I need to find the domain and range, x and y intercepts, horizontal and vertical asymptotes. Can someone help me through it please?

- unicwaan

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- anonymous

Hi
what is your function?

- unicwaan

There should be a picture attached but it's:
X^2 + x - 2 divided by x^2 - 3x - 4

- anonymous

first thing, you should factor the num and denom of your function\[f(x)=\frac{x^2+x-2}{x^2-3x-4}\]have you tried to do so?

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- unicwaan

I did, the numerator factors into (x-1)(x+2) and the deniminator factors into (x-4)(x+1)

- anonymous

very good, so what will be the domain considering the denom?

- unicwaan

The domain would be all reals except x=4 and x=-1?

- anonymous

quite right

- anonymous

let's work on range of the function

- unicwaan

alright

- unicwaan

Is the range all reals?

- anonymous

thats right, do you have a reasoning for that?

- unicwaan

Well by the graph it extends infinitely up and down

- unicwaan

and because quadratic functions' range will always be all reals

- anonymous

that's right, but what if we don't know about the graph

- anonymous

note that quadratics don't cover all reals|dw:1434182772266:dw|

- unicwaan

Oh okay, then I do not a reasoning

- unicwaan

do not have a reasoning*

- anonymous

function is continuous on the interval (-1,4) and\[\lim_{x \to -1^{+}} f(x)=\infty \]\[\lim_{x \to 4^{-}} f(x)=-\infty \]therefore range is all of real numbers

- anonymous

does that make sense?

- unicwaan

So when you plugged in the ordered pair, the output was infinite basically?

- anonymous

thats right

- unicwaan

okay that makes sense

- anonymous

for example\[\lim_{x \to -1^{+}} f(x)=\frac{1 \times -2}{-5 \times 0^{+}}=\infty\]

- anonymous

ok what are x and y intercepts?

- unicwaan

Alright so the y intercept = 1/2 because you plug in 0 for x to find y. The x intercept = -2,1 because they are the zeros of the numerator

- anonymous

very right, thanks

- anonymous

and what about asymptotes

- unicwaan

The horizontal asymptotes is 1 because the degree of each function are the same and therefore, must divide the leading coeeficients. The vertical asymptotes are y= -1,4 because they are the zeros of the denominator.

- anonymous

right, you can put your words in mathematical phrases

- anonymous

now you can graph your function

- unicwaan

Okay so for the asympotes I put dotted lines to represent them, and then I plugged in points according to get this graph|dw:1434183896630:dw| Poor drawing but the basic jist of it :/

- anonymous

we're done

- unicwaan

Great! Thank you so much this really helped! :D

- anonymous

no problem

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