To find the points of intersection between the given curves and lines, we'll follow these steps:
### Intersection of [tex]\( f(x) = x^2 - 1 \)[/tex] and [tex]\( f(x) = 3 \)[/tex]:
1. Set the equations equal to each other:
[tex]\[
x^2 - 1 = 3
\][/tex]
2. Isolate [tex]\( x^2 \)[/tex] by adding 1 to both sides:
[tex]\[
x^2 = 4
\][/tex]
3. Solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[
x = \pm 2
\][/tex]
So, the solutions are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex]. The points of intersection are:
[tex]\[
(2, 3) \quad \text{and} \quad (-2, 3)
\][/tex]
### Intersection of [tex]\( f(x) = x^2 \)[/tex] and [tex]\( f(x) = 3 \)[/tex]:
1. Set the equations equal to each other:
[tex]\[
x^2 = 3
\][/tex]
2. Solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[
x = \pm \sqrt{3}
\][/tex]
So, the solutions are [tex]\( x = \sqrt{3} \)[/tex] and [tex]\( x = -\sqrt{3} \)[/tex]. The points of intersection are:
[tex]\[
(\sqrt{3}, 3) \quad \text{and} \quad (-\sqrt{3}, 3)
\][/tex]
### Summary:
Combining both results, the points of intersection are:
- For [tex]\( f(x) = x^2 - 1 \)[/tex] and [tex]\( f(x) = 3 \)[/tex]:
[tex]\[
(2, 3) \quad \text{and} \quad (-2, 3)
\][/tex]
- For [tex]\( f(x) = x^2 \)[/tex] and [tex]\( f(x) = 3 \)[/tex]:
[tex]\[
(\sqrt{3}, 3) \quad \text{and} \quad (-\sqrt{3}, 3)
\][/tex]
