The curve $C_1$ has equation $$y = x^2(x + 2)$$ (a) Find $ \frac{dy}{dx} $. (b) Sketch $C_1$, showing the coordinates of the points where $C_1$ meets the x-axis. (c) Find the gradient of $C_1$ at each point where $C_1$ meets the x-axis. The curve $C_2$ has equation $$y = (x-k^2)(x-k+2)$$ where $k$ is a constant and $k > 2$ (d) Sketch $C_2$, showing the coordinates of the points where $C_2$ meets the x and y axes.

A-Level Edexcel Maths Pure Graphs of Functions: The curve $C

The curve $C_1$ has equation $$y = x^2(x + 2)$$ (a) Find $ \frac{dy}{dx} $ - Edexcel - A-Level Maths Pure - Question 10 - 2012 - Paper 1

Question 10

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The curve $C_1$ has equation $$y = x^2(x + 2)$$ (a) Find $ \frac{dy}{dx} $. (b) Sketch $C_1$, showing the coordinates of the points where $C_1$ meets the x-axis... show full transcript

Worked Solution & Example Answer:The curve $C_1$ has equation $$y = x^2(x + 2)$$ (a) Find $ \frac{dy}{dx} $ - Edexcel - A-Level Maths Pure - Question 10 - 2012 - Paper 1

Step 1

Find $ \frac{dy}{dx} $

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Answer

To find ( \frac{dy}{dx} ), we first multiply out the expression:

$y = x^3 + 2x^2$

Now, differentiating with respect to x:

$\frac{dy}{dx} = 3x^2 + 4x$

Step 2

Sketch $C_1$, showing the coordinates of the points where $C_1$ meets the x-axis.

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Answer

To find the points where the curve meets the x-axis, we set (y = 0):

$0 = x^2 (x + 2)$

This gives:

(x = 0)
(x = -2)

These points are where the curve intersects the x-axis: (0,0) and (-2,0). The sketch should show these points as intersections.

Step 3

Find the gradient of $C_1$ at each point where $C_1$ meets the x-axis.

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Answer

To find the gradient at the x-axis intersection points, we will evaluate (\frac{dy}{dx}) at the points x = 0 and x = -2:

At (x = 0):

$\frac{dy}{dx} \bigg|_{x=0} = 3(0)^2 + 4(0) = 0$

At (x = -2):

$\frac{dy}{dx} \bigg|_{x=-2} = 3(-2)^2 + 4(-2) = 12 - 8 = 4$

Thus, the gradients are 0 and 4 at (0,0) and (-2,0) respectively.

Step 4

Sketch $C_2$, showing the coordinates of the points where $C_2$ meets the x and y axes.

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Answer

To find where $C_2$ meets the x-axis, we set $y = 0$ :

$0 = (x-k^2)(x-k+2)$

This gives:

(x = k^2)
(x = k - 2)

To find where $C_2$ meets the y-axis, we set (x=0):

$y = (0-k^2)(0-k+2) = k^2(k-2)$

In the sketch, mark these intersections accordingly.

The curve $C_1$ has equation $$y = x^2(x + 2)$$ (a) Find \( \frac{dy}{dx} \) - Edexcel - A-Level Maths Pure - Question 10 - 2012 - Paper 1

Worked Solution & Example Answer:The curve $C_1$ has equation $$y = x^2(x + 2)$$ (a) Find \( \frac{dy}{dx} \) - Edexcel - A-Level Maths Pure - Question 10 - 2012 - Paper 1

Find \( \frac{dy}{dx} \)

Sketch $C_1$, showing the coordinates of the points where $C_1$ meets the x-axis.

Find the gradient of $C_1$ at each point where $C_1$ meets the x-axis.

Sketch $C_2$, showing the coordinates of the points where $C_2$ meets the x and y axes.

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