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To solve the given quadratic equation [tex]\( y = -x^2 + 8x + 2 \)[/tex] and find its vertex and solutions (roots), let's go through the detailed steps.
### 1. Find the Vertex of the Parabola
The vertex form of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] gives us the peak (or trough) of the parabola. The vertex [tex]\((h, k)\)[/tex] can be calculated using the formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
where [tex]\(a\)[/tex], and [tex]\(b\)[/tex] are coefficients from the equation [tex]\( ax^2 + bx + c \)[/tex].
For the given equation [tex]\( y = -x^2 + 8x + 2 \)[/tex]:
- [tex]\(a = -1\)[/tex]
- [tex]\(b = 8\)[/tex]
Using the formula:
[tex]\[ h = -\frac{8}{2(-1)} = -\frac{8}{-2} = 4 \][/tex]
To find [tex]\(k\)[/tex], substitute [tex]\(x = 4\)[/tex] back into the equation:
[tex]\[ k = -4^2 + 8 \cdot 4 + 2 = -16 + 32 + 2 = 18 \][/tex]
Therefore, the vertex is:
[tex]\[ (4, 18) \][/tex]
### 2. Find the Roots of the Quadratic Equation
To find the roots (solutions) of the quadratic equation [tex]\( y = -x^2 + 8x + 2 \)[/tex], set [tex]\( y = 0 \)[/tex]:
[tex]\[ -x^2 + 8x + 2 = 0 \][/tex]
This is a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. To find the roots, use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the given equation:
- [tex]\(a = -1\)[/tex]
- [tex]\(b = 8\)[/tex]
- [tex]\(c = 2\)[/tex]
First, compute the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 8^2 - 4(-1)(2) = 64 + 8 = 72 \][/tex]
Then, apply the quadratic formula:
[tex]\[ x = \frac{-8 \pm \sqrt{72}}{2(-1)} = \frac{-8 \pm 6\sqrt{2}}{-2} \][/tex]
This simplifies to:
[tex]\[ x = 4 \pm 3\sqrt{2} \][/tex]
So, the roots are:
[tex]\[ x = 4 - 3\sqrt{2} \quad \text{and} \quad x = 4 + 3\sqrt{2} \][/tex]
### Summary:
- Vertex: [tex]\((4, 18)\)[/tex]
- Solutions: [tex]\( x = 4 - 3\sqrt{2} \)[/tex] and [tex]\( x = 4 + 3\sqrt{2} \)[/tex]
Thus, the solutions and vertex of the quadratic equation [tex]\( y = -x^2 + 8x + 2 \)[/tex] are:
[tex]\[ \left(4 - 3\sqrt{2}, 0\right) \quad \text{and} \quad \left(4 + 3\sqrt{2}, 0\right) \][/tex]
with the vertex:
[tex]\[ (4, 18) \][/tex]
### 1. Find the Vertex of the Parabola
The vertex form of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] gives us the peak (or trough) of the parabola. The vertex [tex]\((h, k)\)[/tex] can be calculated using the formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
where [tex]\(a\)[/tex], and [tex]\(b\)[/tex] are coefficients from the equation [tex]\( ax^2 + bx + c \)[/tex].
For the given equation [tex]\( y = -x^2 + 8x + 2 \)[/tex]:
- [tex]\(a = -1\)[/tex]
- [tex]\(b = 8\)[/tex]
Using the formula:
[tex]\[ h = -\frac{8}{2(-1)} = -\frac{8}{-2} = 4 \][/tex]
To find [tex]\(k\)[/tex], substitute [tex]\(x = 4\)[/tex] back into the equation:
[tex]\[ k = -4^2 + 8 \cdot 4 + 2 = -16 + 32 + 2 = 18 \][/tex]
Therefore, the vertex is:
[tex]\[ (4, 18) \][/tex]
### 2. Find the Roots of the Quadratic Equation
To find the roots (solutions) of the quadratic equation [tex]\( y = -x^2 + 8x + 2 \)[/tex], set [tex]\( y = 0 \)[/tex]:
[tex]\[ -x^2 + 8x + 2 = 0 \][/tex]
This is a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. To find the roots, use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the given equation:
- [tex]\(a = -1\)[/tex]
- [tex]\(b = 8\)[/tex]
- [tex]\(c = 2\)[/tex]
First, compute the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 8^2 - 4(-1)(2) = 64 + 8 = 72 \][/tex]
Then, apply the quadratic formula:
[tex]\[ x = \frac{-8 \pm \sqrt{72}}{2(-1)} = \frac{-8 \pm 6\sqrt{2}}{-2} \][/tex]
This simplifies to:
[tex]\[ x = 4 \pm 3\sqrt{2} \][/tex]
So, the roots are:
[tex]\[ x = 4 - 3\sqrt{2} \quad \text{and} \quad x = 4 + 3\sqrt{2} \][/tex]
### Summary:
- Vertex: [tex]\((4, 18)\)[/tex]
- Solutions: [tex]\( x = 4 - 3\sqrt{2} \)[/tex] and [tex]\( x = 4 + 3\sqrt{2} \)[/tex]
Thus, the solutions and vertex of the quadratic equation [tex]\( y = -x^2 + 8x + 2 \)[/tex] are:
[tex]\[ \left(4 - 3\sqrt{2}, 0\right) \quad \text{and} \quad \left(4 + 3\sqrt{2}, 0\right) \][/tex]
with the vertex:
[tex]\[ (4, 18) \][/tex]
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