][/tex]
This gives solutions:
[tex]\[ x = 0 \quad \text{or} \quad 3x^2 - 1 = 0 \][/tex]
[tex]\[ 3x^2 - 1 = 0 \implies x^2 = \frac{1}{3} \implies x = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3} \][/tex]
So, the critical points are \( x = 0 \) and \( x = \pm\frac{\sqrt{3}}{3} \).
2. Second derivative \( y'' \) to determine concavity:
[tex]\[ y'' = \frac{d(12x^3 - 4x)}{dx} = 36x^2 - 4 \][/tex]
Evaluate \( y'' \) at the critical points:
- At \( x = 0 \):
[tex]\[ y''(0) = 36(0)^2 - 4 = -4 \quad (\text{concave down}) \][/tex]
- At \( x = \pm\frac{\sqrt{3}}{3} \):
[tex]\[ y''\left(\pm\frac{\sqrt{3}}{3}\right) = 36\left(\frac{(\sqrt{3}/3)^2}{\}} - 4 = 36 \left( \frac{1}{3} \right) - 4 = 12 - 4 = 8 \quad (\text{concave up}) \][/tex]
### 4. Summarizing Findings
- The function \( y = 3x^4 - 2x^2 + 8 \) has its y-intercept at \( (0, 8) \).
- The critical points are at \( x = 0 \), where it has a local maximum due to \( y''(0) < 0 \); and \( x = \pm\frac{\sqrt{3}}{3} \), where it has local minima since \( y''\left(\pm\frac{\sqrt{3}}{3}\right) > 0 \).
- As \( x \to \pm \infty \), \( y \to \infty \) due to the dominance of the \( 3x^4 \) term.
This provides a thorough understanding of the function's behavior.
