To find the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-intercepts for the equation [tex]\( x^2 + y = 16 \)[/tex], we need to find the points where the graph of the equation crosses the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-axes.
### Finding the [tex]\( x \)[/tex]-Intercepts:
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( y = 0 \)[/tex]. To find these, we substitute [tex]\( y = 0 \)[/tex] into the equation [tex]\( x^2 + y = 16 \)[/tex].
1. Substitute [tex]\( y = 0 \)[/tex] into the equation:
[tex]\[
x^2 + 0 = 16
\][/tex]
[tex]\[
x^2 = 16
\][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \pm \sqrt{16}
\][/tex]
[tex]\[
x = \pm 4
\][/tex]
Hence, the [tex]\( x \)[/tex]-intercepts are [tex]\( x = 4 \)[/tex] and [tex]\( x = -4 \)[/tex].
Therefore, the [tex]\( x \)[/tex]-intercepts as ordered pairs are:
[tex]\[
(4, 0) \quad \text{and} \quad (-4, 0)
\][/tex]
### Finding the [tex]\( y \)[/tex]-Intercepts:
The [tex]\( y \)[/tex]-intercepts occur where [tex]\( x = 0 \)[/tex]. To find these, we substitute [tex]\( x = 0 \)[/tex] into the equation [tex]\( x^2 + y = 16 \)[/tex].
1. Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[
0^2 + y = 16
\][/tex]
[tex]\[
y = 16
\][/tex]
Hence, the [tex]\( y \)[/tex]-intercept is [tex]\( y = 16 \)[/tex].
Therefore, the [tex]\( y \)[/tex]-intercept as an ordered pair is:
[tex]\[
(0, 16)
\][/tex]
In conclusion, the intercepts are:
[tex]\[
x\text{-intercepts:} \quad (4, 0) \quad \text{and} \quad (-4, 0)
\][/tex]
[tex]\[
y\text{-intercept:} \quad (0, 16)
\][/tex]
