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17 (a) Write $x^2 + 8x + 3$ in the form $(x + a)^2 - b$ - OCR - GCSE Maths - Question 18 - 2019 - Paper 5

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17 (a) Write $x^2 + 8x + 3$ in the form $(x + a)^2 - b$. (b) Sketch the graph of $y = x^2 + 8x + 3$. Show clearly the coordinates of any turning points and the y-in... show full transcript

Worked Solution & Example Answer:17 (a) Write $x^2 + 8x + 3$ in the form $(x + a)^2 - b$ - OCR - GCSE Maths - Question 18 - 2019 - Paper 5

Step 1

Write $x^2 + 8x + 3$ in the form $(x + a)^2 - b$.

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Answer

To rewrite the quadratic expression in the required form, we can complete the square. Start with the expression:

x2+8x+3x^2 + 8x + 3

  1. Take the coefficient of xx, which is 88, divide it by 22, and square it:

    (82)2=16\left(\frac{8}{2}\right)^2 = 16

  2. Now rewrite the expression as:

x2+8x+1616+3x^2 + 8x + 16 - 16 + 3

This simplifies to:

(x+4)213(x + 4)^2 - 13

Thus, we have:

(x+4)213(x + 4)^2 - 13

Step 2

Sketch the graph of $y = x^2 + 8x + 3$. Show clearly the coordinates of any turning points and the y-intercept.

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Answer

To sketch the graph of the quadratic function y=x2+8x+3y = x^2 + 8x + 3, we can identify its vertex and y-intercept.

  1. Turning Point (Vertex): We have already rewritten the function as:

    y=(x+4)213y = (x + 4)^2 - 13

    The vertex is at the point (4,13)(-4, -13).

  2. Y-Intercept: To find the y-intercept, we set x=0x = 0:

    y=02+8(0)+3=3y = 0^2 + 8(0) + 3 = 3

    Therefore, the y-intercept is at (0,3)(0, 3).

  3. Graphing: Plot the vertex at (4,13)(-4, -13) and the y-intercept at (0,3)(0, 3). Since the parabola opens upwards (the coefficient of x2x^2 is positive), sketch the curve accordingly, ensuring the points are labeled. The graph should reflect the shape of a parabola opening upwards, with the turning point as the minimum point.

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