Curve-Line Intersection Area

7 \includegraphics[max width=\textwidth, alt={}, center]{574b96b2-62f2-41b3-a178-8e68e16429ff-12_631_1031_267_534} The diagram shows the curve with equation \(y = ( 3 x - 2 ) ^ { \frac { 1 } { 2 } }\) and the line \(y = \frac { 1 } { 2 } x + 1\). The curve and the line intersect at points \(A\) and \(B\).

1. Find the coordinates of \(A\) and \(B\).
2. Hence find the area of the region enclosed between the curve and the line.

6 \includegraphics[max width=\textwidth, alt={}, center]{bb7595c9-93ae-49e8-9cc5-9ecc802e6060-08_613_865_262_632} The diagram shows the curve with equation \(y = 5 x ^ { \frac { 1 } { 2 } }\) and the line with equation \(y = 2 x + 2\).
Find the exact area of the shaded region which is bounded by the line and the curve.

3 \includegraphics[max width=\textwidth, alt={}, center]{b2cefbd6-6e89-495a-9f42-60f76c8c5975-2_629_659_715_740} The diagram shows the curve \(y = 3 \sqrt { } x\) and the line \(y = x\) intersecting at \(O\) and \(P\). Find

1. the coordinates of \(P\),
2. the area of the shaded region.

11 A line has equation \(y = 2 x + c\) and a curve has equation \(y = 8 - 2 x - x ^ { 2 }\).

1. For the case where the line is a tangent to the curve, find the value of the constant \(c\).
2. For the case where \(c = 11\), find the \(x\)-coordinates of the points of intersection of the line and the curve. Find also, by integration, the area of the region between the line and the curve.

10 \includegraphics[max width=\textwidth, alt={}, center]{0b047754-84f2-46ea-b441-7c68cef47641-3_812_720_1484_715} The diagram shows the curve \(y = - x ^ { 2 } + 12 x - 20\) and the line \(y = 2 x + 1\). Find, showing all necessary working, the area of the shaded region.

8 \includegraphics[max width=\textwidth, alt={}, center]{17ca6dd2-271b-4b06-8433-354493feaf06-12_485_570_262_790} The diagram shows parts of the graphs of \(y = 3 - 2 x\) and \(y = 4 - 3 \sqrt { } x\) intersecting at points \(A\) and \(B\).

1. Find by calculation the \(x\)-coordinates of \(A\) and \(B\).
2. Find, showing all necessary working, the area of the shaded region.

7 \includegraphics[max width=\textwidth, alt={}, center]{d178603a-f59a-4986-b5ab-b47eceedb2fc-10_503_853_260_641} The diagram shows part of the curve with equation \(y = k \left( x ^ { 3 } - 7 x ^ { 2 } + 12 x \right)\) for some constant \(k\). The curve intersects the line \(y = x\) at the origin \(O\) and at the point \(A ( 2,2 )\).

1. Find the value of \(k\).
2. Verify that the curve meets the line \(y = x\) again when \(x = 5\).
3. Find, showing all necessary working, the area of the shaded region.

14. \begin{figure}[h] \end{figure} The finite region \(R\), which is shown shaded in Figure 3, is bounded by the straight line \(l\) with equation \(y = 4 x + 3\) and the curve \(C\) with equation \(y = 2 x ^ { \frac { 3 } { 2 } } - 2 x + 3 , x \geqslant 0\) The line \(l\) meets the curve \(C\) at the point \(A\) on the \(y\)-axis and \(l\) meets \(C\) again at the point \(B\), as shown in Figure 3.

1. Use algebra to find the coordinates of \(A\) and \(B\).
2. Use integration to find the area of the shaded region \(R\).

11. \begin{figure}[h] \end{figure} The straight line with equation \(y = x + 4\) cuts the curve with equation \(y = - x ^ { 2 } + 2 x + 24\) at the points \(A\) and \(B\), as shown in Figure 2.

1. Use algebra to find the coordinates of the points \(A\) and \(B\). The finite region \(R\) is bounded by the straight line and the curve and is shown shaded in Figure 2.
2. Use calculus to find the exact area of \(R\).

8. \begin{figure}[h] \end{figure} The line with equation \(y = 3 x + 20\) cuts the curve with equation \(y = x ^ { 2 } + 6 x + 10\) at the points \(A\) and \(B\), as shown in Figure 2.

1. Use algebra to find the coordinates of \(A\) and the coordinates of \(B\). The shaded region \(S\) is bounded by the line and the curve, as shown in Figure 2.
2. Use calculus to find the exact area of \(S\).

10. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve \(C\) with equation \(y = 2 x + \frac { 8 } { x ^ { 2 } } - 5 , x > 0\).
The points \(P\) and \(Q\) lie on \(C\) and have \(x\)-coordinates 1 and 4 respectively. The region \(R\), shaded in Figure 1, is bounded by \(C\) and the straight line joining \(P\) and \(Q\).

1. Find the exact area of \(R\).
2. Use calculus to show that \(y\) is increasing for \(x > 2\).

9. \begin{figure}[h] \end{figure} The straight line with equation \(y = x + 4\) cuts the curve with equation \(y = - x ^ { 2 } + 2 x + 24\) at the points \(A\) and \(B\), as shown in Figure 3.

1. Use algebra to find the coordinates of the points \(A\) and \(B\). The finite region \(R\) is bounded by the straight line and the curve and is shown shaded in Figure 3.
2. Use calculus to find the exact area of \(R\).

5. \begin{figure}[h] \end{figure} Figure 2 shows the line with equation \(y = 10 - x\) and the curve with equation \(y = 10 x - x ^ { 2 } - 8\) The line and the curve intersect at the points \(A\) and \(B\), and \(O\) is the origin.

1. Calculate the coordinates of \(A\) and the coordinates of \(B\). The shaded area \(R\) is bounded by the line and the curve, as shown in Figure 2.
2. Calculate the exact area of \(R\).

8. \begin{figure}[h] \end{figure} The line with equation \(y = x + 5\) cuts the curve with equation \(y = x ^ { 2 } - 3 x + 8\) at the points \(A\) and \(B\), as shown in Fig. 2.

1. Find the coordinates of the points \(A\) and \(B\).
2. Find the area of the shaded region between the curve and the line, as shown in Fig. 2.

4 \includegraphics[max width=\textwidth, alt={}, center]{367db494-294e-4b53-b9e8-fd2a69fb6069-2_634_670_1123_740} The diagram shows the curve \(y = 4 - x ^ { 2 }\) and the line \(y = x + 2\).

1. Find the \(x\)-coordinates of the points of intersection of the curve and the line.
2. Use integration to find the area of the shaded region bounded by the line and the curve.

2 Fig. 9 shows a sketch of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - 22 x + 24\) and the line \(y = 6 x + 24\). \begin{figure}[h] \end{figure}

1. Differentiate \(y = x ^ { 3 } - 3 x ^ { 2 } - 22 x + 24\) and hence find the \(x\)-coordinates of the turning points of the curve. Give your answers to 2 decimal places.
2. You are given that the line and the curve intersect when \(x = 0\) and when \(x = - 4\). Find algebraically the \(x\)-coordinate of the other point of intersection.
3. Use calculus to find the area of the region bounded by the curve and the line \(y = 6 x + 24\) for \(- 4 \leqslant x \leqslant 0\), shown shaded on Fig. 9.

1

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-1_650_759_252_762} \captionsetup{labelformat=empty} \caption{Fig. 12}
   \end{figure} Fig. 12 shows part of the curve \(y = x ^ { 4 }\) and the line \(y = 8 x\), which intersect at the origin and the point P .
   (A) Find the coordinates of P , and show that the area of triangle OPQ is 16 square units.
   (B) Find the area of the region bounded by the line and the curve.
2. You are given that \(\mathrm { f } ( x ) = x ^ { 4 }\).
   (A) Complete this identity for \(\mathrm { f } ( x + h )\). $$\mathrm { f } ( x + h ) = ( x + h ) ^ { 4 } = x ^ { 4 } + 4 x ^ { 3 } h + \ldots$$ (B) Simplify \(\frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\).
   (C) Find \(\lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\).
   (D) State what this limit represents.

2 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

1 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

8 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

\includegraphics{figure_2} Figure 2 shows the line with equation \(y = 9 - x\) and the curve with equation \(y = x^2 - 2x + 3\). The line and the curve intersect at the points \(A\) and \(B\), and \(O\) is the origin.

1. Calculate the coordinates of \(A\) and the coordinates of \(B\). [5]

The shaded region \(R\) is bounded by the line and the curve.

1. Calculate the area of \(R\). [7]

\includegraphics{figure_2} Figure 2 shows the line with equation \(y = x + 1\) and the curve with equation \(y = 6x - x^2 - 3\). The line and the curve intersect at the points \(A\) and \(B\), and \(O\) is the origin.

1. Calculate the coordinates of \(A\) and the coordinates of \(B\). [5]

The shaded region \(R\) is bounded by the line and the curve.

1. Calculate the area of \(R\). [7]