The curve C has equation y = \frac{2x^2 + 2x + 3}{x^2 + 2}. Show that, for all x, 1 \leq y \leq \frac{5}{2}. Find the coordinates of the turning points on C. Find the equation of the asymptote of C. Sketch the graph of C, stating the coordinates of any intersections with the y-axis and the asymptote.

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The curve C has equation y = \frac{2x^2 + 2x + 3}{x^2 + 2} - CIE - A-Level Further Maths - Question 9 - 2012 - Paper 1

Question 9

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The curve C has equation y = \frac{2x^2 + 2x + 3}{x^2 + 2}. Show that, for all x, 1 \leq y \leq \frac{5}{2}. Find the coordinates of the turning points on C. Fin... show full transcript

Worked Solution & Example Answer:The curve C has equation y = \frac{2x^2 + 2x + 3}{x^2 + 2} - CIE - A-Level Further Maths - Question 9 - 2012 - Paper 1

Step 1

Show that, for all x, 1 \leq y \leq \frac{5}{2}.

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Answer

To show that for all x, the values of y are restricted between 1 and \frac{5}{2}, we first rewrite the equation of the curve:

y = \frac{2x^2 + 2x + 3}{x^2 + 2}.

Next, we analyze for which values of x the function reaches maximum and minimum values.

Finding Minimum Value: We observe that as x approaches \pm \infty, y tends to 2, thus suggesting y can be minimized. However, to find the actual minimum, we set the derivative y' = 0 and determine turning points: After applying differentiation:

$y' = \frac{(2x^2 + 2x + 3)'(x^2 + 2) - (2x^2 + 2x + 3)(x^2 + 2)'}{(x^2 + 2)^2} = 0$

Solving gives turning points, minimizing at (–1, 1).
Finding Maximum Value: The second turning point emerges from further solving, yielding the maximum approximately 2.5. We validate that \frac{5}{2} is the maximum achieved by the function.

Conclusively:

[ 1 \leq y \leq \frac{5}{2} ]

Step 2

Find the coordinates of the turning points on C.

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Answer

To find the turning points, we previously derived the derivative:

$y' = \frac{(2x^2 + 2x + 3)'(x^2 + 2) - (2x^2 + 2x + 3)(x^2 + 2)'}{(x^2 + 2)^2} = 0.$

After simplifying, we set:

$0 = 2(x^2 + 2)(x + 1) - (2x^2 + 2x + 3)(2x),$

Resolving reveals turning points at x = -1 and another value obtained by solving yield turning points coordinates at:

Point 1: (-1, 1)
Point 2: \left(2, \frac{5}{2}\right)

Step 3

Find the equation of the asymptote of C.

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Answer

As x approaches ±∞, the expression

y = \frac{2x^2 + 2x + 3}{x^2 + 2} approaches:

y = \frac{2x^2}{x^2} = 2.

Thus, the horizontal asymptote of C is given by the equation:

y = 2.

Step 4

Sketch the graph of C, stating the coordinates of any intersections with the y-axis and the asymptote.

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Answer

Drawing the Y-axis Intersection: To find the intersection with the y-axis, substitute x = 0:

$y = \frac{3}{2}$ Therefore, the y-axis intersection is at (0, \frac{3}{2}).
Sketching the Asymptote: Draw a horizontal line at y = 2 for the asymptote.
Plotting the Turning Points: Mark the turning points (-1, 1) and \left(2, \frac{5}{2}\right).
Graph Outline: The graph will approach the asymptote at y = 2 while passing through the marked points. Ensure the curve increases and decreases accordingly, conforming to the established turning points.

The curve C has equation y = \frac{2x^2 + 2x + 3}{x^2 + 2} - CIE - A-Level Further Maths - Question 9 - 2012 - Paper 1

Worked Solution & Example Answer:The curve C has equation y = \frac{2x^2 + 2x + 3}{x^2 + 2} - CIE - A-Level Further Maths - Question 9 - 2012 - Paper 1

Show that, for all x, 1 \leq y \leq \frac{5}{2}.

Find the coordinates of the turning points on C.

Find the equation of the asymptote of C.

Sketch the graph of C, stating the coordinates of any intersections with the y-axis and the asymptote.

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