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$sider the following relation. x=-4|y|+3 1 of 2: Find four points contained in the inverse. Express your values as an integer or simplified fraction.$

sider the following relation. $\newline$ $x=-4|y|+3$ $\newline$ $1$ of $2$ : Find four points contained in the inverse. Express your values as an integer or simplified fraction.

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Q. sider the following relation.

\newline

x=-4|y|+3

\newline

1

2

: Find four points contained in the inverse. Express your values as an integer or simplified fraction.

Interchanging roles of $x$ and $y$ : To find the inverse of the given relation, we need to solve for $y$ in terms of $x$ . This means we will interchange the roles of $x$ and $y$ and then solve for $y$ . The original relation is $x = -4|y| + 3$ .
Expressing relation with interchanged variables: First, we express the relation with $y$ and $x$ interchanged: $y = -4|x| + 3$ .
Considering absolute value: Now, we need to consider the absolute value. Since |x|\
Case \(1: $x$ is non-negative: Case $1$ : If $x$ is non-negative ( $x \geq 0$ ), then $|x| = x$ , and the equation becomes $y = -4x + 3$ .
Finding points for Case $1$ : From Case $1$ , we can find two points by choosing non-negative values for $x$ and calculating the corresponding $y$ values. Let's choose $x = 0$ and $x = 1$ .
Case $2$ : $x$ is negative: For $x = 0$ : $y = -4(0) + 3 = 3$ . So, one point is $(0, 3)$ .
Finding points for Case $2$ : For $x = 1$ : $y = -4(1) + 3 = -1$ . So, another point is $(1, -1)$ .
Finding points for Case $2$ : For $x = 1$ : $y = -4(1) + 3 = -1$ . So, another point is $(1, -1)$ .Case $2$ : If $x$ is negative ( $x < 0$ ), then $|x| = -x$ , and the equation becomes $y = -4(-x) + 3 = 4x + 3$ .
Finding points for Case $2$ : For $x = 1$ : $y = -4(1) + 3 = -1$ . So, another point is $(1, -1)$ .Case $2$ : If $x$ is negative ( $x < 0$ ), then $|x| = -x$ , and the equation becomes $y = -4(-x) + 3 = 4x + 3$ .From Case $2$ , we can find two more points by choosing negative values for $x$ and calculating the corresponding $y$ values. Let's choose $x = -1$ and $y = -4(1) + 3 = -1$ $0$ .
Finding points for Case $2$ : For $x = 1$ : $y = -4(1) + 3 = -1$ . So, another point is $(1, -1)$ .Case $2$ : If $x$ is negative ( $x < 0$ ), then $|x| = -x$ , and the equation becomes $y = -4(-x) + 3 = 4x + 3$ .From Case $2$ , we can find two more points by choosing negative values for $x$ and calculating the corresponding $y$ values. Let's choose $x = -1$ and $y = -4(1) + 3 = -1$ $0$ .For $x = -1$ : $y = -4(1) + 3 = -1$ $2$ . So, another point is $y = -4(1) + 3 = -1$ $3$ .
Finding points for Case $2$ : For $x = 1$ : $y = -4(1) + 3 = -1$ . So, another point is $(1, -1)$ .Case $2$ : If $x$ is negative ( $x < 0$ ), then $|x| = -x$ , and the equation becomes $y = -4(-x) + 3 = 4x + 3$ .From Case $2$ , we can find two more points by choosing negative values for $x$ and calculating the corresponding $y$ values. Let's choose $x = -1$ and $y = -4(1) + 3 = -1$ $0$ .For $x = -1$ : $y = -4(1) + 3 = -1$ $2$ . So, another point is $y = -4(1) + 3 = -1$ $3$ .For $y = -4(1) + 3 = -1$ $0$ : $y = -4(1) + 3 = -1$ $5$ . So, the last point is $y = -4(1) + 3 = -1$ $6$ .