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Home » Use the Pythagorean Theorem: \(c^2 = a^2 + b^2\). - Get link 4share

Use the Pythagorean Theorem: \(c^2 = a^2 + b^2\). - Get link 4share

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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The numerator \( y + 1 \geq 0 \Rightarrow y \geq -1 \).
The denominator \( y - 1 > 0 \Rightarrow y > 1 \), or \( y - 1 < 0 \Ri...
Case 1: \( y > 1 \), both numerator and denominator positive.
Case 2: \( -1 \leq y < 1 \), numerator non-negative, denominator negat...
Case 3: \( y < -1 \), numerator and denominator both negative → positi...
But if \( y < -1 \), \( \frac{y+1}{y-1} > 0 \), valid. However, \( x^2...
But we must check if all such \( y \) are attainable. From \( x^2 = \f...
But when \( y = -1 \), \( x^2 = 0 \Rightarrow x = 0 \), valid.
As \( y \to 1^- \), \( \frac{y+1}{y-1} \to -\infty \), invalid.
When \( y < -1 \), \( \frac{y+1}{y-1} > 0 \), e.g., \( y = -2 \Rightar...

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