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To solve for the [tex]\(x\)[/tex]-intercepts and the coordinates of the vertex for the parabola given by the equation [tex]\(y = -x^2 + 2x - 1\)[/tex], we need to follow these steps:
1. Identify the coefficients of the quadratic equation: [tex]\(a = -1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -1\)[/tex].
2. Calculate the discriminant: The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
3. Determine the [tex]\(x\)[/tex]-intercepts:
- If [tex]\(\Delta > 0\)[/tex], the quadratic equation has two distinct real roots.
- If [tex]\(\Delta = 0\)[/tex], the quadratic equation has exactly one real root.
- If [tex]\(\Delta < 0\)[/tex], the quadratic equation has no real roots.
4. Find the [tex]\(x\)[/tex]-intercepts: The roots of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
5. Calculate the vertex of the parabola: The vertex [tex]\((x_v, y_v)\)[/tex] of a parabola [tex]\(y = ax^2 + bx + c\)[/tex] is found using:
[tex]\[ x_v = -\frac{b}{2a} \][/tex]
[tex]\[ y_v = a(x_v)^2 + b(x_v) + c \][/tex]
Following these steps:
1. Identify coefficients:
[tex]\(a = -1\)[/tex], [tex]\(b = 2\)[/tex], [tex]\(c = -1\)[/tex]
2. Calculate the discriminant:
[tex]\[ \Delta = 2^2 - 4(-1)(-1) = 4 - 4 = 0 \][/tex]
3. Since [tex]\(\Delta = 0\)[/tex], the quadratic equation has exactly one real root.
4. Find the [tex]\(x\)[/tex]-intercept:
[tex]\[ x = \frac{-2 \pm \sqrt{0}}{2(-1)} = \frac{-2}{-2} = 1 \][/tex]
Therefore, the [tex]\(x\)[/tex]-intercept is [tex]\(x = 1.0\)[/tex].
5. Calculate the vertex:
[tex]\[ x_v = -\frac{2}{2(-1)} = 1 \][/tex]
[tex]\[ y_v = -1(1)^2 + 2(1) - 1 = -1 + 2 - 1 = 0 \][/tex]
Therefore, the coordinates of the vertex are [tex]\((1.0, 0.0)\)[/tex].
Results:
- Discriminant: [tex]\(0\)[/tex]
- [tex]\(x\)[/tex]-intercept(s): [tex]\(1.0\)[/tex]
- Vertex: [tex]\((1.0, 0.0)\)[/tex]
So, the [tex]\(x\)[/tex]-intercept is [tex]\(1.0\)[/tex], and the coordinates of the vertex are [tex]\((1.0, 0.0)\)[/tex].
1. Identify the coefficients of the quadratic equation: [tex]\(a = -1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -1\)[/tex].
2. Calculate the discriminant: The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
3. Determine the [tex]\(x\)[/tex]-intercepts:
- If [tex]\(\Delta > 0\)[/tex], the quadratic equation has two distinct real roots.
- If [tex]\(\Delta = 0\)[/tex], the quadratic equation has exactly one real root.
- If [tex]\(\Delta < 0\)[/tex], the quadratic equation has no real roots.
4. Find the [tex]\(x\)[/tex]-intercepts: The roots of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
5. Calculate the vertex of the parabola: The vertex [tex]\((x_v, y_v)\)[/tex] of a parabola [tex]\(y = ax^2 + bx + c\)[/tex] is found using:
[tex]\[ x_v = -\frac{b}{2a} \][/tex]
[tex]\[ y_v = a(x_v)^2 + b(x_v) + c \][/tex]
Following these steps:
1. Identify coefficients:
[tex]\(a = -1\)[/tex], [tex]\(b = 2\)[/tex], [tex]\(c = -1\)[/tex]
2. Calculate the discriminant:
[tex]\[ \Delta = 2^2 - 4(-1)(-1) = 4 - 4 = 0 \][/tex]
3. Since [tex]\(\Delta = 0\)[/tex], the quadratic equation has exactly one real root.
4. Find the [tex]\(x\)[/tex]-intercept:
[tex]\[ x = \frac{-2 \pm \sqrt{0}}{2(-1)} = \frac{-2}{-2} = 1 \][/tex]
Therefore, the [tex]\(x\)[/tex]-intercept is [tex]\(x = 1.0\)[/tex].
5. Calculate the vertex:
[tex]\[ x_v = -\frac{2}{2(-1)} = 1 \][/tex]
[tex]\[ y_v = -1(1)^2 + 2(1) - 1 = -1 + 2 - 1 = 0 \][/tex]
Therefore, the coordinates of the vertex are [tex]\((1.0, 0.0)\)[/tex].
Results:
- Discriminant: [tex]\(0\)[/tex]
- [tex]\(x\)[/tex]-intercept(s): [tex]\(1.0\)[/tex]
- Vertex: [tex]\((1.0, 0.0)\)[/tex]
So, the [tex]\(x\)[/tex]-intercept is [tex]\(1.0\)[/tex], and the coordinates of the vertex are [tex]\((1.0, 0.0)\)[/tex].
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