The diagram shows the curve with equation $y=\mathrm{f}\left( x \right)$. The curve crosses the \(

Question

The diagram shows the curve with equation $y=\mathrm{f}\left( x \right)$. The curve crosses the $x$-axis at $x=1$ and $x=2.5$, crosses the $y$-axis at $y=1$ and has a maximum point at $\left( 4,\ 6 \right)$. The equations of the asymptotes are $x=2$ and $y=3$. Sketch the graph of $y=\mathrm{f}\,\mathrm{ }\!\!'\!\!\mathrm{ }\left( x \right)$, giving the equations of asymptotes, coordinates of turning points and axial intercepts, where possible.[2] The curve $C$ has equation $y=\frac{{{x}^{2}}+kx-1}{x+1}$, where $k$ is a non-zero constant. Find the range of values of $k$ for which $C$ has no stationary points.[4] Given that $y=x+3$ is an asymptote of $C$, show that $k=4$.[2] State a sequence of transformations which transform the graph of $y=\frac{x}{4}-\frac{1}{x}$ onto the graph of $y=\frac{{{x}^{2}}+4x-1}{x+1}$.[3]

Accepted Answer

(a) Asymptotes $x=2$, $y=0$, $x$-intercept $(4,0)$ (b)(i) $k>0$ (b)(ii) $k=4$ (b)(iii) Translate $1$ unit left, scale parallel to the $y$-axis by $4$, then translate $2$ units up

2024 TJC P1 Q7

2024 TJC P1 Q7

Video Solution:

Solution:

Solution locked