Answer :
Certainly! Let's write a detailed, step-by-step solution for the given equation [tex]\( y = 2x^3 - 3x^2 + 6x \)[/tex].
### Step 1: Understanding the Problem
You are given a polynomial equation:
[tex]\[ y = 2x^3 - 3x^2 + 6x \][/tex]
### Step 2: Identify the Terms
The given polynomial is of the third degree (cubic equation), and it consists of the following terms:
- [tex]\( 2x^3 \)[/tex] (cubic term)
- [tex]\( -3x^2 \)[/tex] (quadratic term)
- [tex]\( 6x \)[/tex] (linear term)
### Step 3: Derivatives (If Asked for Slope or Critical Points)
If the task is to find the first derivative [tex]\( \frac{dy}{dx} \)[/tex], which represents the slope of the tangent line to the curve at any point [tex]\( x \)[/tex], it would involve differentiating each term individually.
The first derivative [tex]\( \frac{dy}{dx} \)[/tex] will be:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(6x) \][/tex]
By applying the power rule [tex]\( \frac{d}{dx}(x^n) = n*x^{n-1} \)[/tex]:
[tex]\[ \frac{dy}{dx} = 6x^2 - 6x + 6 \][/tex]
### Step 4: Evaluating [tex]\(y\)[/tex] at Specific Points (If Asked)
To find the value of [tex]\(y\)[/tex] at a specific [tex]\(x\)[/tex], substitute that value of [tex]\(x\)[/tex] into the equation. For example:
- If [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1)^3 - 3(1)^2 + 6(1) = 2 - 3 + 6 = 5 \][/tex]
- If [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0)^3 - 3(0)^2 + 6(0) = 0 \][/tex]
### Step 5: Critical Points (If Asked for Extrema)
To find critical points, set the first derivative to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ 6x^2 - 6x + 6 = 0 \][/tex]
Since we recognize this as a quadratic equation, solve using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 6 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = 6 \)[/tex]:
[tex]\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4(6)(6)}}{2(6)} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 - 144}}{12} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{-108}}{12} \][/tex]
Since the discriminant ([tex]\( -108 \)[/tex]) is negative, there are no real roots, meaning the equation [tex]\( 6x^2 - 6x + 6 = 0 \)[/tex] does not have real solutions, and hence no critical points derived from real solutions.
### Step 6: Understanding the Shape of the Curve
To determine the behavior of the polynomial [tex]\( y = 2x^3 - 3x^2 + 6x \)[/tex]:
- As [tex]\( x \to \infty \)[/tex], the [tex]\( 2x^3 \)[/tex] term dominates, so [tex]\( y \to \infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], the [tex]\( 2x^3 \)[/tex] term again dominates, so [tex]\( y \to -\infty \)[/tex].
### Conclusion
The curve described by the equation [tex]\( y = 2x^3 - 3x^2 + 6x \)[/tex] is a cubic polynomial with no real critical points from the first derivative and trends to [tex]\( \infty \)[/tex] as [tex]\( x \to \infty \)[/tex] and to [tex]\( -\infty \)[/tex] as [tex]\( x \to -\infty \)[/tex]. For any specific [tex]\( x \)[/tex], you can substitute it into the polynomial to find the corresponding [tex]\( y \)[/tex]-value.
### Step 1: Understanding the Problem
You are given a polynomial equation:
[tex]\[ y = 2x^3 - 3x^2 + 6x \][/tex]
### Step 2: Identify the Terms
The given polynomial is of the third degree (cubic equation), and it consists of the following terms:
- [tex]\( 2x^3 \)[/tex] (cubic term)
- [tex]\( -3x^2 \)[/tex] (quadratic term)
- [tex]\( 6x \)[/tex] (linear term)
### Step 3: Derivatives (If Asked for Slope or Critical Points)
If the task is to find the first derivative [tex]\( \frac{dy}{dx} \)[/tex], which represents the slope of the tangent line to the curve at any point [tex]\( x \)[/tex], it would involve differentiating each term individually.
The first derivative [tex]\( \frac{dy}{dx} \)[/tex] will be:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(6x) \][/tex]
By applying the power rule [tex]\( \frac{d}{dx}(x^n) = n*x^{n-1} \)[/tex]:
[tex]\[ \frac{dy}{dx} = 6x^2 - 6x + 6 \][/tex]
### Step 4: Evaluating [tex]\(y\)[/tex] at Specific Points (If Asked)
To find the value of [tex]\(y\)[/tex] at a specific [tex]\(x\)[/tex], substitute that value of [tex]\(x\)[/tex] into the equation. For example:
- If [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1)^3 - 3(1)^2 + 6(1) = 2 - 3 + 6 = 5 \][/tex]
- If [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0)^3 - 3(0)^2 + 6(0) = 0 \][/tex]
### Step 5: Critical Points (If Asked for Extrema)
To find critical points, set the first derivative to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ 6x^2 - 6x + 6 = 0 \][/tex]
Since we recognize this as a quadratic equation, solve using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 6 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = 6 \)[/tex]:
[tex]\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4(6)(6)}}{2(6)} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 - 144}}{12} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{-108}}{12} \][/tex]
Since the discriminant ([tex]\( -108 \)[/tex]) is negative, there are no real roots, meaning the equation [tex]\( 6x^2 - 6x + 6 = 0 \)[/tex] does not have real solutions, and hence no critical points derived from real solutions.
### Step 6: Understanding the Shape of the Curve
To determine the behavior of the polynomial [tex]\( y = 2x^3 - 3x^2 + 6x \)[/tex]:
- As [tex]\( x \to \infty \)[/tex], the [tex]\( 2x^3 \)[/tex] term dominates, so [tex]\( y \to \infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], the [tex]\( 2x^3 \)[/tex] term again dominates, so [tex]\( y \to -\infty \)[/tex].
### Conclusion
The curve described by the equation [tex]\( y = 2x^3 - 3x^2 + 6x \)[/tex] is a cubic polynomial with no real critical points from the first derivative and trends to [tex]\( \infty \)[/tex] as [tex]\( x \to \infty \)[/tex] and to [tex]\( -\infty \)[/tex] as [tex]\( x \to -\infty \)[/tex]. For any specific [tex]\( x \)[/tex], you can substitute it into the polynomial to find the corresponding [tex]\( y \)[/tex]-value.
