Question

What are the poins of discontinuity? Are they all removable?
y=(x+4)/x^2+5x+4

Answer 1

for clarity it's important that you enclose x^2+5x+4 inside parentheses. this would indicate that you're dividing x+4 by the quadratic (x^2+5x+4).

Division by zero is not defined. So, one of the first things you must do here is to determine whether x^2+5x+4 is ever zero.

I would also recommend graphing the given function. Once you have that graph, determine visually whether there's any way in which you could remove one or both of the discontinuities in this graph. Please share your work. Try the Draw utility for this presentation.

Answer 2

is there an easy way out?

Answer 3

the easy way is finding the zeros of the denominator as @mathmale suggested.

Answer 4

would the zeros be -4 and -1?

Answer 5

correct

Answer 6

is 0 one also ?

Answer 7

what happens if you plug in 0?

Answer 8

oh wait never mind

Answer 9

ok now if something is a zero of the top and the bottom, it is a removable discontinuity.

Answer 10

so is -4 the removable?

Answer 11

correct

Answer 12

so would that be the final answer?

Answer 13

???

Answer 14

can you help @zzr0ck3r

Answer 15

Yes if your graph is not defined at the point x = a but the limit at that pt, a exists and is finite, we say it is a removable discontinuity. As mentioned, a limit of the form 0/0 will generally reduce to a finite limit vs k/0 (there is no defined limit, as f(x) --> +/- infinity).

You could use L'Hopital's Rule to find the 'value' of the discontinuity at the point x = -4. by evaluating the limit at x = -4. However the question just wants to you to state where discontinuities exist and are removable.

Answer 16

sorry, I don't know what you are talking about

Answer 17

Go back to what @mathmale has suggested, plot the graph!

You should find that the graph looks like this, with a single-point discontinuity (represented by the small circle). Use what you already have to complete the graph and make your conclusions.