Answer to Question #256169 in Calculus for Alunsina

Question

Find an expression of the function whose graph is the bottom half of the given parabola

y²-x-4y+3=0

Answer 1

$\displaystyle y^2 - x - 4y + 3 = 0\\ y^2 - 4y + 3 = x\\ y^2 - 4y = x - 3\\ y^2 - 4y + 4 = x - 3 + 4 = x + 1\\ (y - 2)^2 = x + 1\\ y = \pm\sqrt{x + 1} + 2\\ \textup{The bottom half of the parabola is}\\ y = -\sqrt{x + 1} + 2\\ \textup{Hence,}\\ \textup{domain}(y) = \{x \in \mathbb{R}: x \geq -1\}\\ \forall x \geq -1,\,\,\, -\sqrt{x + 1} \leq 0, \\ \textup{and so}\,\,\, 2 - \sqrt{x + 1} \leq 2\\ \textup{Thus,}\,\,\, y \leq 2.\,\,\,\forall x \geq 1.\\ \textup{Hence,}\\ \textup{range}(y) = \{y \in \mathbb{R}: y \leq 2\}\\$