finding-slant-asymptotes-precalculus-vertical-horizontal-end-behavior-asymptote

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Good day students in this clip we are going to be going over two examples on how to find the slant asymptotes of rational funtions ok so let us go ahead and write down the instructions for the examples the task is to find the slant asymptote of the following functions so for number 1 we have the rational function given by y=x^3+1 over x^2+x now if you take a look at this rational function what do you notice about the degree of the numerator and the denominator what you notice is that the degree of the numerator is one more than the degree of the denominator so anytime you have a rational function like this if the degree of the numerator is one higher than the degree of the denominator you will always have a slant asymptote if the degrees are thesame as you know we will have a constant which will be the horizontal asymptote just by dividing the coefficients of the leading terms with equal degrees then if the denominator has a bigger degree then you know that you have a horizontal asymptote of y=0 okay so either you have a horizontal asymptote, slant or other types of asymptotes alright so in this case the degree of the numerator is one greater than the denominator so we are going to have a slant asymptote here so in order to find the slant asymptote all you do is you are going to do the long division you divide the numerator by the denominator and the polynomial portion of your resulting answer is going to be your slant asymptote alright so let us go ahead and set up our long division bars we are going to divide these two polynomials alright so the top goes under the division bar so we are going to write it as x^3 there is no second degree term so we are going to use the place holder 0x^2 there is no first degree term so we use a place holder of 0x plus the constant 1 alright so that is going to be divided by x^2 plus x alright so we always look at the first or leading term of the dividend and the divisor so we have x^2 x^2 goes into x^3 x times so this is the x column right here make sure they line up so you have an x x times squared is x to the third and then x times x is x^2 so this is your standard long division algorithm that I am carrying out here now what you do is you just subtract so we are going to subtract this . Alright so if I subtract that this becomes negative x ^2 and then we can bring that down plus zero x and then we repeat thesame procedure again now x^2 goes into x^2 -1 times x^2 will give you -x^2 and then -1 times x will give you -x. and then you are going to subtract again alright so let us go ahead and subtract put a minus there so -x^2 - -x^2 cancels out and then 0x--x will just give you x plus one alright so the result after we divide is going to be x minus one plus the remainder which is x+1 divided by x^2+x okay put the remainder over the divisor ok now what is your slant asymptote it is the resulting polynomial this linear function right here this is going to be your slant asymptote ok so lets write down what the result is the slant asymptote is given by y=x-1 alright so there you have it alright so this is the graph of the function the rational function that we just worked on finding the slant asymptote of x^3+x/x^2+x this is what it looks like and this is the graph of the slant asymptote that we just found y=x-1 the rise over run and the y intercept is -1 so you can see how the graph of the slant asymptote affects the shape of the curve alright so there you have it. alright let us take a look at another example. question number 2 the task is also to find the slant asymptote of the rational function y=(2x^2+3x+5)/(x+1). do you see why we have a slant asymptote in this situation? absolutely because you notice the degree of the numerator is one more than the degree of the denominator and whenever you have that situation you always have a slant asymptote ok alright so in order for us to find what a slant asymptote is we are going to use long division just as we did in the previous

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