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set x=0 to find your y-intercept. set y=0 to find your x-intercept.
look at this funtion f(x)=2(x+1)/x^2 if i do dat,my y intercept turns out to be infinity
am confuse,am doing calculus,trying to plot a graph
anytime you want to find x and y intercepts that's what you would do. but you also have to consider the domain of your function first. notice that the function you listed is a rational function. these type of functions need special attention.
0 doesn't belong to the domain of this function so it doesn't have a y intercept.
my domain is 0
you know dividing by zero is not allowed. for example 5/0 = ? so in your rational function, you must not let the denominator be zero. so x cannot be zero in the first place.
no, your domain is any number except 0.
like you said you were trying to graph.. you got values f(1), f(2), f(2), f(-1), f(-2)... but what happened when you tried f(0)?
yeah sorry lol,(x not =0) u are right so took my limits in the negative and positive.my vertical asymptotes ended up as positive and my horizontal asympto =0
f(0) gave me =2/0
right, 2/0 has no meaning. so you can't graph the point (0, f(0)).
so what happens now am doing graph,of application of Differentiation
whatever x value makes your denominator zero will be your vertical asymptote (MOST of the time)
try put in values close to x=0 like .5, then .1, then .01, .001. do you get y-values when you stick in those numbers?
mine is positive infinity in both left and right hand side
yes, so your graph should look like this:|dw:1332540936986:dw|
so back your question... does the graph have a y-intercept in this case?
no it doesnt
right, not all graphs will have intercepts x or y.