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9231 P11 - Jun 2020 - Q1 - 6 marks
5808
1 Let \(a\) be a positive constant. (a) Sketch the curve with equation \(y=\frac{a x}{x+7}\).
(b) Sketch the curve with equation \(y=\left|\frac{a x}{x+7}\right|\) and find the set of values of \(x\) for which \(\left|\frac{a x}{x+7}\right|>\frac{a}{2}\).
Solution
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Answer:
The curve has vertical asymptote \(x=-7\), horizontal asymptote \(y=a\), and passes through \((0,0)\). For \(x<-7\) the branch lies above \(y=a\); for \(x>-7\) the branch comes from \(-\infty\), passes through \((0,0)\), and approaches \(y=a\) from below.
The modulus graph is found by reflecting the part of the graph below the \(x\)-axis in the \(x\)-axis.
The required set of values is \(x<-7\), \(-77\).
Rewrite the function as \(y=\frac{ax}{x+7}=a-\frac{7a}{x+7}\).
Hence the vertical asymptote is \(x=-7\), and the horizontal asymptote is \(y=a\). Since \(a>0\), the curve crosses the axes at \((0,0)\).
For \(x<-7\), \(\frac{x}{x+7}>1\), so the branch lies above \(y=a\) and tends to \(+\infty\) as \(x\to -7^-\). For \(-70\), the curve is positive and approaches \(y=a\) from below.
For \(y=\left|\frac{ax}{x+7}\right|\), reflect the part of the graph from part (a) that lies below the \(x\)-axis in the \(x\)-axis. The vertical asymptote remains \(x=-7\), the horizontal asymptote remains \(y=a\), and the graph meets the \(x\)-axis at \((0,0)\).
To solve \(\left|\frac{ax}{x+7}\right|>\frac{a}{2}\), use \(a>0\) and divide through by \(a\): \(\left|\frac{x}{x+7}\right|>\frac{1}{2}\).
The boundary points occur when \(\frac{x}{x+7}=\frac{1}{2}\) or \(\frac{x}{x+7}=-\frac{1}{2}\).
Solving \(\frac{x}{x+7}=\frac{1}{2}\) gives \(2x=x+7\), so \(x=7\).
Solving \(\frac{x}{x+7}=-\frac{1}{2}\) gives \(2x=-(x+7)\), so \(3x=-7\), and therefore \(x=-\frac{7}{3}\).
Using the intervals separated by \(x=-7\), \(x=-\frac{7}{3}\), and \(x=7\), the inequality is satisfied for \(x<-7\), \(-77\).
