Hey
How do you do Q21 of 2015 Exam 2 VCAA? Stuck btwn C and D (I got to m^2+4ac<0)
And for Q22, how is g(x)=tan(x) as the examiners report says? It's not symmetrical?
For 21 you simply have to solve the equation you found (m
2+4ac<0)
Which corresponds to D
22. They just chose functions that look similar to the displayed ones. Also I believe |x| (modulus graphs) aren't in the methods study design anymore.
For -f(x), -f(x)=-f(-x) so the graph is symmetrical (even function) and is non-negative for all values of x. So for g(-f(x)), its also symmetrical, as -f(x) is even / symmetrical. And since -f(x) is positive and a straight line starting from 0, you can essentially just take the right hand side of the graph g(x) ie where x>0 and make it symmetrical.
Since -f(x) is just like x for positive x values, g(-f(x)) is just like the right hand side for the graph, and since -f(-x)=-f(x), it's symmetrical.
Sorry for the long and confusing explanation, but hope this makes a bit of sense