Answer ALL the questions

  1. (a)

    Express 3x2−8x−33x^2 - 8x - 3 in the form a(x+b)2+ca(x + b)^2 + c, where a,ba, b and cc are constants to be determined.

  2. (b)

    Hence, find the turning point and the value of xx where it occurs.

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Worked solution

1(a)

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

3(x2−83x)−33\left(x^2 - \frac{8}{3}x\right) - 3

2. Complete the square inside the bracket

3[(x−43)2−(43)2]−33\left[ \left(x - \frac{4}{3}\right)^2 - \left(\frac{4}{3}\right)^2 \right] - 3
=3[(x−43)2−169]−3= 3\left[ \left(x - \frac{4}{3}\right)^2 - \frac{16}{9} \right] - 3

3. Expand and simplify to reach the final form

3(x−43)2−3(169)−33\left(x - \frac{4}{3}\right)^2 - 3\left(\frac{16}{9}\right) - 3
=3(x−43)2−163−93= 3\left(x - \frac{4}{3}\right)^2 - \frac{16}{3} - \frac{9}{3}
=3(x−43)2−253= 3\left(x - \frac{4}{3}\right)^2 - \frac{25}{3}

Answer

3(x−43)2−2533\left(x - \frac{4}{3}\right)^2 - \frac{25}{3}

1(b)

1. State the completed square form from part (a)

From part (a),
3x2−8x−3=3(x−43)2−2533x^2 - 8x - 3 = 3\left(x - \frac{4}{3}\right)^2 - \frac{25}{3}

2. Identify the turning point and the corresponding $x$-value

The minimum value is −253-\frac{25}{3}, which occurs when x−43=0  ⟹  x=43x - \frac{4}{3} = 0 \implies x = \frac{4}{3}.
Turning point is (43,−253)\left(\frac{4}{3}, -\frac{25}{3}\right).

Answer

Turning point: (43,−253)\left(\frac{4}{3}, -\frac{25}{3}\right), x=43x = \frac{4}{3}

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