GCSE OCR Maths Quadratic and cubic graphs: 20 (a) Write $x^2

20 (a) Write $x^2 - 6x + 11$ in the form $(x - a)^2 + b$ - OCR - GCSE Maths - Question 20 - 2019 - Paper 1

Question 20

20 (a) Write $x^2 - 6x + 11$ in the form $(x - a)^2 + b$. (b) Sketch the graph of $y = x^2 - 6x + 11$. Show clearly the coordinates of any turning points.

Worked Solution & Example Answer:20 (a) Write $x^2 - 6x + 11$ in the form $(x - a)^2 + b$ - OCR - GCSE Maths - Question 20 - 2019 - Paper 1

Step 1

(a) Write $x^2 - 6x + 11$ in the form $(x - a)^2 + b$.

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Answer

To express the quadratic equation $x^2 - 6x + 11$ in the form $(x - a)^2 + b$ , we will complete the square.

Start with the initial expression:
$x^2 - 6x + 11$
Take the coefficient of $x$ , which is $-6$ , divide it by 2 to get $-3$ , and then square it to find $( -3 )^2 = 9$ .
Rewrite the expression by adding and subtracting this square: $x^2 - 6x + 9 - 9 + 11$
This simplifies to: $(x - 3)^2 + 2$

Thus, we can express it as:
$y = (x - 3)^2 + 2$ .

Step 2

(b) Sketch the graph of $y = x^2 - 6x + 11$. Show clearly the coordinates of any turning points.

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Answer

The turning point of the graph can be determined from the completed square form, which is $(x - 3)^2 + 2$ . The vertex of the parabola occurs at $(3, 2)$ , which is the minimum point since the parabola opens upwards.

Turning Point Coordinates: The coordinates of the turning point are:
- Turning Point: $(3, 2)$
Y-intercept: To find the y-intercept, set $x = 0$ :
$y = 0^2 - 6(0) + 11 = 11$
So, the y-intercept is $(0, 11)$ .
X-intercepts: Set $y = 0$ to find x-intercepts: $0 = x^2 - 6x + 11$ The discriminant is $(-6)^2 - 4(1)(11) = 36 - 44 = -8$ , which indicates there are no real x-intercepts.
Sketching the Graph:
- Start by plotting the turning point $(3, 2)$ and the y-intercept $(0, 11)$ .
- Draw a symmetrical U-shaped curve opening upwards as it does not intercept the x-axis.
- Clearly label the turning point and the y-intercept on the graph.

20 (a) Write $x^2 - 6x + 11$ in the form $(x - a)^2 + b$ - OCR - GCSE Maths - Question 20 - 2019 - Paper 1

Worked Solution & Example Answer:20 (a) Write $x^2 - 6x + 11$ in the form $(x - a)^2 + b$ - OCR - GCSE Maths - Question 20 - 2019 - Paper 1

(a) Write $x^2 - 6x + 11$ in the form $(x - a)^2 + b$.

(b) Sketch the graph of $y = x^2 - 6x + 11$. Show clearly the coordinates of any turning points.

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