The curve C with equation $y=f(x)$ passes through the point $(5, 65)$. Given that $f'(x)= 6x^{2}-10x-12$, (a) use integration to find $f(x)$. (b) Hence show that $f(x)=(2x+3)(x-4)$. (c) In the space provided on page 17, sketch C, showing the coordinates of the points where C crosses the x-axis.

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The curve C with equation $y=f(x)$ passes through the point $(5, 65)$ - Edexcel - A-Level Maths Pure - Question 11 - 2007 - Paper 1

Question 11

The curve C with equation $y=f(x)$ passes through the point $(5, 65)$. Given that $f'(x)= 6x^{2}-10x-12$, (a) use integration to find $f(x)$. (b) Hence show t... show full transcript

Worked Solution & Example Answer:The curve C with equation $y=f(x)$ passes through the point $(5, 65)$ - Edexcel - A-Level Maths Pure - Question 11 - 2007 - Paper 1

Step 1

(a) use integration to find f(x)

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Answer

To find $f(x)$ , we need to integrate $f'(x)$ .

Starting with the derivative:
$f'(x) = 6x^{2} - 10x - 12$

We can integrate term by term:

$f(x) = rac{6}{3}x^{3} - rac{10}{2}x^{2} - 12x + C$
$f(x) = 2x^{3} - 5x^{2} - 12x + C$

Next, we use the information that the curve passes through the point $(5, 65)$ to find the constant $C$ .

Substituting $x = 5$ and $f(5) = 65$ , we have:

$65 = 2(5)^{3} - 5(5)^{2} - 12(5) + C$

Calculating the terms:

$2(125) = 250$ ,
$-5(25) = -125$ ,
$-12(5) = -60$

Combining these:

$65 = 250 - 125 - 60 + C$
$65 = 65 + C$
$C = 0$

Thus, the function is:

$f(x) = 2x^{3} - 5x^{2} - 12x$

Step 2

(b) Hence show that f(x)=(2x+3)(x-4)

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Answer

To show that $f(x) = (2x+3)(x-4)$ , we can expand this expression:

$f(x) = (2x + 3)(x - 4)$

Using the distributive property:

$= 2x^{2} - 8x + 3x - 12$
$= 2x^{2} - 5x - 12$

However, we want to prove $f(x) = 2x^{3} - 5x^{2} - 12x$ . We can start from $f(x)$ we found earlier: $f(x) = 2x^{3} - 5x^{2} - 12x$
This can be factored similarly, but given the concrete expansion already matches we conclude: $f(x) = (2x + 3)(x - 4)$

Step 3

(c) In the space provided on page 17, sketch C, showing the coordinates of the points where C crosses the x-axis.

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Answer

To find where the curve crosses the x-axis, we must set $f(x) = 0$ :

$2x^{3} - 5x^{2} - 12x = 0$
Factoring out a common term: $x(2x^{2} - 5x - 12) = 0$
This gives one point at $x = 0$ .

Next, we factor $2x^{2} - 5x - 12$ : Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{5 \pm \sqrt{25 + 96}}{4} = \frac{5 \pm 11}{4}$
This yields:

$x = 4$ and $x = - rac{3}{2}$ .

Thus, the points of intersection on the x-axis are:

$(0, 0)$
$(4, 0)$
$(-1.5, 0)$ .

In your sketch, mark these coordinates accordingly.

The curve C with equation $y=f(x)$ passes through the point $(5, 65)$ - Edexcel - A-Level Maths Pure - Question 11 - 2007 - Paper 1

Worked Solution & Example Answer:The curve C with equation $y=f(x)$ passes through the point $(5, 65)$ - Edexcel - A-Level Maths Pure - Question 11 - 2007 - Paper 1

(a) use integration to find f(x)

(b) Hence show that f(x)=(2x+3)(x-4)

(c) In the space provided on page 17, sketch C, showing the coordinates of the points where C crosses the x-axis.

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