2.12.6 Quadratic Functions, SPM Practice (Paper 1)

Question 17:

Find the minimum value of the function f (x) = 2x²+ 6x + 5. State the value of xthat makes f (x) a minimum value.

Solution:

By completing the square for f (x) in the form of f (x) = a(x + p)² + q to find the minimum value of f (x).

\begin{array}{l} f (x) = 2 x^{2} + 6 x + 5 \\ = 2 [x^{2} + 3 x + \frac{5}{2}] \\ = 2 [x^{2} + 3 x + {(3 \times \frac{1}{2})}^{2} - {(3 \times \frac{1}{2})}^{2} + \frac{5}{2}] \end{array}

\begin{array}{l} = 2 [{(x + \frac{3}{2})}^{2} - \frac{9}{4} + \frac{5}{2}] \\ = 2 [{(x + \frac{3}{2})}^{2} + \frac{1}{4}] \\ = 2 {(x + \frac{3}{2})}^{2} + \frac{1}{2} \end{array}

Since a = 2 > 0,

Therefore f (x) has a minimum value when

x = - \frac{3}{2} .

The minimum value of f (x) = ½.

Question 18:

The quadratic function f (x) = –x²+ 4x + k², where k is a constant, has a maximum value of 8.

Find the possible values of k.

Solution:

f (x) = –x²+ 4x + k²

f (x) = –(x²– 4x) + k² ← [completing the square for f (x) in

the form of f (x) = a(x + p)²+ q]

f (x) = –[x²– 4x + (–2)² – (–2)²] + k²

f (x) = –[(x– 2)² – 4] + k²

f (x) = –(x– 2)² + 4 + k²

Given the maximum value is 8.

Therefore, 4 + k²= 8

k²= 4

k= ±2

Question 19:
Find the maximum value of the function 5 – x – 2x² , and the corresponding value of x.

Solution:

\begin{array}{l} 5 - x - 2 x^{2} \\ = - 2 x^{2} - x + 5 \\ = - 2 [x^{2} + \frac{1}{2} x - \frac{5}{2}] \\ = - 2 [x^{2} + \frac{1}{2} x + {(\frac{1}{4})}^{2} - {(\frac{1}{4})}^{2} - \frac{5}{2}] \\ = - 2 [{(x + \frac{1}{4})}^{2} - \frac{1}{16} - \frac{5}{2}] \\ = - 2 [{(x + \frac{1}{4})}^{2} - \frac{41}{16}] \\ = - 2 {(x + \frac{1}{4})}^{2} + 5 \frac{1}{8} \end{array}

\begin{array}{l} 5 - x - 2 x^{2} has a maximum value when \\ 2 {(x + \frac{1}{4})}^{2} = 0 \\ x = - \frac{1}{4} \\ The maximum value of 5 - x - 2 x^{2} is 5 \frac{1}{8} . \end{array}