The curve \(C\) has equation \(y=\dfrac{kx^2-kx+68}{x-2}\), where \(k\) is a constant.
If the equation of the oblique asymptote of \(C\) is \(y=2x+2\), state the value of \(k\).[1]
Use the value of \(k\) found in part (a)(i) for the rest of the question.
Sketch \(C\), showing clearly the coordinates of any axial intercepts and stationary points, and the equations of any asymptotes.[3]
Find the range of values of \(r\), where \(r>0\), such that the equation \[ (x-8)^2+\left(\frac{kx^2-kx+68}{x-2}-10\right)^2=r^2 \] has no real solutions for \(x\).[2]
The graph of \(y=\mathrm{f}(x)\) is transformed onto the graph of \(y=\mathrm{g}(x)\) in a sequence of three transformations. The graph of \(y=\mathrm{f}(x)\) is first transformed onto the graph of \(y=\mathrm{f}(x)+1\), followed by a reflection in the \(x\)-axis and is finally transformed onto the graph of \(y=\mathrm{g}(x)\) by transformation \(T\). A point \(P\) with coordinates \((a,b)\) lies on the graph of \(y=\mathrm{f}(x)\). \(P\) is mapped onto a point \(Q\) with coordinates \((a-1,m)\) where \(Q\) lies on the graph of \(y=\mathrm{g}(x)\). Find \(m\) in terms of \(b\) and describe \(T\).[3]
Solution:
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Answer:(a)(i) \(k=2\) (ii) Maximum \((-4,-18)\), minimum \((8,30)\), \(y\)-intercept \((0,-34)\), no \(x\)-intercept; asymptotes \(x=2\), \(y=2x+2\). (iii) \(0<r<20\) (b) Intended: \(m=-b-1\), translation 1 unit left; not uniquely determined without a restriction on \(T\).