Curve sketching with asymptotes

7 A curve has equation \(y = \frac { 5 } { ( x + 2 ) ( 4 - x ) }\).

1. Write down the value of \(y\) when \(x = 0\).
2. Write down the equations of the three asymptotes.
3. Sketch the curve.
4. Find the values of \(x\) for which \(\frac { 5 } { ( x + 2 ) ( 4 - x ) } = 1\) and hence solve the inequality $$\frac { 5 } { ( x + 2 ) ( 4 - x ) } < 1 .$$

1 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model.

7 A curve has equation \(y = \frac { 5 x - 9 } { ( 2 x - 3 ) ( 2 x + 7 ) }\).

1. Write down the equations of the two vertical asymptotes and the one horizontal asymptote.
2. Describe the behaviour of the curve for large positive and large negative values of \(x\), justifying your answers.
3. Sketch the curve.
4. Solve the inequality \(\frac { 5 x - 9 } { ( 2 x - 3 ) ( 2 x + 7 ) } \leqslant 0\).

Express \(\frac { 2 x ^ { 2 } - x + 5 } { x ^ { 2 } - 1 }\) in the form \(2 + \frac { A } { x - 1 } + \frac { B } { x + 1 }\), where \(A\) and \(B\) are integers to be found. The curve \(C\) has equation \(y = \frac { 2 x ^ { 2 } - x + 5 } { x ^ { 2 } - 1 }\). Show that there are two distinct values of \(x\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\). Sketch \(C\), stating the equations of the asymptotes and giving the coordinates of any points of intersection with the coordinate axes and with the asymptotes. You do not need to find the coordinates of the turning points.

The curve \(C\) has equation \(y = \frac { 5 ( x - 1 ) ( x + 2 ) } { ( x - 2 ) ( x + 3 ) }\).

1. Express \(y\) in the form \(P + \frac { Q } { x - 2 } + \frac { R } { x + 3 }\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) for exactly one value of \(x\) and find the corresponding value of \(y\).
3. Write down the equations of all the asymptotes of \(C\).
4. Find the set of values of \(k\) for which the line \(y = k\) does not intersect \(C\).

10 The curve \(C\) has equation $$y = \frac { x ^ { 2 } + 2 x - 3 } { ( \lambda x + 1 ) ( x + 4 ) }$$ where \(\lambda\) is a constant.

1. Find the equations of the asymptotes of \(C\) for the case where \(\lambda = 0\).
2. Find the equations of the asymptotes of \(C\) for the case where \(\lambda\) is not equal to any of \(- 1,0 , \frac { 1 } { 4 } , \frac { 1 } { 3 }\).
3. Sketch \(C\) for the case where \(\lambda = - 1\). Show, on your diagram, the equations of the asymptotes and the coordinates of the points of intersection of \(C\) with the coordinate axes.

9 The curve \(C\) has equation $$y = \frac { 3 x - 9 } { ( x - 2 ) ( x + 1 ) }$$

1. Find the equations of the asymptotes of \(C\).
   \includegraphics[max width=\textwidth, alt={}, center]{9221f480-4af6-44be-a535-d2ceb0f8b5d2-14_61_1566_513_328}
2. Show that there is no point on \(C\) for which \(\frac { 1 } { 3 } < y < 3\).
3. Find the coordinates of the turning points of \(C\).
4. Sketch \(C\).

9 The curve \(C\) has equation $$y = \frac { 3 x - 9 } { ( x - 2 ) ( x + 1 ) }$$

1. Find the equations of the asymptotes of \(C\).
   \includegraphics[max width=\textwidth, alt={}, center]{a0987277-06e9-451b-ae18-bb7de9e7661c-14_61_1566_513_328}
2. Show that there is no point on \(C\) for which \(\frac { 1 } { 3 } < y < 3\).
3. Find the coordinates of the turning points of \(C\).
4. Sketch \(C\).

10 The curve \(C\) has equation \(y = \frac { 4 x ^ { 2 } - 3 x } { x ^ { 2 } + 1 }\). Verify that the equation of \(C\) may be written in the form \(y = - \frac { 1 } { 2 } + \frac { ( 3 x - 1 ) ^ { 2 } } { 2 \left( x ^ { 2 } + 1 \right) }\) and also in the form \(y = \frac { 9 } { 2 } - \frac { ( x + 3 ) ^ { 2 } } { 2 \left( x ^ { 2 } + 1 \right) }\). Hence show that \(- \frac { 1 } { 2 } \leqslant y \leqslant \frac { 9 } { 2 }\). Without differentiating, write down the coordinates of the turning points of \(C\). State the equation of the asymptote of \(C\). Sketch the graph of \(C\), stating the coordinates of the intersections with the coordinate axes and the asymptote.

1. The curve \(C\) with equation

$$y = \frac { p - 3 x } { ( 2 x - q ) ( x + 3 ) } \quad x \in \mathbb { R } , x \neq - 3 , x \neq 2$$ where \(p\) and \(q\) are constants, passes through the point \(\left( 3 , \frac { 1 } { 2 } \right)\) and has two vertical asymptotes
with equations \(x = 2\) and \(x = - 3\) with equations \(x = 2\) and \(x = - 3\)

1. 1. Explain why you can deduce that \(q = 4\)
   2. Show that \(p = 15\) \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{91a2f26a-add2-4b58-997d-2ae229548217-38_616_889_842_587} \captionsetup{labelformat=empty} \caption{Figure 4}
      \end{figure} Figure 4 shows a sketch of part of the curve \(C\). The region \(R\), shown shaded in Figure 4, is bounded by the curve \(C\), the \(x\)-axis and the line with equation \(x = 3\)
2. Show that the exact value of the area of \(R\) is \(a \ln 2 + b \ln 3\), where \(a\) and \(b\) are rational constants to be found.

10

1. Write down the equation of the horizontal asymptote of \(C\) 10
2. Find the value of \(m\) and the value of \(p\)
   

   10 10	Hence, or otherwise, write down the coordinates of the \(y\)-intercept of \(C\)
   	Without using calculus, show that the line \(y = - 1\) does not intersect \(C\)

5 A curve has equation $$y = \frac { x } { x ^ { 2 } - 1 }$$

1. Write down the equations of the three asymptotes to the curve.
2. Sketch the curve.
   (You are given that the curve has no stationary points.)
3. Solve the inequality $$\frac { x } { x ^ { 2 } - 1 } > 0$$

6 A curve has equation $$y = \frac { ( x - 1 ) ( x - 3 ) } { x ( x - 2 ) }$$

1. 1. Write down the equations of the three asymptotes of this curve.
   2. State the coordinates of the points at which the curve intersects the \(x\)-axis.
   3. Sketch the curve.
      (You are given that the curve has no stationary points.)
2. Hence, or otherwise, solve the inequality $$\frac { ( x - 1 ) ( x - 3 ) } { x ( x - 2 ) } < 0$$

7 A graph has equation $$y = \frac { x - 4 } { x ^ { 2 } + 9 }$$

1. Explain why the graph has no vertical asymptote and give the equation of the horizontal asymptote.
2. Show that, if the line \(y = k\) intersects the graph, the \(x\)-coordinates of the points of intersection of the line with the graph must satisfy the equation $$k x ^ { 2 } - x + ( 9 k + 4 ) = 0$$
3. Show that this equation has real roots if \(- \frac { 1 } { 2 } \leqslant k \leqslant \frac { 1 } { 18 }\).
4. Hence find the coordinates of the two stationary points on the curve.
   (No credit will be given for methods involving differentiation.)

9 A curve has equation $$y = \frac { x } { x - 1 }$$

1. Find the equations of the asymptotes of this curve.
2. Given that the line \(y = - 4 x + c\) intersects the curve, show that the \(x\)-coordinates of the points of intersection must satisfy the equation $$4 x ^ { 2 } - ( c + 3 ) x + c = 0$$
3. It is given that the line \(y = - 4 x + c\) is a tangent to the curve.
   1. Find the two possible values of \(c\).
      (No credit will be given for methods involving differentiation.)
   2. For each of the two values found in part (c)(i), find the coordinates of the point where the line touches the curve.

8 A curve has equation $$y = \frac { x ^ { 2 } } { ( x - 1 ) ( x - 5 ) }$$

1. Write down the equations of the three asymptotes to the curve.
2. Show that the curve has no point of intersection with the line \(y = - 1\).
3. 1. Show that, if the curve intersects the line \(y = k\), then the \(x\)-coordinates of the points of intersection must satisfy the equation $$( k - 1 ) x ^ { 2 } - 6 k x + 5 k = 0$$
   2. Show that, if this equation has equal roots, then $$k ( 4 k + 5 ) = 0$$
4. Hence find the coordinates of the two stationary points on the curve.

7

1. 1. Write down the equations of the two asymptotes of the curve \(y = \frac { 1 } { x - 3 }\).
   2. Sketch the curve \(y = \frac { 1 } { x - 3 }\), showing the coordinates of any points of intersection with the coordinate axes.
   3. On the same axes, again showing the coordinates of any points of intersection with the coordinate axes, sketch the line \(y = 2 x - 5\).
2. 1. Solve the equation $$\frac { 1 } { x - 3 } = 2 x - 5$$
   2. Find the solution of the inequality $$\frac { 1 } { x - 3 } < 2 x - 5$$ □
      \includegraphics[max width=\textwidth, alt={}, center]{763d89e4-861a-4754-a93c-d0902987673f-08_367_197_2496_155}

8 A curve has equation \(y = \frac { 1 } { x ^ { 2 } - 4 }\).

1. 1. Write down the equations of the three asymptotes of the curve.
   2. Sketch the curve, showing the coordinates of any points of intersection with the coordinate axes.
2. Hence, or otherwise, solve the inequality $$\frac { 1 } { x ^ { 2 } - 4 } < - 2$$

5 The curve \(C\) has equation \(y = \frac { x } { ( x + 1 ) ( x - 2 ) }\).
The line \(L\) has equation \(y = - \frac { 1 } { 2 }\).

1. Write down the equations of the asymptotes of \(C\).
2. The line \(L\) intersects the curve \(C\) at two points. Find the \(x\)-coordinates of these two points.
3. Sketch \(C\) and \(L\) on the same axes.
   (You are given that the curve \(C\) has no stationary points.)
4. Solve the inequality $$\frac { x } { ( x + 1 ) ( x - 2 ) } \leqslant - \frac { 1 } { 2 }$$