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.
