F(x) = (x^2+x-2)/(x^2-3x-4) domain and range x- and y-intercepts horizontal and vertical asymptotes

The domain and range are from negative infinity to positive infinity in both cases because these are polynomials.  They are in rational form, which causes other problems, such as vertical asymptotes and the like, but domain and range are easy.  The x-intercept is found when y = 0 in the numerator of the function.  Think of it like this.  When we have a fraction that is 0/4, the value of that fraction is 0, no matter what the denominator is, right?  so if the equation was y=(0/4)x, then y = 0.  Same idea here.  This function equals 0 when the numerator is 0.  We set the numerator equal to 0 and solve for x by factoring.  .  Factor that to get x-intercepts of 1 and -2.  Now on to the y intercept.  The y intercept exists when x is set to 0.  Doing that we end up with simply -2/-4 which is 1/2.  The vertical asympotes are where the denominator of the function is undefined, or set to equal 0.  So we set the denominator of the function equal to 0 and factor to solve for x.  .  Factoring that we get vertical asymptotes at x = 4 and x = -1.  Lastly, is the horizontal asymptotes.  There is a set of rules for this as far as a rational function goes.  If the degree of the numerator is HIGHER than the degree of the denominator, then the asymptote is oblique.  If the degree of the numerator is EQUAL TO the degree of the denominator, then the asymptote is equal to the leading coefficients in fraction form (for lack of a better way to word it).  And if the degree of the numerator is lower than the degree of the denominator, the asymptote is always 0.  Our degrees are equal, so our leading coefficients, both 1, in fraction form is 1/1 which is just 1.  So that's all your info.  Read this carefully to understand the differences between all these things that happen in rational functions.

128/3reason being 5 and 1/3 = 16/316/3 • 2/3 = 32/31 and 1/3 = 4/3 32/3 • 4/3 this will equal 128/3

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