Photo AI

The curve C has equation y = (x + 3)(x - 1)^2 - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

Question 1

The curve C has equation y = (x + 3)(x - 1)^2. (a) Sketch C showing clearly the coordinates of the points where the curve meets the coordinate axes. (b) Show that... show full transcript

Worked Solution & Example Answer:The curve C has equation y = (x + 3)(x - 1)^2 - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

Step 1

Sketch C showing clearly the coordinates of the points where the curve meets the coordinate axes.

96%

114 rated

Answer

To sketch the curve defined by the equation $y = (x + 3)(x - 1)^2$ , we need to find points where the curve meets the axes.

Finding x-intercepts: Set $y = 0$ :
- $(x + 3)(x - 1)^2 = 0$ implies $x + 3 = 0$ or $(x - 1)^2 = 0$
- Thus, $x = -3$ and $x = 1$ .
- The x-intercepts are at points $(-3, 0)$ and $(1, 0)$ .
Finding y-intercept: Set $x = 0$ :
- $y = (0 + 3)(0 - 1)^2 = 3(1) = 3$ , which gives the point $(0, 3)$ .

In sketching, ensure to indicate these points correctly, along with the shape of the curve (which opens upward with a local minimum at $(1, 0)$ ).

Step 2

Show that the equation of C can be written in the form y = x^3 + x^2 - 5x + k, where k is a positive integer, and state the value of k.

99%

104 rated

Answer

Expand the equation $y = (x + 3)(x - 1)^2$ :

First expand $(x - 1)^2 = x^2 - 2x + 1$ .
Now express the entire equation: $y = (x + 3)(x^2 - 2x + 1)$ $= x(x^2 - 2x + 1) + 3(x^2 - 2x + 1)$ $= x^3 - 2x^2 + x + 3x^2 - 6x + 3$ $= x^3 + x^2 - 5x + 3$ .

Thus, we confirm the form $y = x^3 + x^2 - 5x + k$ , where $k = 3$ .

Step 3

Find the x-coordinates of these two points.

96%

101 rated

Answer

To find the x-coordinates where the gradient of the tangent to C is equal to 3, we first find the derivative of $y$ :

Differentiate $y = x^3 + x^2 - 5x + 3$ : $\frac{dy}{dx} = 3x^2 + 2x - 5$ .
Set the derivative equal to 3: $3x^2 + 2x - 5 = 3$ $\Rightarrow 3x^2 + 2x - 8 = 0$ .
Solve using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 3$ , $b = 2$ , and $c = -8$ :
- Calculate the discriminant: $b^2 - 4ac = 2^2 - 4(3)(-8) = 4 + 96 = 100$ .
- Substitute: $x = \frac{-2 \pm 10}{6}$ .
- This gives two solutions: $x_1 = \frac{8}{6} = \frac{4}{3}$ $x_2 = \frac{-12}{6} = -2$ .

Therefore, the x-coordinates of the points are $\frac{4}{3}$ and $-2$ .

The curve C has equation y = (x + 3)(x - 1)^2 - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

Worked Solution & Example Answer:The curve C has equation y = (x + 3)(x - 1)^2 - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

Sketch C showing clearly the coordinates of the points where the curve meets the coordinate axes.

Show that the equation of C can be written in the form y = x^3 + x^2 - 5x + k, where k is a positive integer, and state the value of k.

Find the x-coordinates of these two points.

Join the A-Level students using SimpleStudy...