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f(x) = x² - 8x + 19 (a) Express f(x) in the form (x + a)² + b, where a and b are constants - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 1

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f(x)-=-x²---8x-+-19--(a)-Express-f(x)-in-the-form-(x-+-a)²-+-b,-where-a-and-b-are-constants-Edexcel-A-Level Maths Pure-Question 7-2017-Paper 1.png

f(x) = x² - 8x + 19 (a) Express f(x) in the form (x + a)² + b, where a and b are constants. The curve C with equation y = f(x) crosses the y-axis at the point P a... show full transcript

Worked Solution & Example Answer:f(x) = x² - 8x + 19 (a) Express f(x) in the form (x + a)² + b, where a and b are constants - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 1

Step 1

Express f(x) in the form (x + a)² + b

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Answer

To express the quadratic function in the form (x + a)² + b, we can complete the square.

  1. Start with the given function: f(x)=x28x+19f(x) = x² - 8x + 19

  2. To complete the square for the expression x28xx² - 8x, take half of the coefficient of xx, which is 4-4, and square it to get 1616.

  3. Rewrite the function: f(x)=(x28x+16)+1916f(x) = (x² - 8x + 16) + 19 - 16

  4. This results in: f(x)=(x4)2+3f(x) = (x - 4)² + 3

Thus, we have a=4a = -4 and b=3b = 3.

Step 2

Sketch the graph of C showing the coordinates of point P and the coordinates of point Q

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Answer

The curve C is a parabola that opens upwards, with its vertex representing the minimum point Q.

  1. The coordinates of point P, where the curve crosses the y-axis, can be found by evaluating f(0)f(0): f(0)=(04)2+3=16+3=19f(0) = (0 - 4)² + 3 = 16 + 3 = 19 Therefore, P(0, 19).

  2. The coordinates of point Q, which is the vertex, are (4, 3).

  3. When sketching the graph, ensure to mark:

    • Point P at (0, 19)
    • Point Q at (4, 3)

The graph should be a U-shaped curve that is symmetrical about the vertical line x = 4.

Step 3

Find the distance PQ, writing your answer as a simplified surd

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Answer

To find the distance between points P(0, 19) and Q(4, 3), we can use the distance formula:

d = ext{PQ} = rac{ ext{distance between } P ext{ and } Q}{ ext{distance formula}}

d = ext{PQ} = rac{(x_2 - x_1)² + (y_2 - y_1)²}{ ext{where } (x_1, y_1) = (0, 19) ext{ and } (x_2, y_2) = (4, 3)}

  1. Substitute the coordinates into the formula: d = ext{PQ} = rac{(4 - 0)² + (3 - 19)²}{ ext{which simplifies to}} d = ext{PQ} = rac{4² + (-16)²}{ ext{which gives us}} d = ext{PQ} = rac{16 + 256} = rac{272} d = PQ = rac{16 ext{√}17}{7}

Final distance, written as a simplified surd: d = rac{4 ext{√}17}{7}.

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