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 Equation of a Straight Line: The curve $C

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 - 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 $\frac{dy}{dx}$ (b) Sketch $C_1$, showing the coordinates of the points where $C_1$ meets the x-axis - 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 differentiate the equation of the curve:

Expand the equation: $y = x^3 + 2x^2$
Differentiate 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, set (y = 0):

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

This gives:

(x = 0)
(x + 2 = 0 \Rightarrow x = -2)

Thus, the coordinates of the points where $C_1$ meets the x-axis are ((0, 0)) and ((-2, 0)).

In the sketch of the curve, ensure the curve:

Passes through the origin (0, 0)
Touches the x-axis at ((-2, 0)) without crossing it.

Step 3

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

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Answer

Using the derivative found in part (a), we evaluate (\frac{dy}{dx}) at the points ((0, 0)) and ((-2, 0)):

At (x = 0): $\frac{dy}{dx} = 3(0)^2 + 4(0) = 0$
At (x = -2): $\frac{dy}{dx} = 3(-2)^2 + 4(-2) = 12 - 8 = 4$

This gives us gradients of 0 at the origin and 4 at ((-2, 0)).

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

First, identify where the curve meets the axes:

Y-axis Intercept: Set (x = 0): $y = (0 - k^2)(0 - k + 2) = (0 - k^2)(2 - k) = k^2(k - 2)$

Coordinates: ((0, k^2(k - 2))).

X-axis Intercepts: Set (y = 0): $0 = (x - k^2)(x - k + 2)$ This yields: (x = k^2) and (x = k - 2)

Coordinates: ((k^2, 0)) and ((k - 2, 0)).

In the sketch, appropriately mark the intercepts on both axes.

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 - 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 \(\frac{dy}{dx}\) (b) Sketch $C_1$, showing the coordinates of the points where $C_1$ meets the x-axis - 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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