Express y=−2x2−12x+1y = -2x^2 - 12x + 1 in the form y=a−2(x+b)2y = a - 2(x + b)^2 and hence state the maximum value of yy, and its corresponding value of xx.

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Worked solution

3

1. Factor out the coefficient of $x^2$ from the $x$ terms

y=−2(x2+6x)+1y = -2(x^2 + 6x) + 1

2. Complete the square inside the bracket

y=−2[(x+62)2−(62)2]+1y = -2\left[\left(x + \frac{6}{2}\right)^2 - \left(\frac{6}{2}\right)^2\right] + 1
y=−2[(x+3)2−9]+1y = -2[(x + 3)^2 - 9] + 1

3. Expand and simplify to the required form

y=−2(x+3)2+18+1y = -2(x + 3)^2 + 18 + 1
y=19−2(x+3)2y = 19 - 2(x + 3)^2

4. State the maximum value and corresponding $x$

Maximum value of yy is 1919.
This occurs when x+3=0  ⟹  x=−3x + 3 = 0 \implies x = -3.

Answer

y=19−2(x+3)2y = 19 - 2(x + 3)^2; Maximum value =19= 19, corresponding x=−3x = -3

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