# Thread: functions

1. ## functions

1) The function f(x) has an x-intercept at "a" and a y-intercept at "b". Where are the x and y intercepts of f(1/2x).

My solution:
y = mx + b => 0 = mx + b => -b/m = x
y = mx +b => y = m(0) + b => y = b

Is it correct?

2) Simply using identities. Should contain only one trigonometric function and should not contain any fraction

a) (csc(x) cos^2(x)) / (1 + csc(x))
b) (1/(sec(x) + 1)) - (1/(sec(x) - 1))

2. Originally Posted by mwok
1) The function f(x) has an x-intercept at "a" and a y-intercept at "b". Where are the x and y intercepts of f(1/2x).

My solution:
y = mx + b => 0 = mx + b => -b/m = x
y = mx +b => y = m(0) + b => y = b

Is it correct?
[snip]
I assume f(1/2x) means f((1/2)x) = f(x/2) and not f(1/(2x)).

Why do you have lines?? You don't have to be a line to have an x-intercept and a y-intercept!

x-intercept has coordinates (a, 0).
y-intercept has coordinates (0, b).

If f(x) ---> f(x/2) then the genreal point (A, B) on y = f(x) transforms to (2A, B).

Therefore .....