Answer :
Absolutely, let's explore how we would approach solving the function [tex]\( -2x^3 + 4x^2 + 5x \)[/tex].
### Understanding the Polynomial
The function given to us is:
[tex]\[ f(x) = -2x^3 + 4x^2 + 5x \][/tex]
This is a cubic polynomial in one variable [tex]\( x \)[/tex].
### Step-by-Step Solution
1. Identify the terms:
The function has three terms:
- [tex]\(-2x^3\)[/tex]: This is the cubic term, where [tex]\( a_3 = -2 \)[/tex].
- [tex]\(4x^2\)[/tex]: This is the quadratic term, where [tex]\( a_2 = 4 \)[/tex].
- [tex]\(5x\)[/tex]: This is the linear term, where [tex]\( a_1 = 5 \)[/tex].
2. Determine the degree:
The highest power of [tex]\( x \)[/tex] in the polynomial is 3. Thus, the degree of the polynomial is 3.
3. Finding roots:
To find the roots, or zeros, of the polynomial, we need to solve:
[tex]\[ -2x^3 + 4x^2 + 5x = 0 \][/tex]
Factoring out [tex]\( x \)[/tex]:
[tex]\[ x(-2x^2 + 4x + 5) = 0 \][/tex]
This gives us one root immediately:
[tex]\[ x = 0 \][/tex]
For the quadratic term, we solve:
[tex]\[ -2x^2 + 4x + 5 = 0 \][/tex]
We can use the quadratic formula for this ( [tex]\(ax^2 + bx + c = 0\)[/tex] ):
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = -2 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 5 \)[/tex].
Calculate the discriminant:
[tex]\[ b^2 - 4ac = 4^2 - 4(-2)(5) = 16 + 40 = 56 \][/tex]
Thus, the roots are:
[tex]\[ x = \frac{-4 \pm \sqrt{56}}{-4} \][/tex]
Simplify [tex]\(\sqrt{56}\)[/tex]:
[tex]\[ \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \][/tex]
Now, we substitute back in:
[tex]\[ x = \frac{-4 \pm 2\sqrt{14}}{-4} = \frac{-4 \pm 2\sqrt{14}}{-4} = 1 \mp \frac{\sqrt{14}}{2} \][/tex]
So, the other two roots are:
[tex]\[ x = 1 + \frac{\sqrt{14}}{2}, \quad x = 1 - \frac{\sqrt{14}}{2} \][/tex]
4. Graphical representation:
The general shape of the graph of [tex]\( -2x^3 + 4x^2 + 5x \)[/tex] can be understood by considering the leading term [tex]\(-2x^3\)[/tex]. Since the coefficient is negative, the graph will start from negative infinity as [tex]\( x \to -\infty \)[/tex] and will end at negative infinity as [tex]\( x \to \infty \)[/tex], producing an overall 'N' shape.
### Key Takeaways
- The polynomial [tex]\( -2x^3 + 4x^2 + 5x \)[/tex] is cubic.
- The roots found are [tex]\( 0 \)[/tex], [tex]\( 1 + \frac{\sqrt{14}}{2} \)[/tex], and [tex]\( 1 - \frac{\sqrt{14}}{2} \)[/tex].
- The shape of the graph is determined by the leading term, which is cubic and negative.
This concisely explores the function and its attributes. If you need further properties like maxima, minima, or inflection points, do let me know!
### Understanding the Polynomial
The function given to us is:
[tex]\[ f(x) = -2x^3 + 4x^2 + 5x \][/tex]
This is a cubic polynomial in one variable [tex]\( x \)[/tex].
### Step-by-Step Solution
1. Identify the terms:
The function has three terms:
- [tex]\(-2x^3\)[/tex]: This is the cubic term, where [tex]\( a_3 = -2 \)[/tex].
- [tex]\(4x^2\)[/tex]: This is the quadratic term, where [tex]\( a_2 = 4 \)[/tex].
- [tex]\(5x\)[/tex]: This is the linear term, where [tex]\( a_1 = 5 \)[/tex].
2. Determine the degree:
The highest power of [tex]\( x \)[/tex] in the polynomial is 3. Thus, the degree of the polynomial is 3.
3. Finding roots:
To find the roots, or zeros, of the polynomial, we need to solve:
[tex]\[ -2x^3 + 4x^2 + 5x = 0 \][/tex]
Factoring out [tex]\( x \)[/tex]:
[tex]\[ x(-2x^2 + 4x + 5) = 0 \][/tex]
This gives us one root immediately:
[tex]\[ x = 0 \][/tex]
For the quadratic term, we solve:
[tex]\[ -2x^2 + 4x + 5 = 0 \][/tex]
We can use the quadratic formula for this ( [tex]\(ax^2 + bx + c = 0\)[/tex] ):
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = -2 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 5 \)[/tex].
Calculate the discriminant:
[tex]\[ b^2 - 4ac = 4^2 - 4(-2)(5) = 16 + 40 = 56 \][/tex]
Thus, the roots are:
[tex]\[ x = \frac{-4 \pm \sqrt{56}}{-4} \][/tex]
Simplify [tex]\(\sqrt{56}\)[/tex]:
[tex]\[ \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \][/tex]
Now, we substitute back in:
[tex]\[ x = \frac{-4 \pm 2\sqrt{14}}{-4} = \frac{-4 \pm 2\sqrt{14}}{-4} = 1 \mp \frac{\sqrt{14}}{2} \][/tex]
So, the other two roots are:
[tex]\[ x = 1 + \frac{\sqrt{14}}{2}, \quad x = 1 - \frac{\sqrt{14}}{2} \][/tex]
4. Graphical representation:
The general shape of the graph of [tex]\( -2x^3 + 4x^2 + 5x \)[/tex] can be understood by considering the leading term [tex]\(-2x^3\)[/tex]. Since the coefficient is negative, the graph will start from negative infinity as [tex]\( x \to -\infty \)[/tex] and will end at negative infinity as [tex]\( x \to \infty \)[/tex], producing an overall 'N' shape.
### Key Takeaways
- The polynomial [tex]\( -2x^3 + 4x^2 + 5x \)[/tex] is cubic.
- The roots found are [tex]\( 0 \)[/tex], [tex]\( 1 + \frac{\sqrt{14}}{2} \)[/tex], and [tex]\( 1 - \frac{\sqrt{14}}{2} \)[/tex].
- The shape of the graph is determined by the leading term, which is cubic and negative.
This concisely explores the function and its attributes. If you need further properties like maxima, minima, or inflection points, do let me know!
