Translations and Reflections

Sol A Interchanging $x$ and $y$ gives the equation $x=y^3$ or equivalently $y=\sqrt[3]{x}$.

Sol B Replacing $y$ by $-y$ gives $-y=x^2+5$ or $y=-x^2-5$, and then replacing $y$ by $y-3$ gives $y-3=-x^2-5$ or $y=-x^2-2$.

Sol C Replacing $x$ by $-x$ gives $y=e^{-x}$, and then replacing $x$ by $x-4$ gives $y=e^{-(x-4)}$ or $y=e^{4-x}$.

Sol 1 Replacing $x$ by $x+5$ and $y$ by $y-4$ gives $\frac{(x+5)^2}{9}+\frac{(y-4)^2}{16}=1$.

Sol 2 Replacing $x$ by $-x$ gives $y=\sqrt{-x}$, and then replacing $x$ by $x+3$ gives $y=\sqrt{-(x+3)}$ or $y=\sqrt{-x-3}$.

Sol 3 Replacing $x$ by $x+3$ gives $y=\sqrt{x+3}$, and then replacing $x$ by $-x$ gives $y=\sqrt{-x+3}$ or $y=\sqrt{3-x}$.

Sol 4 Shift 5 units to the left to get $y=\ln(x+5)$, and then reflect in the y-axis to get $y=\ln(-x+5)$ or $y=\ln(5-x)$. Then reflect in the x-axis to get $-y=\ln(5-x)$ or $y=-\ln(5-x)$, and then shift 9 units up to get $y-9=-\ln(5-x)$ or $y=9-\ln(5-x)$. (This is not the only solution.)

Sol 5 Interchanging $x$ and $y$ gives $x=y^2-5$, and then replacing $y$ by $y+4$ gives $x=(y+4)^2-5$.

Sol 6 Replacing $y$ by $y+4$ gives $y+4=x^2-5$ or $y=x^2-9$, and then interchanging $x$ and $y$ gives $x=y^2-9$.

Sol 7 Interchanging $x$ and $y$ gives the equation $x=\sqrt{y}$, and then replacing $x$ by $x-3$ gives $x-3=\sqrt{y}$ or $x=3+\sqrt{y}$. Next replacing $y$ by $y-4$ gives the equation $x=3+\sqrt{y-4}$.

Sol 8 Since $\frac{x-4}{x-3}=\frac{(x-3)-1}{x-3}=1-\frac{1}{x-3}$, we can shift 3 units to the right to get $y=\frac{1}{x-3}$, reflect in the x-axis to get $-y=\frac{1}{x-3}$ or $y=-\frac{1}{x-3}$, and then shift 1 unit up to get $y-1=-\frac{1}{x-3}$ or $y=1-\frac{1}{x-3}$. (This is not the only solution.)

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