Answer :
To find the intercepts of the function [tex]\( x = -y^2 + 25 \)[/tex], let's break it down into two parts: finding the [tex]\(x\)[/tex]-intercepts and the [tex]\(y\)[/tex]-intercepts.
### Finding the [tex]\(x\)[/tex]-Intercepts
The [tex]\(x\)[/tex]-intercepts occur where the graph of the equation crosses the [tex]\(x\)[/tex]-axis. This happens when [tex]\( y = 0 \)[/tex].
1. Set [tex]\( y = 0 \)[/tex] in the equation [tex]\( x = -y^2 + 25 \)[/tex]:
[tex]\[ x = - (0)^2 + 25 \][/tex]
Simplify the equation:
[tex]\[ x = 25 \][/tex]
So, the [tex]\( x \)[/tex]-intercept is [tex]\( (25, 0) \)[/tex].
### Finding the [tex]\(y\)[/tex]-Intercepts
The [tex]\(y\)[/tex]-intercepts occur where the graph of the equation crosses the [tex]\(y\)[/tex]-axis. This happens when [tex]\( x = 0 \)[/tex].
1. Set [tex]\( x = 0 \)[/tex] in the equation [tex]\( x = -y^2 + 25 \)[/tex]:
[tex]\[ 0 = -y^2 + 25 \][/tex]
2. Solve for [tex]\( y \)[/tex]:
[tex]\[ -y^2 = -25 \][/tex]
[tex]\[ y^2 = 25 \][/tex]
[tex]\[ y = \pm \sqrt{25} \][/tex]
[tex]\[ y = \pm 5 \][/tex]
So, the [tex]\( y \)[/tex]-intercepts are [tex]\( (0, 5) \)[/tex] and [tex]\( (0, -5) \)[/tex].
### Summary:
- The [tex]\( x \)[/tex]-intercept is [tex]\( (25, 0) \)[/tex].
- The [tex]\( y \)[/tex]-intercepts are [tex]\( (0, 5) \)[/tex] and [tex]\( (0, -5) \)[/tex].
Thus, the intercepts for the given function [tex]\( x = -y^2 + 25 \)[/tex] are:
- [tex]\(x\)[/tex]-intercepts: [tex]\( (25, 0) \)[/tex]
- [tex]\(y\)[/tex]-intercepts: [tex]\( (0, 5) \)[/tex] and [tex]\( (0, -5) \)[/tex]
### Finding the [tex]\(x\)[/tex]-Intercepts
The [tex]\(x\)[/tex]-intercepts occur where the graph of the equation crosses the [tex]\(x\)[/tex]-axis. This happens when [tex]\( y = 0 \)[/tex].
1. Set [tex]\( y = 0 \)[/tex] in the equation [tex]\( x = -y^2 + 25 \)[/tex]:
[tex]\[ x = - (0)^2 + 25 \][/tex]
Simplify the equation:
[tex]\[ x = 25 \][/tex]
So, the [tex]\( x \)[/tex]-intercept is [tex]\( (25, 0) \)[/tex].
### Finding the [tex]\(y\)[/tex]-Intercepts
The [tex]\(y\)[/tex]-intercepts occur where the graph of the equation crosses the [tex]\(y\)[/tex]-axis. This happens when [tex]\( x = 0 \)[/tex].
1. Set [tex]\( x = 0 \)[/tex] in the equation [tex]\( x = -y^2 + 25 \)[/tex]:
[tex]\[ 0 = -y^2 + 25 \][/tex]
2. Solve for [tex]\( y \)[/tex]:
[tex]\[ -y^2 = -25 \][/tex]
[tex]\[ y^2 = 25 \][/tex]
[tex]\[ y = \pm \sqrt{25} \][/tex]
[tex]\[ y = \pm 5 \][/tex]
So, the [tex]\( y \)[/tex]-intercepts are [tex]\( (0, 5) \)[/tex] and [tex]\( (0, -5) \)[/tex].
### Summary:
- The [tex]\( x \)[/tex]-intercept is [tex]\( (25, 0) \)[/tex].
- The [tex]\( y \)[/tex]-intercepts are [tex]\( (0, 5) \)[/tex] and [tex]\( (0, -5) \)[/tex].
Thus, the intercepts for the given function [tex]\( x = -y^2 + 25 \)[/tex] are:
- [tex]\(x\)[/tex]-intercepts: [tex]\( (25, 0) \)[/tex]
- [tex]\(y\)[/tex]-intercepts: [tex]\( (0, 5) \)[/tex] and [tex]\( (0, -5) \)[/tex]
