Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Let's look at the key features of the quadratic function [tex]\(h(x) = (x-1)^2 - 9\)[/tex].
### [tex]\(x\)[/tex]-intercepts
To find the [tex]\(x\)[/tex]-intercepts, we need to solve for [tex]\(x\)[/tex] when [tex]\(h(x) = 0\)[/tex]:
[tex]\[ (x-1)^2 - 9 = 0 \][/tex]
Rearrange and solve for [tex]\(x\)[/tex]:
[tex]\[ (x-1)^2 = 9 \][/tex]
Take the square root of both sides:
[tex]\[ x-1 = \pm 3 \][/tex]
This gives us two solutions:
[tex]\[ x-1 = 3 \quad \text{or} \quad x-1 = -3 \][/tex]
Solving these:
[tex]\[ x = 4 \quad \text{or} \quad x = -2 \][/tex]
So, the [tex]\(x\)[/tex]-intercepts are at [tex]\(x = 4\)[/tex] and [tex]\(x = -2\)[/tex].
### [tex]\(y\)[/tex]-intercept
The [tex]\(y\)[/tex]-intercept occurs when [tex]\(x = 0\)[/tex]:
[tex]\[ h(0) = (0-1)^2 - 9 = 1 - 9 = -8 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is at [tex]\((0, -8)\)[/tex].
### Vertex
The given function [tex]\(h(x)\)[/tex] is in vertex form [tex]\(h(x) = (x-1)^2 - 9\)[/tex]. The vertex form of a quadratic equation is [tex]\(y = a(x-h)^2 + k\)[/tex], where [tex]\((h, k)\)[/tex] is the vertex. Here, [tex]\((h, k) = (1, -9)\)[/tex].
So, the vertex of [tex]\(h(x)\)[/tex] is at [tex]\((1, -9)\)[/tex].
### Axis of Symmetry
The axis of symmetry for a quadratic in vertex form [tex]\(y = a(x-h)^2 + k\)[/tex] is [tex]\(x = h\)[/tex]. Therefore, the axis of symmetry is:
[tex]\[ x = 1 \][/tex]
### Summary
- [tex]\(x\)[/tex]-intercepts: [tex]\((4, 0)\)[/tex] and [tex]\((-2, 0)\)[/tex]
- [tex]\(y\)[/tex]-intercept: [tex]\((0, -8)\)[/tex]
- Vertex: [tex]\((1, -9)\)[/tex]
- Axis of Symmetry: [tex]\(x = 1\)[/tex]
Using this information, you can plot these points on the provided graph. Remember to mark:
- The [tex]\(x\)[/tex]-intercepts at [tex]\((4, 0)\)[/tex] and [tex]\((-2, 0)\)[/tex]
- The [tex]\(y\)[/tex]-intercept at [tex]\((0, -8)\)[/tex]
- The vertex at [tex]\((1, -9)\)[/tex]
- The axis of symmetry with a dashed vertical line at [tex]\(x = 1\)[/tex]
These points and the line will give you a clear picture of the function's key features.
### [tex]\(x\)[/tex]-intercepts
To find the [tex]\(x\)[/tex]-intercepts, we need to solve for [tex]\(x\)[/tex] when [tex]\(h(x) = 0\)[/tex]:
[tex]\[ (x-1)^2 - 9 = 0 \][/tex]
Rearrange and solve for [tex]\(x\)[/tex]:
[tex]\[ (x-1)^2 = 9 \][/tex]
Take the square root of both sides:
[tex]\[ x-1 = \pm 3 \][/tex]
This gives us two solutions:
[tex]\[ x-1 = 3 \quad \text{or} \quad x-1 = -3 \][/tex]
Solving these:
[tex]\[ x = 4 \quad \text{or} \quad x = -2 \][/tex]
So, the [tex]\(x\)[/tex]-intercepts are at [tex]\(x = 4\)[/tex] and [tex]\(x = -2\)[/tex].
### [tex]\(y\)[/tex]-intercept
The [tex]\(y\)[/tex]-intercept occurs when [tex]\(x = 0\)[/tex]:
[tex]\[ h(0) = (0-1)^2 - 9 = 1 - 9 = -8 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is at [tex]\((0, -8)\)[/tex].
### Vertex
The given function [tex]\(h(x)\)[/tex] is in vertex form [tex]\(h(x) = (x-1)^2 - 9\)[/tex]. The vertex form of a quadratic equation is [tex]\(y = a(x-h)^2 + k\)[/tex], where [tex]\((h, k)\)[/tex] is the vertex. Here, [tex]\((h, k) = (1, -9)\)[/tex].
So, the vertex of [tex]\(h(x)\)[/tex] is at [tex]\((1, -9)\)[/tex].
### Axis of Symmetry
The axis of symmetry for a quadratic in vertex form [tex]\(y = a(x-h)^2 + k\)[/tex] is [tex]\(x = h\)[/tex]. Therefore, the axis of symmetry is:
[tex]\[ x = 1 \][/tex]
### Summary
- [tex]\(x\)[/tex]-intercepts: [tex]\((4, 0)\)[/tex] and [tex]\((-2, 0)\)[/tex]
- [tex]\(y\)[/tex]-intercept: [tex]\((0, -8)\)[/tex]
- Vertex: [tex]\((1, -9)\)[/tex]
- Axis of Symmetry: [tex]\(x = 1\)[/tex]
Using this information, you can plot these points on the provided graph. Remember to mark:
- The [tex]\(x\)[/tex]-intercepts at [tex]\((4, 0)\)[/tex] and [tex]\((-2, 0)\)[/tex]
- The [tex]\(y\)[/tex]-intercept at [tex]\((0, -8)\)[/tex]
- The vertex at [tex]\((1, -9)\)[/tex]
- The axis of symmetry with a dashed vertical line at [tex]\(x = 1\)[/tex]
These points and the line will give you a clear picture of the function's key features.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.