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The curve C has equation y = (x + 1)(x + 3)^2 (a) Sketch C, showing the coordinates of the points at which C meets the axes - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 2
Question 2
The curve C has equation y = (x + 1)(x + 3)^2 (a) Sketch C, showing the coordinates of the points at which C meets the axes. (b) Show that \( \frac{dy}{dx} = 3x^... show full transcript
Worked Solution & Example Answer:The curve C has equation y = (x + 1)(x + 3)^2 (a) Sketch C, showing the coordinates of the points at which C meets the axes - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 2
Step 1
Sketch C, showing the coordinates of the points at which C meets the axes.
Answer
To sketch the curve, we need to identify where it intersects the axes.
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Finding x-intercepts:
Set ( y = 0 ):
( (x + 1)(x + 3)^2 = 0 )
Hence, the x-intercepts are at ( x = -1 ) and ( x = -3 ). -
Finding y-intercept:
Set ( x = 0 ):
( y = (0 + 1)(0 + 3)^2 = 9 ).
The y-intercept is at (0, 9).
The sketch should show these points clearly.
Step 2
Step 3
Find the equation of the tangent to C at A, giving your answer in the form \( y = mx + c \).
Answer
At point A ( x = -5 ), first we find ( y ):
( y = (-5 + 1)(-5 + 3)^2 = -4(4) = -16 ).
Therefore, A is (-5, -16).
Next, we calculate ( \frac{dy}{dx} ) at ( x = -5 ):
( \frac{dy}{dx} = 3(-5)^2 + 14(-5) + 15 = 75 - 70 + 15 = 20 ).
Thus, the slope (m) is 20.
Using point-slope form:
( y - (-16) = 20(x - (-5)) )
Simplifying gives:
( y = 20x + 84 ).
Therefore, the equation of the tangent is ( y = 20x + 84 ).
Step 4
Find the x-coordinate of B.
Answer
Since the tangents to C at A and B are parallel, they have the same slope (20).
The tangent at B will also be given by ( \frac{dy}{dx} = 20 ).
From ( \frac{dy}{dx} = 3x^2 + 14x + 15 = 20 ):
( 3x^2 + 14x - 5 = 0 ).
Using the quadratic formula:
( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-14 \pm \sqrt{196 + 60}}{6} = \frac{-14 \pm \sqrt{256}}{6} = \frac{-14 \pm 16}{6} ).
Therefore, the possible x-coordinates are:
( \frac{2}{6} = \frac{1}{3} ) and ( -5 ).
Since -5 is the point A, the x-coordinate of point B must be ( \frac{1}{3} ).