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Sagot :
Sure, let's work through the problem step-by-step.
We start with the quadratic equation:
[tex]\[ y = 5x^2 - x - 3 \][/tex]
Here, we want to understand the behavior of this quadratic equation and possibly find its key features such as roots, vertex, and also discuss its general shape. Let's go through these one by one.
### 1. Standard Form and Identifying Coefficients:
The equation is already in standard form [tex]\( ax^2 + bx + c \)[/tex]:
- [tex]\( a = 5 \)[/tex]
- [tex]\( b = -1 \)[/tex]
- [tex]\( c = -3 \)[/tex]
### 2. Roots of the Equation:
To find the roots, we need to solve:
[tex]\[ 5x^2 - x - 3 = 0 \][/tex]
The roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 5 \cdot (-3)}}{2 \cdot 5} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 60}}{10} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{61}}{10} \][/tex]
So, the two roots are:
[tex]\[ x_1 = \frac{1 + \sqrt{61}}{10} \][/tex]
[tex]\[ x_2 = \frac{1 - \sqrt{61}}{10} \][/tex]
### 3. Vertex of the Parabola:
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x_{\text{vertex}} = \frac{-b}{2a} \][/tex]
For our equation:
[tex]\[ x_{\text{vertex}} = \frac{-(-1)}{2 \cdot 5} = \frac{1}{10} \][/tex]
Substitute [tex]\( x_{\text{vertex}} \)[/tex] back into the original equation to find [tex]\( y_{\text{vertex}} \)[/tex]:
[tex]\[ y = 5\left(\frac{1}{10}\right)^2 - \left(\frac{1}{10}\right) - 3 \][/tex]
[tex]\[ y = 5 \cdot \frac{1}{100} - \frac{1}{10} - 3 \][/tex]
[tex]\[ y = \frac{5}{100} - \frac{10}{100} - 3 \][/tex]
[tex]\[ y = \frac{5 - 10 - 300}{100} \][/tex]
[tex]\[ y = \frac{-305}{100} \][/tex]
[tex]\[ y = -3.05 \][/tex]
So, the vertex is at:
[tex]\[ \left(\frac{1}{10}, -3.05\right) \][/tex]
### 4. General Shape of the Parabola:
Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a = 5 \)[/tex]) is positive, the parabola opens upwards.
### Summary:
- The roots of the quadratic equation are given by [tex]\( x_1 = \frac{1 + \sqrt{61}}{10} \)[/tex] and [tex]\( x_2 = \frac{1 - \sqrt{61}}{10} \)[/tex].
- The vertex of the parabola is at [tex]\( \left(\frac{1}{10}, -3.05\right) \)[/tex].
- The parabola opens upwards, indicating that the vertex is a minimum point.
This gives a comprehensive understanding of the quadratic equation [tex]\( y = 5x^2 - x - 3 \)[/tex].
We start with the quadratic equation:
[tex]\[ y = 5x^2 - x - 3 \][/tex]
Here, we want to understand the behavior of this quadratic equation and possibly find its key features such as roots, vertex, and also discuss its general shape. Let's go through these one by one.
### 1. Standard Form and Identifying Coefficients:
The equation is already in standard form [tex]\( ax^2 + bx + c \)[/tex]:
- [tex]\( a = 5 \)[/tex]
- [tex]\( b = -1 \)[/tex]
- [tex]\( c = -3 \)[/tex]
### 2. Roots of the Equation:
To find the roots, we need to solve:
[tex]\[ 5x^2 - x - 3 = 0 \][/tex]
The roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 5 \cdot (-3)}}{2 \cdot 5} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 60}}{10} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{61}}{10} \][/tex]
So, the two roots are:
[tex]\[ x_1 = \frac{1 + \sqrt{61}}{10} \][/tex]
[tex]\[ x_2 = \frac{1 - \sqrt{61}}{10} \][/tex]
### 3. Vertex of the Parabola:
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x_{\text{vertex}} = \frac{-b}{2a} \][/tex]
For our equation:
[tex]\[ x_{\text{vertex}} = \frac{-(-1)}{2 \cdot 5} = \frac{1}{10} \][/tex]
Substitute [tex]\( x_{\text{vertex}} \)[/tex] back into the original equation to find [tex]\( y_{\text{vertex}} \)[/tex]:
[tex]\[ y = 5\left(\frac{1}{10}\right)^2 - \left(\frac{1}{10}\right) - 3 \][/tex]
[tex]\[ y = 5 \cdot \frac{1}{100} - \frac{1}{10} - 3 \][/tex]
[tex]\[ y = \frac{5}{100} - \frac{10}{100} - 3 \][/tex]
[tex]\[ y = \frac{5 - 10 - 300}{100} \][/tex]
[tex]\[ y = \frac{-305}{100} \][/tex]
[tex]\[ y = -3.05 \][/tex]
So, the vertex is at:
[tex]\[ \left(\frac{1}{10}, -3.05\right) \][/tex]
### 4. General Shape of the Parabola:
Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a = 5 \)[/tex]) is positive, the parabola opens upwards.
### Summary:
- The roots of the quadratic equation are given by [tex]\( x_1 = \frac{1 + \sqrt{61}}{10} \)[/tex] and [tex]\( x_2 = \frac{1 - \sqrt{61}}{10} \)[/tex].
- The vertex of the parabola is at [tex]\( \left(\frac{1}{10}, -3.05\right) \)[/tex].
- The parabola opens upwards, indicating that the vertex is a minimum point.
This gives a comprehensive understanding of the quadratic equation [tex]\( y = 5x^2 - x - 3 \)[/tex].
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