# 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).

$\begin{array}{l}f\left(x\right)=2{x}^{2}+6x+5\\ =2\left[{x}^{2}+3x+\frac{5}{2}\right]\\ =2\left[{x}^{2}+3x+{\left(3×\frac{1}{2}\right)}^{2}-{\left(3×\frac{1}{2}\right)}^{2}+\frac{5}{2}\right]\end{array}$
$\begin{array}{l}=2\left[{\left(x+\frac{3}{2}\right)}^{2}-\frac{9}{4}+\frac{5}{2}\right]\\ =2\left[{\left(x+\frac{3}{2}\right)}^{2}+\frac{1}{4}\right]\\ =2{\left(x+\frac{3}{2}\right)}^{2}+\frac{1}{2}\end{array}$

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

### 1 thought on “2.12.6 Quadratic Functions, SPM Practice (Paper 1)”

1. Good questions helps me to clear my problem