To find the solutions to the equation [tex]\( x^2 - 1 = 3 \)[/tex], we can follow these steps to solve it algebraically:
1. Write down the equation:
[tex]\[
x^2 - 1 = 3
\][/tex]
2. Isolate the quadratic term by adding 1 to both sides:
[tex]\[
x^2 - 1 + 1 = 3 + 1
\][/tex]
[tex]\[
x^2 = 4
\][/tex]
3. Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[
x = \pm \sqrt{4}
\][/tex]
4. Simplify the square root:
[tex]\[
x = \pm 2
\][/tex]
This gives us two solutions:
[tex]\[
x = 2 \quad \text{and} \quad x = -2
\][/tex]
Now, let's interpret this graphically. To find the solutions graphically, we can compare two functions:
- [tex]\( y = x^2 - 1 \)[/tex]: a parabola opening upwards with its vertex at [tex]\((0, -1)\)[/tex].
- [tex]\( y = 3 \)[/tex]: a horizontal line.
We need to find the points where these two graphs intersect.
5. Sketch the graph of [tex]\( y = x^2 - 1 \)[/tex]:
- It’s a parabola that opens upwards.
- The vertex of the parabola is at the point [tex]\((0, -1)\)[/tex].
6. Sketch the graph of [tex]\( y = 3 \)[/tex]:
- This is a horizontal line that crosses the y-axis at [tex]\( y = 3 \)[/tex].
To find the intersection points:
- Look for the points [tex]\( x \)[/tex] where [tex]\( y = x^2 - 1 \)[/tex] intersects [tex]\( y = 3 \)[/tex].
7. Set up the equation for the intersection points:
[tex]\[
x^2 - 1 = 3
\][/tex]
Solve the equation [tex]\( x^2 - 1 = 3 \)[/tex], which we already solved algebraically and found that the solutions are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex]. Thus, the graph of [tex]\( y = x^2 - 1 \)[/tex] will intersect the line [tex]\( y = 3 \)[/tex] at [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
In conclusion, the graph needed to find the solutions to [tex]\( x^2 - 1 = 3 \)[/tex] comprises the parabola [tex]\( y = x^2 - 1 \)[/tex] and the horizontal line [tex]\( y = 3 \)[/tex]. Their points of intersection, [tex]\( (-2, 3) \)[/tex] and [tex]\( (2, 3) \)[/tex], give the solutions to the equation, [tex]\( x = -2 \)[/tex] and [tex]\( x = 2 \)[/tex].
