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19 (a) Write $x^2 - 10x + 22$ in the form $(x - a)^2 - b.$ (b) Sketch the graph of $y = x^2 - 10x + 22$ - OCR - GCSE Maths - Question 19 - 2020 - Paper 1

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19-(a)-Write-$x^2---10x-+-22$-in-the-form-$(x---a)^2---b.$--(b)-Sketch-the-graph-of-$y-=-x^2---10x-+-22$-OCR-GCSE Maths-Question 19-2020-Paper 1.png

19 (a) Write $x^2 - 10x + 22$ in the form $(x - a)^2 - b.$ (b) Sketch the graph of $y = x^2 - 10x + 22$. Show clearly the coordinates of any turning points and the ... show full transcript

Worked Solution & Example Answer:19 (a) Write $x^2 - 10x + 22$ in the form $(x - a)^2 - b.$ (b) Sketch the graph of $y = x^2 - 10x + 22$ - OCR - GCSE Maths - Question 19 - 2020 - Paper 1

Step 1

Write $x^2 - 10x + 22$ in the form $(x - a)^2 - b.$

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Answer

To express the quadratic in the desired form, we first complete the square:

  1. Start with the expression: x2−10x+22x^2 - 10x + 22.

  2. Take half of the coefficient of xx, which is −10-10, to get −5-5, and square it to get 2525.

  3. Rewrite the quadratic by adding and subtracting 2525:

    x2−10x+25−25+22x^2 - 10x + 25 - 25 + 22

    = (x−5)2−3(x - 5)^2 - 3. Therefore, we have a=5a = 5 and b=3b = 3.

Step 2

Sketch the graph of $y = x^2 - 10x + 22$. Show clearly the coordinates of any turning points and the value of the y-intercept.

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Answer

To sketch the graph:

  1. Identify the Turning Point: The vertex form of the quadratic is (x−5)2−3(x - 5)^2 - 3. The turning point (vertex) is at (5,−3)(5, -3).

  2. Calculate the y-intercept: Set x=0x = 0:

    y=(0)2−10(0)+22=22.y = (0)^2 - 10(0) + 22 = 22.
    The y-intercept is at (0,22)(0, 22).

  3. Sketching the Graph:

    • Draw the parabola opening upwards with the vertex at (5,−3)(5, -3) and the y-intercept at (0,22)(0, 22). The graph will be symmetric about the line x=5x = 5, as it is a standard upwards-opening quadratic function.

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