A curve $C$ has equation $y=\frac{1}{\sqrt{b{{x}^{2}}-2bx}}$, where $b

Question

A curve $C$ has equation $y=\frac{1}{\sqrt{b{{x}^{2}}-2bx}}$, where $b<0$. Sketch $C$ and give in terms of $b$ where appropriate, the equations of any asymptotes and the line of symmetry, and coordinates of turning point.[5]

Accepted Answer

Line of symmetry:$x=1$. Turning minimum point is at $\left( 1,\,\frac{1}{\sqrt{-b}} \right)$. For vertical asymptotes, $x=0$ or $x=2$.

2022 DHS Promo Q4

2022 DHS Promo Q4

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