Question #256171

Find the domain and range of the bottom half of the given parabola

y2-x-4y+3=0


Expert's answer

y2−x−4y+3=0y2−4y+3=xy2−4y=x−3y2−4y+4=x−3+4=x+1(y−2)2=x+1y=±x+1+2The bottom half of the parabola isy=−x+1+2Hence,domain(y)={x∈R:x≥−1}∀x≥−1,   −x+1≤0,and so   2−x+1≤2Thus,   y≤2.   ∀x≥1.Hence,range(y)={y∈R:y≤2}\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\}\\


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