### Functions and graphs - (2.1-2.4)

#### Recognise the intuitive concepts of functions, domains and co-domains, independent and dependent variables

• In each of the following, y is a function of x. Find the largest domain of real numbers for each function.
(a) $$y = {(x - 2)^2}$$
(b) $$y = \frac{x}{{{{(2x + 1)}^2}}}$$
• (a) Consider $$y = {(x - 2)^2}$$.
For any real number x, the corresponding value of y is real according to the expression.
The largest domain of real numbers for the function is the set of all real numbers.

(b) Consider $$y = \frac{x}{{{{(2x + 1)}^2}}}$$.
For any real number x except $$x = - \frac{1}{2}$$, the corresponding value of y is real according to the expression.
The largest domain of real numbers for the function is the set of all real numbers except $$\underline{\underline { - \frac{1}{2}}}$$.

#### Find the maximum and minimum values of quadratic functions by the algebraic method

• Find the maximum / minimum value of each of the following functions.
(a) $$y = - 2{x^2} + 7x - 6$$
(b) $$y = 3{x^2} - 8x + 5$$
• $$\begin{array}{c}a)\qquad y = - 2{x^2} + 7x - 6\\ = - 2({x^2} - \frac{7}{2}x + 3)\\ = - 2{[{x^2} - \frac{7}{2}x + {(\frac{7}{4})^2} - {(\frac{7}{4})^2} + 3]^{^{^{}}}}\\ = - 2{[{(x - \frac{7}{4})^2} - \frac{{49}}{{16}} + 3]^{^{^{}}}}\\ = - 2{[{(x - \frac{7}{4})^2} - \frac{1}{{16}}]^{^{^{}}}}\\ = - 2{(x - \frac{7}{4})^2} + \frac{1}{8}\end{array}$$

For any real number x,
$$- {\rm{ }}2{(x - \frac{7}{4})^2} \le 0$$
$$y = - 2{(x - \frac{7}{4})^2} + \frac{1}{8} \le \frac{1}{8}$$
Therefore, the maximum value of y is $$\frac{1}{8}$$.

$$\begin{array}{c}b)\qquad y = 3{x^2} - 8x + 5\\ = 3({x^2} - \frac{8}{3}x + \frac{5}{3})\\ = 3{[{x^2} - \frac{8}{3}x + {(\frac{4}{3})^2} - {(\frac{4}{3})^2} + \frac{5}{3}]^{^{^{}}}}\\ = 3{[{(x - \frac{4}{3})^2} - \frac{{16}}{9} + \frac{5}{3}]^{^{^{}}}}\\ = 3{[{(x - \frac{4}{3})^2} - \frac{1}{9}]^{^{^{}}}}\\ = 3{(x - \frac{4}{3})^2} - \frac{1}{3}\end{array}$$

For any real number x,
$$3{(x - \frac{4}{3})^2} \ge 0$$
$$y = 3{(x - \frac{4}{3})^2} - \frac{1}{3} \ge - \frac{1}{3}$$
Therefore, the minimum value of y is $$- \frac{1}{3}$$.