Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
