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Sagot :
Certainly! Let's solve the given quadratic equation step-by-step.
The given quadratic equation is:
[tex]\[ y = -4x^2 - 16x - 1 \][/tex]
### 1. Understanding the Equation
This is a standard quadratic equation in the form:
[tex]\[ y = ax^2 + bx + c \][/tex]
where:
- [tex]\( a = -4 \)[/tex]
- [tex]\( b = -16 \)[/tex]
- [tex]\( c = -1 \)[/tex]
### 2. Vertex of the Parabola
The vertex of a parabola defined by [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = \frac{-b}{2a} \][/tex]
Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = \frac{-(-16)}{2(-4)} \][/tex]
[tex]\[ x = \frac{16}{-8} \][/tex]
[tex]\[ x = -2 \][/tex]
To find the corresponding [tex]\( y \)[/tex]-coordinate, substitute [tex]\( x = -2 \)[/tex] back into the original equation:
[tex]\[ y = -4(-2)^2 - 16(-2) - 1 \][/tex]
[tex]\[ y = -4(4) + 32 - 1 \][/tex]
[tex]\[ y = -16 + 32 - 1 \][/tex]
[tex]\[ y = 15 \][/tex]
Thus, the vertex of the parabola is at [tex]\((-2, 15)\)[/tex].
### 3. Intercepts
#### i. y-intercept
To find the y-intercept, substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = -4(0)^2 - 16(0) - 1 \][/tex]
[tex]\[ y = -1 \][/tex]
So, the y-intercept is [tex]\( (0, -1) \)[/tex].
#### ii. x-intercepts
To find the x-intercepts, set [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = -4x^2 - 16x - 1 \][/tex]
This is a quadratic equation and can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = -4 \)[/tex], [tex]\( b = -16 \)[/tex], and [tex]\( c = -1 \)[/tex]:
[tex]\[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(-4)(-1)}}{2(-4)} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{256 - 16}}{-8} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{240}}{-8} \][/tex]
[tex]\[ x = \frac{16 \pm 4\sqrt{15}}{-8} \][/tex]
This simplifies to:
[tex]\[ x = \frac{16}{-8} \pm \frac{4\sqrt{15}}{-8} \][/tex]
[tex]\[ x = -2 \pm \frac{\sqrt{15}}{2} \][/tex]
Thus, the x-intercepts are:
[tex]\[ x_1 = -2 + \frac{\sqrt{15}}{2} \][/tex]
[tex]\[ x_2 = -2 - \frac{\sqrt{15}}{2} \][/tex]
### Summary
- The vertex of the parabola is [tex]\((-2, 15)\)[/tex].
- The y-intercept is [tex]\((0, -1)\)[/tex].
- The x-intercepts are [tex]\(\left( -2 + \frac{\sqrt{15}}{2}, 0 \right)\)[/tex] and [tex]\(\left( -2 - \frac{\sqrt{15}}{2}, 0 \right)\)[/tex].
The quadratic equation [tex]\( y = -4x^2 - 16x - 1 \)[/tex] defines a downward-opening parabola with these intercepts and vertex.
The given quadratic equation is:
[tex]\[ y = -4x^2 - 16x - 1 \][/tex]
### 1. Understanding the Equation
This is a standard quadratic equation in the form:
[tex]\[ y = ax^2 + bx + c \][/tex]
where:
- [tex]\( a = -4 \)[/tex]
- [tex]\( b = -16 \)[/tex]
- [tex]\( c = -1 \)[/tex]
### 2. Vertex of the Parabola
The vertex of a parabola defined by [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = \frac{-b}{2a} \][/tex]
Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = \frac{-(-16)}{2(-4)} \][/tex]
[tex]\[ x = \frac{16}{-8} \][/tex]
[tex]\[ x = -2 \][/tex]
To find the corresponding [tex]\( y \)[/tex]-coordinate, substitute [tex]\( x = -2 \)[/tex] back into the original equation:
[tex]\[ y = -4(-2)^2 - 16(-2) - 1 \][/tex]
[tex]\[ y = -4(4) + 32 - 1 \][/tex]
[tex]\[ y = -16 + 32 - 1 \][/tex]
[tex]\[ y = 15 \][/tex]
Thus, the vertex of the parabola is at [tex]\((-2, 15)\)[/tex].
### 3. Intercepts
#### i. y-intercept
To find the y-intercept, substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = -4(0)^2 - 16(0) - 1 \][/tex]
[tex]\[ y = -1 \][/tex]
So, the y-intercept is [tex]\( (0, -1) \)[/tex].
#### ii. x-intercepts
To find the x-intercepts, set [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = -4x^2 - 16x - 1 \][/tex]
This is a quadratic equation and can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = -4 \)[/tex], [tex]\( b = -16 \)[/tex], and [tex]\( c = -1 \)[/tex]:
[tex]\[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(-4)(-1)}}{2(-4)} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{256 - 16}}{-8} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{240}}{-8} \][/tex]
[tex]\[ x = \frac{16 \pm 4\sqrt{15}}{-8} \][/tex]
This simplifies to:
[tex]\[ x = \frac{16}{-8} \pm \frac{4\sqrt{15}}{-8} \][/tex]
[tex]\[ x = -2 \pm \frac{\sqrt{15}}{2} \][/tex]
Thus, the x-intercepts are:
[tex]\[ x_1 = -2 + \frac{\sqrt{15}}{2} \][/tex]
[tex]\[ x_2 = -2 - \frac{\sqrt{15}}{2} \][/tex]
### Summary
- The vertex of the parabola is [tex]\((-2, 15)\)[/tex].
- The y-intercept is [tex]\((0, -1)\)[/tex].
- The x-intercepts are [tex]\(\left( -2 + \frac{\sqrt{15}}{2}, 0 \right)\)[/tex] and [tex]\(\left( -2 - \frac{\sqrt{15}}{2}, 0 \right)\)[/tex].
The quadratic equation [tex]\( y = -4x^2 - 16x - 1 \)[/tex] defines a downward-opening parabola with these intercepts and vertex.
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