I need to find the domain and range, x and y intercepts, horizontal and vertical asymptotes. Can someone help me through it please?
Hi what is your function?
There should be a picture attached but it's: X^2 + x - 2 divided by x^2 - 3x - 4
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?
I did, the numerator factors into (x-1)(x+2) and the deniminator factors into (x-4)(x+1)
very good, so what will be the domain considering the denom?
The domain would be all reals except x=4 and x=-1?
quite right
let's work on range of the function
alright
Is the range all reals?
thats right, do you have a reasoning for that?
Well by the graph it extends infinitely up and down
and because quadratic functions' range will always be all reals
that's right, but what if we don't know about the graph
note that quadratics don't cover all reals|dw:1434182772266:dw|
Oh okay, then I do not a reasoning
do not have a reasoning*
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
does that make sense?
So when you plugged in the ordered pair, the output was infinite basically?
thats right
okay that makes sense
for example\[\lim_{x \to -1^{+}} f(x)=\frac{1 \times -2}{-5 \times 0^{+}}=\infty\]
ok what are x and y intercepts?
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
very right, thanks
and what about asymptotes
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.
right, you can put your words in mathematical phrases
now you can graph your function
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 :/
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