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Sagot :
Sure, let's walk through the given quadratic equation step by step to better understand its components and characteristics.
Given the quadratic equation:
[tex]\[ y = 6x^2 - 15x - 9 \][/tex]
### Step 1: Identify the coefficients
The equation is in the standard form [tex]\( y = ax^2 + bx + c \)[/tex]. Here, the coefficients are:
- [tex]\( a = 6 \)[/tex]
- [tex]\( b = -15 \)[/tex]
- [tex]\( c = -9 \)[/tex]
### Step 2: Determine the direction of the parabola
The coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a \)[/tex]) dictates the direction in which the parabola opens:
- If [tex]\( a > 0 \)[/tex], the parabola opens upwards.
- If [tex]\( a < 0 \)[/tex], the parabola opens downwards.
Since [tex]\( a = 6 \)[/tex] (which is positive), the parabola opens upwards.
### Step 3: Find the vertex
The vertex of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values for [tex]\( b \)[/tex] and [tex]\( a \)[/tex]:
[tex]\[ x = -\frac{-15}{2 \cdot 6} = \frac{15}{12} = \frac{5}{4} \][/tex]
To find the y-coordinate of the vertex, we substitute [tex]\( x = \frac{5}{4} \)[/tex] back into the equation:
[tex]\[ y = 6\left(\frac{5}{4}\right)^2 - 15\left(\frac{5}{4}\right) - 9 \][/tex]
[tex]\[ y = 6 \cdot \frac{25}{16} - 15 \cdot \frac{5}{4} - 9 \][/tex]
[tex]\[ y = \frac{150}{16} - \frac{75}{4} - 9 \][/tex]
[tex]\[ y = \frac{75}{8} - \frac{150}{8} - \frac{72}{8} \][/tex]
[tex]\[ y = \frac{75 - 150 - 72}{8} \][/tex]
[tex]\[ y = \frac{-147}{8} \][/tex]
[tex]\[ y = -18.375 \][/tex]
So, the vertex of the parabola is at [tex]\( \left( \frac{5}{4}, -18.375 \right) \)[/tex].
### Step 4: Find the x-intercepts (roots)
The x-intercepts occur where [tex]\( y = 0 \)[/tex]. We solve the equation [tex]\( 6x^2 - 15x - 9 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 6 \cdot (-9)}}{2 \cdot 6} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{225 + 216}}{12} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{441}}{12} \][/tex]
[tex]\[ x = \frac{15 \pm 21}{12} \][/tex]
This leads to two solutions:
[tex]\[ x = \frac{15 + 21}{12} = \frac{36}{12} = 3 \][/tex]
[tex]\[ x = \frac{15 - 21}{12} = \frac{-6}{12} = -\frac{1}{2} \][/tex]
So, the x-intercepts are [tex]\( x = 3 \)[/tex] and [tex]\( x = -\frac{1}{2} \)[/tex].
### Step 5: Find the y-intercept
The y-intercept occurs where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 6(0)^2 - 15(0) - 9 \][/tex]
[tex]\[ y = -9 \][/tex]
So, the y-intercept is [tex]\( y = -9 \)[/tex].
### Conclusion
To summarize the characteristics of the quadratic equation [tex]\( y = 6x^2 - 15x - 9 \)[/tex]:
- The parabola opens upwards.
- The vertex is at [tex]\( \left( \frac{5}{4}, -18.375 \right) \)[/tex].
- The x-intercepts are [tex]\( x = 3 \)[/tex] and [tex]\( x = -\frac{1}{2} \)[/tex].
- The y-intercept is [tex]\( y = -9 \)[/tex].
Given the quadratic equation:
[tex]\[ y = 6x^2 - 15x - 9 \][/tex]
### Step 1: Identify the coefficients
The equation is in the standard form [tex]\( y = ax^2 + bx + c \)[/tex]. Here, the coefficients are:
- [tex]\( a = 6 \)[/tex]
- [tex]\( b = -15 \)[/tex]
- [tex]\( c = -9 \)[/tex]
### Step 2: Determine the direction of the parabola
The coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a \)[/tex]) dictates the direction in which the parabola opens:
- If [tex]\( a > 0 \)[/tex], the parabola opens upwards.
- If [tex]\( a < 0 \)[/tex], the parabola opens downwards.
Since [tex]\( a = 6 \)[/tex] (which is positive), the parabola opens upwards.
### Step 3: Find the vertex
The vertex of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values for [tex]\( b \)[/tex] and [tex]\( a \)[/tex]:
[tex]\[ x = -\frac{-15}{2 \cdot 6} = \frac{15}{12} = \frac{5}{4} \][/tex]
To find the y-coordinate of the vertex, we substitute [tex]\( x = \frac{5}{4} \)[/tex] back into the equation:
[tex]\[ y = 6\left(\frac{5}{4}\right)^2 - 15\left(\frac{5}{4}\right) - 9 \][/tex]
[tex]\[ y = 6 \cdot \frac{25}{16} - 15 \cdot \frac{5}{4} - 9 \][/tex]
[tex]\[ y = \frac{150}{16} - \frac{75}{4} - 9 \][/tex]
[tex]\[ y = \frac{75}{8} - \frac{150}{8} - \frac{72}{8} \][/tex]
[tex]\[ y = \frac{75 - 150 - 72}{8} \][/tex]
[tex]\[ y = \frac{-147}{8} \][/tex]
[tex]\[ y = -18.375 \][/tex]
So, the vertex of the parabola is at [tex]\( \left( \frac{5}{4}, -18.375 \right) \)[/tex].
### Step 4: Find the x-intercepts (roots)
The x-intercepts occur where [tex]\( y = 0 \)[/tex]. We solve the equation [tex]\( 6x^2 - 15x - 9 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 6 \cdot (-9)}}{2 \cdot 6} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{225 + 216}}{12} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{441}}{12} \][/tex]
[tex]\[ x = \frac{15 \pm 21}{12} \][/tex]
This leads to two solutions:
[tex]\[ x = \frac{15 + 21}{12} = \frac{36}{12} = 3 \][/tex]
[tex]\[ x = \frac{15 - 21}{12} = \frac{-6}{12} = -\frac{1}{2} \][/tex]
So, the x-intercepts are [tex]\( x = 3 \)[/tex] and [tex]\( x = -\frac{1}{2} \)[/tex].
### Step 5: Find the y-intercept
The y-intercept occurs where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 6(0)^2 - 15(0) - 9 \][/tex]
[tex]\[ y = -9 \][/tex]
So, the y-intercept is [tex]\( y = -9 \)[/tex].
### Conclusion
To summarize the characteristics of the quadratic equation [tex]\( y = 6x^2 - 15x - 9 \)[/tex]:
- The parabola opens upwards.
- The vertex is at [tex]\( \left( \frac{5}{4}, -18.375 \right) \)[/tex].
- The x-intercepts are [tex]\( x = 3 \)[/tex] and [tex]\( x = -\frac{1}{2} \)[/tex].
- The y-intercept is [tex]\( y = -9 \)[/tex].
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