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Sagot :
Certainly! Let's analyze and solve the quadratic function [tex]\( y = -3x^2 + 18x - 5 \)[/tex] step by step.
### 1. Finding the Roots (Solutions)
The roots of the quadratic function are the values of [tex]\( x \)[/tex] where the function [tex]\( y \)[/tex] equals zero. To find the roots, we solve the equation:
[tex]\[ -3x^2 + 18x - 5 = 0 \][/tex]
The solutions to this quadratic equation are:
[tex]\[ x = 3 - \frac{\sqrt{66}}{3} \][/tex]
and
[tex]\[ x = 3 + \frac{\sqrt{66}}{3} \][/tex]
So, the roots are:
[tex]\[ x \approx 0.878 \][/tex]
and
[tex]\[ x \approx 5.122 \][/tex]
### 2. Finding the Vertex
The vertex of a quadratic function in the form [tex]\( y = ax^2 + bx + c \)[/tex] is given by the formula:
[tex]\[ x_{\text{vertex}} = -\frac{b}{2a} \][/tex]
For the function [tex]\( y = -3x^2 + 18x - 5 \)[/tex]:
- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 18 \)[/tex]
Thus,
[tex]\[ x_{\text{vertex}} = -\frac{18}{2(-3)} = -\frac{18}{-6} = 3 \][/tex]
To find the corresponding [tex]\( y \)[/tex]-value, we substitute [tex]\( x = 3 \)[/tex] back into the function:
[tex]\[ y_{\text{vertex}} = -3(3)^2 + 18(3) - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = -3(9) + 54 - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = -27 + 54 - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = 22 \][/tex]
Thus, the vertex of the function is at:
[tex]\[ (3, 22) \][/tex]
### 3. Calculating the First Derivative
The first derivative of the function [tex]\( y = -3x^2 + 18x - 5 \)[/tex] represents the slope of the tangent line to the curve at any point [tex]\( x \)[/tex]. It is found using the power rule for differentiation:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(-3x^2 + 18x - 5) \][/tex]
[tex]\[ \frac{dy}{dx} = -6x + 18 \][/tex]
### 4. Calculating the Second Derivative
The second derivative provides information about the concavity of the function. For the given function, we differentiate the first derivative:
[tex]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-6x + 18) \][/tex]
[tex]\[ \frac{d^2y}{dx^2} = -6 \][/tex]
The second derivative is a constant negative value, [tex]\( -6 \)[/tex], indicating that the function is concave down everywhere.
### Summary of the Results
- Roots of the function: [tex]\( x \approx 0.878 \)[/tex] and [tex]\( x \approx 5.122 \)[/tex]
- Vertex of the function: [tex]\((3, 22)\)[/tex]
- First derivative: [tex]\( 18 - 6x \)[/tex]
- Second derivative: [tex]\( -6 \)[/tex]
These results give us a comprehensive understanding of the behavior of the quadratic function [tex]\( y = -3x^2 + 18x - 5 \)[/tex].
### 1. Finding the Roots (Solutions)
The roots of the quadratic function are the values of [tex]\( x \)[/tex] where the function [tex]\( y \)[/tex] equals zero. To find the roots, we solve the equation:
[tex]\[ -3x^2 + 18x - 5 = 0 \][/tex]
The solutions to this quadratic equation are:
[tex]\[ x = 3 - \frac{\sqrt{66}}{3} \][/tex]
and
[tex]\[ x = 3 + \frac{\sqrt{66}}{3} \][/tex]
So, the roots are:
[tex]\[ x \approx 0.878 \][/tex]
and
[tex]\[ x \approx 5.122 \][/tex]
### 2. Finding the Vertex
The vertex of a quadratic function in the form [tex]\( y = ax^2 + bx + c \)[/tex] is given by the formula:
[tex]\[ x_{\text{vertex}} = -\frac{b}{2a} \][/tex]
For the function [tex]\( y = -3x^2 + 18x - 5 \)[/tex]:
- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 18 \)[/tex]
Thus,
[tex]\[ x_{\text{vertex}} = -\frac{18}{2(-3)} = -\frac{18}{-6} = 3 \][/tex]
To find the corresponding [tex]\( y \)[/tex]-value, we substitute [tex]\( x = 3 \)[/tex] back into the function:
[tex]\[ y_{\text{vertex}} = -3(3)^2 + 18(3) - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = -3(9) + 54 - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = -27 + 54 - 5 \][/tex]
[tex]\[ y_{\text{vertex}} = 22 \][/tex]
Thus, the vertex of the function is at:
[tex]\[ (3, 22) \][/tex]
### 3. Calculating the First Derivative
The first derivative of the function [tex]\( y = -3x^2 + 18x - 5 \)[/tex] represents the slope of the tangent line to the curve at any point [tex]\( x \)[/tex]. It is found using the power rule for differentiation:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(-3x^2 + 18x - 5) \][/tex]
[tex]\[ \frac{dy}{dx} = -6x + 18 \][/tex]
### 4. Calculating the Second Derivative
The second derivative provides information about the concavity of the function. For the given function, we differentiate the first derivative:
[tex]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-6x + 18) \][/tex]
[tex]\[ \frac{d^2y}{dx^2} = -6 \][/tex]
The second derivative is a constant negative value, [tex]\( -6 \)[/tex], indicating that the function is concave down everywhere.
### Summary of the Results
- Roots of the function: [tex]\( x \approx 0.878 \)[/tex] and [tex]\( x \approx 5.122 \)[/tex]
- Vertex of the function: [tex]\((3, 22)\)[/tex]
- First derivative: [tex]\( 18 - 6x \)[/tex]
- Second derivative: [tex]\( -6 \)[/tex]
These results give us a comprehensive understanding of the behavior of the quadratic function [tex]\( y = -3x^2 + 18x - 5 \)[/tex].
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