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Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8$, $-0.5 \leq x \leq 2.2$ The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 3

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Question 10

Figure-2-shows-a-sketch-of-part-of-the-curve-with-equation--$y-=-4x^3-+-9x^2---30x---8$,--$-0.5-\leq-x-\leq-2.2$--The-curve-has-a-turning-point-at-the-point-A-Edexcel-A-Level Maths Pure-Question 10-2017-Paper 3.png

Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8$, $-0.5 \leq x \leq 2.2$ The curve has a turning point at the point A. (a) ... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8$, $-0.5 \leq x \leq 2.2$ The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 3

Step 1

Using calculus, show that the x coordinate of A is 1.

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Answer

To find the turning point, we first need to differentiate the given equation:

dydx=12x2+18x−30\frac{dy}{dx} = 12x^2 + 18x - 30

Next, we set the derivative equal to zero to find the critical points:

12x2+18x−30=012x^2 + 18x - 30 = 0

We can simplify by dividing the entire equation by 6:

2x2+3x−5=02x^2 + 3x - 5 = 0

Now, we apply the quadratic formula:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting in our values, where a=2a=2, b=3b=3, and c=−5c=-5:

x=−3±32−4⋅2⋅(−5)2⋅2x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2}

x=−3±9+404x = \frac{-3 \pm \sqrt{9 + 40}}{4}

x=−3±494x = \frac{-3 \pm \sqrt{49}}{4}

x=−3±74x = \frac{-3 \pm 7}{4}

Calculating the two potential solutions:

  1. First solution: x=44=1x = \frac{4}{4} = 1
  2. Second solution: x=−104=−2.5x = \frac{-10}{4} = -2.5 (not within the given range)

Thus, the x-coordinate of the turning point A is 1.

Step 2

Use integration to find the finite region R, giving your answer to 2 decimal places.

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Answer

To find the area of the shaded region R between the curve and the x-axis, we evaluate the definite integral of the curve from the x-coordinate of point B (x = 2) to the x-coordinate of point C (x = -\frac{1}{4}$):

Area=∫−142(4x3+9x2−30x−8) dx\text{Area} = \int_{-\frac{1}{4}}^{2} (4x^3 + 9x^2 - 30x - 8) \, dx

Calculating the integral:

=[x4+3x3−15x2−8x]−142= \left[ x^4 + 3x^3 - 15x^2 - 8x \right]_{-\frac{1}{4}}^{2}

Now substituting the bounds:

At x=2x = 2:

=(24+3(2)3−15(2)2−8(2))= (2^4 + 3(2)^3 - 15(2)^2 - 8(2))
=(16+24−60−16)=−36= (16 + 24 - 60 - 16) = -36

At x=−14x = -\frac{1}{4}:

=(−14)4+3(−14)3−15(−14)2−8(−14)= \left(-\frac{1}{4}\right)^4 + 3 \left(-\frac{1}{4}\right)^3 - 15 \left(-\frac{1}{4}\right)^2 - 8 \left(-\frac{1}{4}\right)
=1256−364−1516+2= \frac{1}{256} - \frac{3}{64} - \frac{15}{16} + 2

Converting to common denominators and simplifying: =1256−12256−240256+512256= \frac{1}{256} - \frac{12}{256} - \frac{240}{256} + \frac{512}{256} =1−12−240+512256=261256= \frac{1 - 12 - 240 + 512}{256} = \frac{261}{256}

Combining both results gives us:

Area=−36−261256\text{Area} = -36 - \frac{261}{256} Area=−36+1.019=−34.981\text{Area} = -36 + 1.019 = -34.981

To find the total area in the region R, we consider the absolute value:

Area=34.98\text{Area} = 34.98

Thus, rounding to two decimal places, we find:

Area≈34.98.\text{Area} \approx 34.98.

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