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The curve C has equation y = (x + 3)(x - 1)^2 - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 1
Question 2
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. (4) b) Show t... show full transcript
Worked Solution & Example Answer:The curve C has equation y = (x + 3)(x - 1)^2 - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 1
Step 1
Sketch C showing clearly the coordinates of the points where the curve meets the coordinate axes.
Answer
To sketch the curve C, start by finding where it meets the x-axis and y-axis.
Finding the intercepts:
-
Y-Intercept (x = 0):
The y-intercept can be calculated as:
. Evaluate the expression:
y = (0 + 3)(0 - 1)^2 = 3(1) = 3.
Thus, the y-intercept is at (0, 3).
-
X-Intercepts (y = 0): Set y = 0 and solve:
(x + 3)(x - 1)^2 = 0
This gives:
- x + 3 = 0 => x = -3
- (x - 1)^2 = 0 => x = 1
Thus, the x-intercepts are at (-3, 0) and (1, 0).
Sketching the Curve:
- Plot the points: (0, 3), (-3, 0), and (1, 0).
- The shape of the curve is a cubic function, indicating the behavior after the intercepts, showing a minimum point.
- The general form will look like a 'U' shape reaching a minimum at (1, 0) and passing through the other intercept points.
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.
Answer
To rewrite the equation, expand the given equation:
y = (x + 3)(x - 1)^2.
-
First, expand (x - 1)^2:
(x - 1)(x - 1) = x^2 - 2x + 1.
-
Now multiply by (x + 3):
y = (x + 3)(x^2 - 2x + 1).
-
Use the distributive property to expand:
y = x^3 - 2x^2 + x + 3x^2 - 6x + 3
= x^3 + x^2 - 5x + 3.
Thus, we can see that k = 3, which is a positive integer.
Step 3
Find the x-coordinates of these two points.
Answer
To find the x-coordinates where the gradient of the tangent to C is equal to 3, follow these steps:
-
Differentiate the equation y = x^3 + x^2 - 5x + 3:
rac{dy}{dx} = 3x^2 + 2x - 5.
-
Set the derivative equal to 3:
3x^2 + 2x - 5 = 3
3x^2 + 2x - 8 = 0. -
Solve the quadratic equation using the quadratic formula:
x = rac{-b ext{ ± } ext{√}(b^2 - 4ac)}{2a}
where a = 3, b = 2, c = -8.Calculate.
b^2 - 4ac = 2^2 - 4(3)(-8)
= 4 + 96 = 100.Thus,
x = rac{-2 ext{ ± } 10}{6}.
-
From here, solve for x:
x = rac{8}{6} = rac{4}{3} ext{ and } x = rac{-12}{6} = -2.
Therefore, the x-coordinates of these two points are x = -2 and x = rac{4}{3}.