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Given that f(x) = x² - 6x + 18, x ≥ 0, (a) express f(x) in the form (x - α)² + b, where α and b are integers - Edexcel - A-Level Maths Pure - Question 4 - 2005 - Paper 2
Question 4
Given that f(x) = x² - 6x + 18, x ≥ 0, (a) express f(x) in the form (x - α)² + b, where α and b are integers. The curve C with equation y = f(x), x ≥ 0, meets the... show full transcript
Worked Solution & Example Answer:Given that f(x) = x² - 6x + 18, x ≥ 0, (a) express f(x) in the form (x - α)² + b, where α and b are integers - Edexcel - A-Level Maths Pure - Question 4 - 2005 - Paper 2
Step 1
express f(x) in the form (x - α)² + b
Answer
To express f(x) in the form (x - α)² + b, we need to complete the square:
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Start with the quadratic: f(x) = x² - 6x + 18.
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Take the coefficient of x, which is -6, halve it to get -3, and then square it to find 9.
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Rewrite the function: f(x) = (x² - 6x + 9) + 18 - 9.
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This simplifies to: f(x) = (x - 3)² + 9.
Thus, α = 3 and b = 9.
Step 2
Sketch the graph of C, showing the coordinates of P and Q
Answer
The graph of C is a U-shaped parabola based on the equation:
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The minimum point Q is at (3, 9), which is the vertex of the parabola.
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The function meets the y-axis at point P, which can be found by substituting x = 0 into the function:
f(0) = 0² - 6(0) + 18 = 18.
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Therefore, P = (0, 18).
In the sketch, plot points P (0, 18) and Q (3, 9) correctly in the first quadrant.
Step 3
Find the x-coordinate of R
Answer
To find the x-coordinate of R where the line y = 41 intersects the curve:
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Set f(x) equal to 41:
x² - 6x + 18 = 41.
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Rearranging gives:
x² - 6x - 23 = 0.
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Now use the quadratic formula to find x:
x = \frac{-b \pm \sqrt{b² - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)² - 4(1)(-23)}}{2(1)} = \frac{6 \pm \sqrt{36 + 92}}{2} = \frac{6 \pm \sqrt{128}}{2} = \frac{6 \pm 8\sqrt{2}}{2} = 3 \pm 4\sqrt{2}.
Thus, the x-coordinate of R in the form p + q√2 is 3 + 4√2.