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

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19-(a)-Write-$x^2---10x-+-22$-in-the-form-$(x---a)^2---b$-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 $(x - a)^2 - b$ - 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 rewrite the quadratic expression x210x+22x^2 - 10x + 22 in the form (xa)2b(x - a)^2 - b, we will complete the square.

  1. Start with the expression: x210x+22x^2 - 10x + 22

  2. Take the coefficient of xx, which is 10-10, halve it to get 5-5, and then square it to get 2525. We will now add and subtract this value: x210x+2525+22x^2 - 10x + 25 - 25 + 22

  3. This can be rearranged as: (x5)23(x - 5)^2 - 3

Thus, we have: (x5)23(x - 5)^2 - 3

Step 2

Sketch the graph of $y = x^2 - 10x + 22$

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Answer

To sketch the graph of y=x210x+22y = x^2 - 10x + 22, we will first identify key features such as turning points and the y-intercept.

  1. Vertex/Turning Point: From our completed square form (x5)23(x - 5)^2 - 3, we can see that the turning point (vertex) is at (5,3)(5, -3). This is where the minimum value of the function occurs.

  2. Y-Intercept: To find the y-intercept, we set x=0x = 0: y=0210(0)+22=22y = 0^2 - 10(0) + 22 = 22 Thus, the y-intercept is at (0,22)(0, 22).

  3. Graph Characteristics: The graph is a parabola opening upwards since the coefficient of x2x^2 is positive. It will cross the y-axis at (0,22)(0, 22) and have its minimum point at (5,3)(5, -3).

  4. Sketch: Draw a standard Cartesian plane and plot the points (5,3)(5, -3) and (0,22)(0, 22). Then, sketch the parabola, ensuring that it opens upwards and passes through the y-intercept.

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