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2020 EJC Promo Q3

2020 EJC Promo Q3

Promo
  1. Sketch the curve with equation
    \(y=\frac{{{x}^{2}}+9x-5}{x+10}\),
    indicating clearly the equations of any asymptotes, the coordinates of any points where the curve crosses the axes and of any turning points.[4]
  2. Hence, find the range of values of \(k\), where \(k\) is real, such that the equation
    \(\frac{{{x}^{2}}+9x-5}{x+10}=kx+10k-11\)
    has no solution. Show your reasoning clearly.[2]

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Answer:(i) Asymptotes: \(y=x-1\) and \(x=-10\); Intercepts: \(y=-\frac{1}{2}\), \(x=-9.52\) and \(x=0.525\); Turning points: \((-12.2,-15.5)\) and \((-7.76,-6.53)\) (ii) \(RHS\) gives \(y=k(x+10)-11\) or \(y+11=k(x+10)\), which is a straight line passing through \((-10,-11)\) and gradient \(k\). Hence, for equation to have no solutions, \(k\le 1\).

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