f(x)= ((x-4)(x+5)/((13x+4)(4-x)) What graphical feature occurs at x=4? (select all that apply) hole intercept vertical asymptote
can you factor out the offending zero?
what do you mean by "offending zero"?
ldaniel... 4 is just an intercept when y=0 alongside 5. However, y=0 behaves a like horizontal aymptote which rule the possibility of it being a vertical asymptote out. Hole? yes f(x) is discontinuous there but this is simply because of the horizontal asymptote feature it shows at x=4 when y=0
so its a hole and a intercept or just a hole?
its an x-intercept
its just a hole, you cant intercept something if there is nothing there to intercept
an offending zero is a value that makes the bottom of a fraction equal zero\[\frac1n\]if n=0, the expression is undefined
so its just a hole, not a intercept and not a vertical asymptote, right?
\[f(x)= \frac{(x-4)(x+5)}{(13x+4)(4-x)}\] \[f(x)= \frac{-\cancel{(4-x)}(x+5)}{(13x+4)\cancel{(4-x)}}\] \[f(x)= \frac{-(x+5)}{13x+4}\] we can "remove" the offending zero; and when something is removed, it creates a hole.
if it cant be removed, it becomes an asymptote
and, if the value of x doesnt create an offending zero, then its just the value of the function at the stated x ... but in this case, x=4 was offending, and could be removed
if it could be removed then f(x) is continuous...so where the hole come in?
the comes in from the fact that the function is undefined at x=4 to begin with; just becasue we can construct an equivalent (but not the same) function has no bearing on the parts that make it up
we are given f(x) ; we can construct an equivalent g(x) to study the behaviour of f(x) with, but g(x) does not excuse the inherent flaws of f(x)
think of g(x) as lipstick on a pig :)
consider it this way; a map gives a representation of a road, but a map is not a road. when a bridge across the road is out, the map still gives us an equivalent view - but no matter what the map looks like, the bridge is still out.
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