(i)

\begin{align*} f(3) & = 2(3)^2 + 1 \\ & = 19 \end{align*}

(ii)

\begin{align*} f(-2) & = 2(-2)^2 + 1 \\ & = 9 \end{align*}

(iii)

\begin{align*} f(x) & = 2x^2 + 1 \\ \\ \text{Since } x^2 \ge 0, & \phantom{0} 2x^2 + 1 \ge 1 \\ \\ \therefore f(x) & \ge 1 \end{align*}

(a)

\begin{align*} x^2 + 12x & = \left(x + {12 \over 2}\right)^2 - \left(12 \over 2\right)^2 \\ & = (x + 6)^2 - 36 \end{align*}

(b)

\begin{align*} x^2 - 4x + 1 & = \left(x - {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 + 1 \\ & = (x - 2)^2 -3 \end{align*}

(c)

\begin{align*} 3x^2 + 3x - 2 & = 3(x^2 + x) - 2 \\ & = 3 \left[ \left(x + {1 \over 2}\right)^2 - \left( {1 \over 2}\right)^2 \right] - 2 \\ & = 3 \left[ \left(x + {1 \over 2}\right)^2 - {1 \over 4} \right] - 2 \\ & = 3 \left( x + {1 \over 2} \right)^2 - {3 \over 4} - 2 \\ & = 3 \left(x + {1 \over 2} \right)^2 - {11 \over 4} \end{align*}

(d)

\begin{align*} 3x^2 - 6x + 2 & = 3(x^2 - 2x) + 2 \\ & = 3 \left[ \left( x - {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 \right] + 2 \\ & = 3 [ (x - 1)^2 - 1 ] + 2 \\ & = 3(x - 1)^2 - 3 + 2 \\ & = 3(x - 1)^2 - 1 \end{align*}

(e)

\begin{align*} -3x^2 + 6x - 1 & = -3(x^2 - 2x) - 1 \\ & = -3 \left[ \left(x - {2 \over 2}\right)^2 - \left(2 \over 2\right)^2 \right] - 1 \\ & = -3 [ (x - 1)^2 - 1] - 1 \\ & = -3(x - 1)^2 + 3 - 1 \\ & = -3(x - 1)^2 + 2 \end{align*}

(f)

\begin{align*} -x^2 - 4x + 1 & = -(x^2 + 4x) + 1 \\ & = - \left[ \left( x + {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 \right] + 1 \\ & = - [ (x + 2)^2 - 4] + 1 \\ & = -(x + 2)^2 + 4 + 1 \\ & = -(x + 2)^2 + 5 \end{align*}

(g)

\begin{align*} -{1 \over 2}x^2 - 3x + 5 & = -{1 \over 2} \left( x^2 + 6x \right) + 5 \\ & = -{1 \over 2} \left[ \left(x + {6 \over 2} \right)^2 - \left(6 \over 2\right)^2 \right] + 5 \\ & = -{1 \over 2} [ (x + 3)^2 - 9] + 5 \\ & = -{1 \over 2} (x + 3)^2 + {9 \over 2} + 5 \\ & = -{1 \over 2} (x + 3)^2 + {19 \over 2} \end{align*}

(h)

\begin{align*} -2x^2 + 5x + 3 & = -2 \left( x^2 - {5 \over 2}x \right) + 3 \\ & = -2 \left[ \left(x - {5 \over 4}\right)^2 - \left(5 \over 4\right)^2 \right] + 3 \\ & = -2 \left[ \left(x - {5 \over 4}\right)^2 - {25 \over 16} \right] + 3 \\ & = -2 \left( x - {5 \over 4} \right)^2 + {25 \over 8} + 3 \\ & = -2 \left( x - {5 \over 4} \right)^2 + {49 \over 8} \end{align*}

For a quadratic expression in the form a(x - h)² + k,

• If a > 0, the expression has a minimum value of k, when x = h
• If a < 0, the expression has a maximum value of k, when x = h

(a)

$$ \text{Minimum value} = -7, \text{ when } x = -1$$

(b)

$$ \text{Maximum value} = 4, \text{ when } x = 1$$

(c)

$$ \text{Maximum value} = {4 \over 5}, \text{ when } x = {1 \over 2}$$

(d)

$$ \text{Minimum value} = 4.5, \text{ when } x = -3$$

\begin{align} \text{For all real } & \text{values of } x, \\ (x + 3)^2 & \ge 0 \\ (x + 3)^2 + 1 & \ge 1 \\ \\ \therefore (x + 3)^2 + 1 \text{ is } & \text{positive for all real values of } x \end{align}

(a)

\begin{align*} f(x) & = 2x^2 - 4x - 5 \\ & = 2(x^2 - 2x) - 5 \\ & = 2 \left[ \left(x - {2 \over 2} \right)^2 - \left( 2 \over 2 \right)^2 \right] - 5 \\ & = 2 [ (x - 1)^2 - 1 ] - 5 \\ & = 2(x - 1)^2 - 2 - 5 \\ & = 2(x - 1)^2 - 7 \\ \\ \text{Minimum value} & = -7, \text{ when } x = 1 \end{align*}

(b)

\begin{align*} f(x) & = -x^2 - 2x + 3 \\ & = -(x^2 + 2x) + 3 \\ & = - \left[ \left( x + {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 \right] + 3 \\ & = - [ (x + 1)^2 - 1 ] + 3 \\ & = -(x + 1)^2 + 1 + 3 \\ & = -(x + 1)^2 + 4 \\ \\ \text{Maximum value} & = 4, \text{ when } x = -1 \end{align*}

(c)

\begin{align*} f(x) & = 2 - 3x + 3x^2 \\ & = 3x^2 - 3x + 2 \\ & = 3(x^2 - x) + 2 \\ & = 3 \left[ \left(x - {1 \over 2} \right)^2 - \left(1 \over 2\right)^2 \right] + 2 \\ & = 3 \left[ \left(x - {1 \over 2} \right)^2 - {1 \over 4} \right] + 2 \\ & = 3 \left(x - {1 \over 2}\right)^2 - {3 \over 4} + 2 \\ & = 3\left(x - {1 \over 2} \right)^2 + {5 \over 4} \\ \\ \text{Minimum value} & = {5 \over 4}, \text{ when } x = {1 \over 2} \end{align*}

(d)

\begin{align*} f(x) & = 3x + 4 - 2x^2 \\ & = -2x^2 + 3x + 4 \\ & = -2 \left( x^2 - {3 \over 2}x \right) + 4 \\ & = -2 \left[ \left(x - {3 \over 4}\right)^2 - \left(3 \over 4\right)^2 \right] + 4 \\ & = -2 \left[ \left(x - {3 \over 4}\right)^2 - {9 \over 16} \right] + 4 \\ & = -2 \left(x - {3 \over 4} \right)^2 + {9 \over 8} + 4 \\ & = -2 \left(x - {3 \over 4} \right)^2 +{41 \over 8} \\ \\ \text{Maximum value} & = {41 \over 8}, \text{ when } x = {3 \over 4} \end{align*}

\begin{align*} f(x) & = x^2 -2x + 3 \\ & = \left(x - {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 + 3 \\ & = (x - 1)^2 - 1 + 3 \\ & = (x - 1)^2 + 2 \\ \\ \text{For all real } & \text{values of } x, \\ (x - 1)^2 & \ge 0 \\ (x - 1)^2 + 2 & \ge 2 \\ \\ \therefore f(x) \ge 2 \text{ for} & \text{ all real values of } x \end{align*}

\begin{align*} 4x - x^2 - 7 & = -x^2 + 4x - 7 \\ & = -(x^2 - 4x) - 7 \\ & = -\left[ \left(x - {4 \over 2}\right)^2 - \left(4 \over 2\right)^2 \right] - 7 \\ & = -[ (x - 2)^2 - 4] - 7 \\ & = -(x - 2)^2 + 4 - 7 \\ & = -(x - 2)^2 - 3 \\ \\ \text{For all real val} & \text{ues of } x, \\ (x - 2)^2 & \ge 0 \\ -(x - 2)^2 & \le 0 \\ -(x - 2)^2 - 3 & \le -3 \\ \\ \therefore 4x - x^2 - 7 & \text{ is always negative} \end{align*}

\begin{align*} 4x^2 - 12x + 10 & = 4(x^2 - 3x) + 10 \\ & = 4 \left[ \left( x - {3 \over 2} \right)^2 - \left(3 \over 2\right)^2 \right] + 10 \\ & = 4 \left[ \left( x - {3 \over 2} \right)^2 - {9 \over 4} \right] + 10 \\ & = 4 \left(x - {3 \over 2} \right)^2 - 9 + 10 \\ & = 4 \left(x - {3 \over 2} \right)^2 + 1 \\ \\ \text{For all real val} & \text{ues of } x, \\ \left(x - {3 \over 2}\right)^2 & \ge 0 \\ 4\left(x - {3 \over 2}\right)^2 & \ge 0 \\ 4\left(x - {3 \over 2}\right)^2 + 1 & \ge 1 \\ \\ \therefore 4x^2 - 12x + 10 & \text{ is always positive for all real values of } x \end{align*}

By factorisation:
\begin{align*} 3x^2 - 10x - 8 & = 0 \\ (x - 4)(3x + 2) & = 0 \\ \\ x = 4 \phantom{3} \text{ or } \phantom{3} & 3x = -2 \\ & \phantom{3} x = -{2 \over 3} \end{align*}

By completing the square:
\begin{align*} 3x^2 - 10x - 8 & = 0 \\ 3 \left( x^2 - {10 \over 3}x \right) - 8 & = 0 \\ 3 \left[ \left(x - {10 \over 6} \right)^2 - \left(10 \over 6\right)^2 \right] - 8 & = 0 \\ 3 \left[ \left(x - {5 \over 3} \right)^2 - {25 \over 9} \right] - 8 & = 0 \\ 3 \left(x - {5 \over 3} \right)^2 - {25 \over 3} - 8 & = 0 \\ 3 \left(x - {5 \over 3} \right)^2 - {49 \over 3} & = 0 \\ 3 \left(x - {5 \over 3} \right)^2 & = {49 \over 3} \\ \left(x - {5 \over 3} \right)^2 & = {49 \over 9} \\ x - {5 \over 3} & = \pm \sqrt{49 \over 9} \\ x - {5 \over 3} & = \pm {7 \over 3} \\ x & = \pm {7 \over 3} + {5 \over 3} \\ x & = 4 \text{ or } -{2 \over 3} \end{align*}

By quadratic formula:
\begin{align*} 3x^2 - & 10x - 8 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-10) \pm \sqrt{ (-10)^2 - 4(3)(-8)} \over 2(3)} \\ & = {10 \pm \sqrt{196} \over 6} \\ & = {10 \pm 14 \over 6} \\ & = 4 \text{ or } - {2 \over 3} \end{align*}

Preferred method is factorisation, since it is the fastest method.