Graphing Reciprocal Functionss 1 and 2

Hey guys eejay here, im sorry for not updating the blog.

in last week's class, we learned how to Graph Reciprocal Functions I.

Basic Shape: f(x) = 1/x Shifted Shape: f(x) = (1/ x-h ) + k *read h as opposite, k as is*

There will be an asymptote at the value for x that makes the function undefined - this is called a vertical asymptote. In basic form the VA is always at x = 0

There will also be an asymptote at the value for y that is no longer able to occur due to the unacceptable value for the x (VA) - this is called a horizontal asymptote. Basic form the HA is always at y = 0

If the graph is shifted, we have to read the horizontal shifts ( h values ) as opposite of what is given and vertical shifts ( k values ) as is.

If the graph is shifted, the VA will always be at x = h and the HA will always be at y = k

*ASYMPTOTES HAS TO BE DASHED LINES

Next thing we learned was Graphing Reciprocal Functions II: Reciprocal Trig. Functions.

f(x) = csc x = 1/sin x → there will be undefined values when sin x = 0, assuming the period is 2π, the vertical asymptotes will occur at x = 0, π, 2π... etc

• To graph the basic sin x shape, place the asymptotes at the x - intercepts which will become undefined once reciprocated, and then flip the remaining curves OR create a table of values.

f(x) = sec x = 1/cos x → there will be undefined values when cos x = 0, assuming the period is 2π, vertical asymptotes will occur at x = π/2, 3π/2... etc

• To graph the basic cos x shape, place the asymptotes at the x - intercepts which will become undefined once reciprocated, and the flip the remaining curves OR create a tables of values.

f(x) = cot x = 1/tan x = cos x/ sin x → there will be undefined values when tan x = 0, assuming the period is π, vertical asymptotes will occur at x = 0, π, 2π...etc

• To graph cot x we can create a table of values using quadrantals and π/4's. The asymptotes of cot x will occur at different x - values if the b -values (period) changes.