Certainly! To solve the equation [tex]\(x^2 - 1 = 0\)[/tex] by graphing, we can follow these steps:
1. Rewrite the equation in standard form:
The given equation is [tex]\(x^2 - 1 = 0\)[/tex], which can be rewritten as:
[tex]\[
y = x^2 - 1
\][/tex]
2. Identify the vertex of the parabola:
The equation [tex]\(y = x^2 - 1\)[/tex] is a quadratic equation in the form [tex]\(y = ax^2 + bx + c\)[/tex]. Here, [tex]\(a = 1\)[/tex], [tex]\(b = 0\)[/tex], and [tex]\(c = -1\)[/tex]. The vertex of a parabola defined by a quadratic equation [tex]\(y = ax^2 + bx + c\)[/tex] is at:
[tex]\[
x = -\frac{b}{2a} = -\frac{0}{2 \cdot 1} = 0
\][/tex]
Therefore, the vertex is at [tex]\((0, -1)\)[/tex].
3. Plot the vertex:
The vertex of the parabola [tex]\(y = x^2 - 1\)[/tex] is the point [tex]\((0, -1)\)[/tex].
4. Find additional points:
To plot the parabola, we need points on either side of the vertex:
- For [tex]\(x = 1\)[/tex]:
[tex]\[
y = 1^2 - 1 = 0
\][/tex]
So, the point [tex]\((1, 0)\)[/tex] is on the parabola.
- For [tex]\(x = -1\)[/tex]:
[tex]\[
y = (-1)^2 - 1 = 0
\][/tex]
So, the point [tex]\((-1, 0)\)[/tex] is on the parabola.
- For [tex]\(x = 2\)[/tex]:
[tex]\[
y = 2^2 - 1 = 3
\][/tex]
So, the point [tex]\((2, 3)\)[/tex] is on the parabola.
- For [tex]\(x = -2\)[/tex]:
[tex]\[
y = (-2)^2 - 1 = 3
\][/tex]
So, the point [tex]\((-2, 3)\)[/tex] is on the parabola.
5. Graph the parabola:
Plot the points on a graph: [tex]\((0, -1)\)[/tex], [tex]\((1, 0)\)[/tex], [tex]\((-1, 0)\)[/tex], [tex]\((2, 3)\)[/tex], and [tex]\((-2, 3)\)[/tex], and sketch the parabola through these points.
6. Find the solutions from the graph:
The solutions to the equation [tex]\(x^2 - 1 = 0\)[/tex] are the [tex]\(x\)[/tex]-coordinates where the graph intersects the x-axis. From the graph, it's clear that the parabola intersects the x-axis at the points [tex]\((1, 0)\)[/tex] and [tex]\((-1, 0)\)[/tex].
Therefore, the solutions to the equation [tex]\(x^2 - 1 = 0\)[/tex] are:
[tex]\[
x = 1, -1
\][/tex]
