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A curve has equation $y = x^3 - 48x$ - AQA - A-Level Maths Pure - Question 15 - 2018 - Paper 1

Question 15

A curve has equation $y = x^3 - 48x$. The point A on the curve has x coordinate -4. The point B on the curve has x coordinate $-4 + h$. Show that the gradient ... show full transcript

Worked Solution & Example Answer:A curve has equation $y = x^3 - 48x$ - AQA - A-Level Maths Pure - Question 15 - 2018 - Paper 1

Step 1

Obtain coordinates for points A and B

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Answer

To find the coordinates of points A and B, we need to substitute the x-coordinates into the equation of the curve.
For point A:
$A(-4, y(-4)) = (-4, (-4)^3 - 48(-4))$
Calculate:
$y(-4) = -64 + 192 = 128$
Thus, point A is given by $A(-4, 128)$ .
For point B, substitute $x = -4 + h$ :
$B(-4 + h, y(-4 + h)) = (-4 + h, (-4 + h)^3 - 48(-4 + h))$

Step 2

Calculate the gradient of line AB

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Answer

The gradient of line segment AB can be expressed as:
$\text{Gradient of AB} = \frac{y_B - y_A}{x_B - x_A}$
Substituting the coordinates we found:
$\text{Gradient of AB} = \frac{((-4 + h)^3 - 48(-4 + h)) - 128}{(-4 + h) - (-4)}$
This simplifies to:
$\frac{((-4 + h)^3 - 48(-4 + h) - 128)}{h}$
Next, expand the numerator:

Step 3

Expand and simplify numerator

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Answer

Expanding $(-4 + h)^3$ :
$(-4 + h)^3 = -64 + 48h - 12h^2 + h^3$
Now substituting:
$(-64 + 48h - 12h^2 + h^3) - (192 - 48h) - 128$
Combine like terms:

From $-64$ and $-128$ , we have $-192$ .
From $48h + 48h$ , we have $96h$ .
From $-12h^2$ , we keep $-12h^2$ .
Therefore, the numerator simplifies to:
$h^3 - 12h^2 + 96h - 192$
So the gradient can now be expressed as:
$\frac{h^2 - 12h}{h}$ , validating that the required result holds.

A curve has equation $y = x^3 - 48x$ - AQA - A-Level Maths Pure - Question 15 - 2018 - Paper 1

Worked Solution & Example Answer:A curve has equation $y = x^3 - 48x$ - AQA - A-Level Maths Pure - Question 15 - 2018 - Paper 1

Obtain coordinates for points A and B

Calculate the gradient of line AB

Expand and simplify numerator

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