Functions and graphs  (2.12.4)
Recognise the intuitive concepts of functions, domains and codomains, independent and dependent variables

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Theory

ExamplesIn 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) \(y = \frac{x}{{{{(2x + 1)}^2}}}\)

Solutions(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 \(\underline{\underline {  \frac{1}{2}}} \).
Recognise the notation of functions and use tabular, algebraic and graphical methods to represent functions

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Solutions
Understand the features of the graphs of quadratic functions

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

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Theory

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