Which equation has a graph that is a parabola with a vertex at [tex]\((-1, -1)\)[/tex]?

A. [tex]\(y = (x-1)^2 + 1\)[/tex]
B. [tex]\(y = (x-1)^2 - 1\)[/tex]
C. [tex]\(y = (x+1)^2 + 1\)[/tex]
D. [tex]\(y = (x+1)^2 - 1\)[/tex]



Answer :

To determine which equation represents a parabola with a vertex at [tex]\((-1, -1)\)[/tex], we need to consider the general form of the equation of a parabola in vertex form, which is:

[tex]\[ y = (x - h)^2 + k \][/tex]

Here, [tex]\((h, k)\)[/tex] represents the vertex of the parabola. For our specific problem, the vertex given is [tex]\((-1, -1)\)[/tex]. Therefore, we can substitute [tex]\(h = -1\)[/tex] and [tex]\(k = -1\)[/tex] into the general form of the equation:

[tex]\[ y = (x - (-1))^2 + (-1) \][/tex]
[tex]\[ y = (x + 1)^2 - 1 \][/tex]

Now let's compare this with the given options:

1. [tex]\(y = (x-1)^2 + 1\)[/tex]
2. [tex]\(y = (x-1)^2 - 1\)[/tex]
3. [tex]\(y = (x+1)^2 + 1\)[/tex]
4. [tex]\(y = (x+1)^2 - 1\)[/tex]

We need to find the equation that matches the form [tex]\(y = (x + 1)^2 - 1\)[/tex]. Evaluating the options:

1. [tex]\(y = (x - 1)^2 + 1\)[/tex]
- Vertex: [tex]\((1, 1)\)[/tex]

2. [tex]\(y = (x - 1)^2 - 1\)[/tex]
- Vertex: [tex]\((1, -1)\)[/tex]

3. [tex]\(y = (x + 1)^2 + 1\)[/tex]
- Vertex: [tex]\((-1, 1)\)[/tex]

4. [tex]\(y = (x + 1)^2 - 1\)[/tex]
- Vertex: [tex]\((-1, -1)\)[/tex]

From these evaluations, option 4, [tex]\(y = (x + 1)^2 - 1\)[/tex], is the equation with a vertex at [tex]\((-1, -1)\)[/tex].

Thus, the correct equation is:

[tex]\[ y = (x + 1)^2 - 1 \][/tex]

This corresponds to option 4.