Given the ration function f (x) = x+1/x2+3x-10 state the domain give all asymptotes and holes and sketch the graph. now I know how to factor it out but how do I determine which is the Vertical Asymptote and which is the Horizontal Asymptote and which are holes. and How do I graph it on my calculator? Need help big quiz tomorrow and my teacher sux at teaching
Here are the rules you need to know:Originally Posted by thebest
Vertical asymptotes:
vertical asymptotes occur at values of x for which our rational function is undefined, that is, any x that is not in our domain, will form a vertical line that the graph can't attain.
Horizontal asymptotes:
- if the highest power in the numerator is equal to the highest power in the denominator, we have a horizontal asymptote. this asymptote will have the y-value obtained by the ratio of the leading coeffcients. that is, we take the coefficient of the highest power of x in the numerator and divide it by the coefficient of the highest power in the denominator.
- if the highest power in the numerator is greater than the highest power in the denominator, then there is no horizontal asymptote (we would get what is called a slant asymptote, but you never asked about that, so forget about it for the moment).
- if the highest power in the numerator is less than the highest power in the denominator, then we get a horizontal asymptote at y = 0
Holes:
these occur if the numerator and denominator have the same zeroes, in which they can cancel and remove a vertical asymptote
so now let's find the asymptotes and such of your graph. i will then post the graph below. if you need help sketching the graph, say so.
f(x) = (x+1)/(x^2+3x-10)
begin by factorizing the bottom by foiling
=> f(x) = (x + 1)/[(x + 5)(x - 2)]
now what is the domain of f(x)?
the domain is the set of x-values for which the function is defined, so it will be all x's EXCEPT the ones that make the denominator zero. so set the denominator to zero to find these bad x's
(x + 5)(x - 2) = 0
=> x = -5, x = 2
so the domain is all real x except -5 and 2, that is the set {x : x in the set of real numbers and x not= -5 and x not= 2} or in interval notation.
dom(f) = (-infinity, -5)U(-5,2)U(2,infinity)
now since these are x values we can't attain, they also form the lines for the vertical asymptotes. so the vertical asymptoes are:
x = -5, and x = 2
the degree of the numerator is smaller than the degree of the denominator, so the horizontal asymptote is y = 0
the x-intercepts, is where y = 0
=> for x-intercept:
0 = (x + 1)/(x^2+3x-10)
since the bottom can't be zero, we get zero if:
x + 1 = 0
=> x = -1
so (-1,0) is our x-intercept
the y-intercept occurs when x = 0
so for y-intercept:
y = (0 + 1)/(0 + 0 - 10) = -1/10
now we have almost everything to draw the graph, but i assume you can take it from here (tell me if you can't).
i have attached the graph below. the vertical lines you see are the vertical asymptotes
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