Horizontal / Vertical Asymptotes - Part 1 of 2

- WE WILL CONTINUE OUR STUDY OF GRAPH SKETCHING. WE'LL BE LOOKING AT ASYMPTOTES AND RATIONAL FUNCTIONS TODAY. OUR GOALS WILL BE TO FIND LIMITS INVOLVING INFINITY, DETERMINE THE ASYMPTOTES OF A FUNCTION'S GRAPH AND ALSO GRAPH RATIONAL FUNCTIONS. A RATIONAL FUNCTION IS A FUNCTION F THAT CAN BE DESCRIBED BY F OF X IS EQUAL TO P OF X DIVIDED BY Q OF X WHERE P OF X AND Q OF X ARE POLYNOMIALS WITH Q OF X NOT THE ZERO POLYNOMIAL. THE DOMAIN OF F CONSISTS OF ALL INPUTS X FOR WHICH Q OF X DOES NOT EQUAL ZERO. HERE'S A BASIC EXAMPLE OF A RATIONAL FUNCTION. NOTICE BOTH THE NUMERATOR AND DENOMINATOR ARE POLYNOMIALS AND MANY TIMES WHEN WE WANT TO GRAPH THESE FUNCTIONS IT IS HELPFUL TO LOOK AT THEM IN FACTORED FORM. LET'S TAKE A LOOK AT SOME OF THE CHARACTERISTICS OF RATIONAL FUNCTIONS. MANY RATIONAL FUNCTIONS HAVE VERTICAL AND HORIZONTAL ASYMPTOTES FOR EXAMPLE THIS GRAPH HAS TWO VERTICAL ASYMPTOTES ONE AT X = -1 ANOTHER AT X = 3. REMEMBER A VERTICAL ASYMPTOTE IS JUST A VERTICAL LINE THAT THE GRAPH APPROACHES BUT NEVER TOUCHES. THIS GRAPH ALSO HAS A HORIZONTAL ASYMPTOTE AT Y = 1. AGAIN, A HORIZONTAL ASYMPTOTE IS A HORIZONTAL LINE THAT A GRAPH APPROACHES IT MAY OR MAY NOT CROSS A HORIZONTAL ASYMPTOTE. THE LINE X = "A" IS A VERTICAL ASYMPTOTE IF ANY OF THE FOLLOWING LIMIT STATEMENTS ARE TRUE. IF THE LIMIT AS X APPROACHES "A" FROM THE LEFT OF F OF X IS EQUAL TO INFINITY OR IF THE SAME LIMIT IS EQUAL TO NEGATIVE INFINITY OR IF THE LIMIT AS X APPROACHES "A" FROM THE RIGHT OR THE POSITIVE SIDE OF F OF X EQUALS INFINITY OR NEGATIVE INFINITY THEN WE HAVE A VERTICAL ASYMPTOTE. SO WHAT THEY'RE TRYING TO TELL US HERE IS FOR EXAMPLE IF WE LOOK AT THE VALUE X = 3 AS X APPROACHES 3 FROM THE RIGHT THE LIMIT WOULD BE POSITIVE INFINITY THEREFORE WE HAVE A VERTICAL ASYMPTOTE OR AS WE APPROACH 3 FROM THE LEFT THAT LIMIT WOULD BE EQUAL TO NEGATIVE INFINITY. AGAIN, VERIFYING THERE'S A VERTICAL ASYMPTOTE THERE. IT IS ALSO IS IMPORTANT TO NOTE THAT FROM YOUR COLLEGE ALGEBRA DAYS VERTICAL ASYMPTOTES ALSO OCCUR AT THE ZEROS OF THE DENOMINATOR. THEY ARE NOT COMMON FACTORS WITH THE NUMERATOR. SO FOR EXAMPLE, IF YOU TAKE A LOOK AT THIS DENOMINATOR DOWN HERE THE X = -1 AND X = 3 ARE ZEROS OF THE DENOMINATOR AND THOSE ARE ALSO WHERE WE FIND THE VERTICAL ASYMPTOTES. THE LINE Y = B IS A HORIZONTAL ASYMPTOTE IF EITHER OF THE FOLLOWING LIMIT STATEMENTS ARE TRUE. THE LIMIT AS X APPROACHES POSITIVE INFINITY OF F OF X IS EQUAL TO B OR THE LIMIT AS X APPROACHES NEGATIVE INFINITY OF F OF X IS EQUAL TO B. [ AUDIO SKIP ] ON THIS PROBLEM AS X APPROACHES POSITIVE INFINITY NOTICE WE WOULD MOVING TOWARD THE RIGHT THE Y VALUE IS EQUAL 1. SO Y = 1 IS A HORIZONTAL ASYMPTOTE. THE SAME IS TRUE AS WE APPROACH X EQUALS NEGATIVE INFINITY ON THE GRAPH WE APPROACH Y = 1. NOW IN GENERAL TO FIND LIMITS AT INFINITY IT MAY BE HELPFUL TO DIVIDE ALL OVER THE TERMS BY THE HIGHEST POWER OF X IN THE DENOMINATOR AND THEN FIND THAT LIMIT. THE GRAPH OF A RATIONAL FUNCTION MAY OR MAY NOT CROSS A HORIZONTAL ASYMPTOTE. HORIZONTAL ASYMPTOTES OCCUR WHEN THE DEGREE OF THE NUMERATOR IS LESS THAN OR EQUAL TO THE DEGREE OF THE DENOMINATOR. REMEMBER THE DEGREE OF A POLYNOMIAL IN ONE VARIABLE IS THE HIGHEST POWER OF THAT VARIABLE. HERE ARE SOME SHORTCUTS OF FINDING LIMITS AT INFINITY. IF THE DEGREE OF THE NUMERATOR IS HIGHER THAN THE DEGREE OF THE DENOMINATOR, THERE IS NO HORIZONTAL ASYMPTOTE. IF THE DEGREE OF THE NUMERATOR AND DENOMINATOR ARE EQUAL, THE HORIZONTAL ASYMPTOTE IS THE RATIO OF THE LEADING COEFFICIENTS. THIS SHOULD READ RATIO. IF THE DEGREE OF THE DENOMINATOR IS HIGHER THAN THE NUMERATOR THE HORIZONTAL ASYMPTOTE WILL BE Y = 0. OKAY. LET'S GO AHEAD AND OUT THIS INTO PRACTICE. LET'S FIND THE HORIZONTAL AND VERTICAL ASYMPTOTES AND THEN SKETCH THE GRAPH. OKAY, SO TO FIND THE HORIZONTAL ASYMPTOTE WE NEED TO FIND THE LIMIT AS X APPROACHES EITHER INFINITY OR NEGATIVE INFINITY. I'M GOING TO CHOOSE POSITIVE INFINITY OF 2X - 5/X - 3. NOW IF YOU TAKE A LOOK AT THE SHORTCUTS WE CAN VERIFY THIS LIMIT BECAUSE THE DEGREE OF THE TOP AND BOTTOM ARE THE SAME. THE LIMIT IS EQUAL TO THE RATIO OF THE LEADING COEFFICIENTS OR 2/1. THIS LIMIT IS EQUAL TO 2. OR IF WE WANT TO VERIFY BY DIVIDING BY THE HIGHEST POWER OF X IN THE DENOMINATOR WE WOULD DIVIDE EVERY TERM BY X. LET'S JUST SHOW THAT. SO NOTICE HOW I'M DIVIDING EVERY TERM BY X AND NOW I'M GOING TO SIMPLIFY. NOTICE HOW HERE WE HAVE AN X/X AND HERE WE HAVE X/X. NOW THE LIMIT AS X APPROACHES INFINITY OF THIS IS GOING TO APPROACH ZERO. WE HAVE A FIXED NUMERATOR AND AN INCREASING DENOMINATOR AND THE SAME THING WITH THIS. THIS APPROACHES ZERO THEREFORE OUR LIMIT WOULD JUST BE EQUAL TO 2/1 OR 2 WHICH WE'VE ALREADY DISCOVERED. LET'S GO AHEAD AND SKETCH THAT IN. Y = 2. LET'S GO AHEAD AND FIND THE VERTICAL ASYMPTOTE. NOW AGAIN, REMEMBERING BACK TO YOU COLLEGE ALGEBRA DAYS, SINCE 3 IS A ZERO DENOMINATOR WE ARE GOING TO HAVE A VERTICAL ASYMPTOTE THERE. LET'S GO BACK AND LABEL THIS FOR A HORIZONTAL ASYMPTOTE AND NOW OUR WORK FOR OUR VERTICAL ASYMPTOTE. WE KNOW THERE'S GOING TO BE A VERTICAL ASYMPTOTE AT X = 3 BUT TO VERIFY IT USING LIMITS WE REALLY HAVE TO FIND THE LIMIT AS X APPROACHES 3 TO EITHER THE RIGHT OR THE LEFT. I'M GO AHEAD AND FIND FROM THE POSITIVE SIDE OR THE RIGHT SIDE AND IF I DID THAT THIS LIMIT DOES EQUAL INFINITY. WE'RE VERIFY THAT WITH THE GRAPH IN JUST A MOMENT BUT LET'S GO AHEAD AND SKETCH THE VERTICAL ASYMPTOTE IN AT X = 3. OKAY. LET'S USE SOME TECHNOLOGY HERE TO HELP US. WE'RE GOING TO GO AHEAD AND LOOK AT THE GRAPH OF THIS ON OUR GRAPHING CALCULATOR. LET'S GO AHEAD AND HIT Y =. WE DO NEED TO PUT OUR NUMERATOR AND DENOMINATOR IN A SET OF PARENTHESES. HIT GRAPH. NOW I WANT TO TAKE A MOMENT AND VERIFY THIS LIMIT OVER HERE THE LIMIT AS X APPROACHES 3 FROM THE RIGHT. IF WE WERE ON THE GRAPH AND WE APPROACHED 3 FROM THE RIGHT SIDE NOTICE HOW THE GRAPH GOES UP. IT IS APPROACHING POSITIVE INFINITY. I COULD ALSO VERIFY THIS AS X APPROACHES 3 FROM THE LEFT AND WE CAN SEE FROM THE GRAPH THAT WOULD EQUAL NEGATIVE INFINITY. NOW TO MAKE A NICE ACCURATE GRAPH OF THIS FUNCTION WE DO NEED TO FIND SOME ADDITIONAL POINTS. THE WAY WE COULD DO THIS IS BY HITTING 2nd GRAPH AND FINDING SOME CONVENIENT POINTS TO PLOT. FOR EXAMPLE, WE COULD PLOT THE POINT (2, 1) (0, 1.667) AND SO ON. FIND ENOUGH POINTS TO MAKE A NICE ACCURATE GRAPH. IF WE DID THAT, THE GRAPH WOULD LOOK SOMETHING LIKE THIS. AGAIN NOTICE THIS IS ENLARGED BUT WE DO HAVE OUR VERTICAL ASYMPTOTE HERE AT X = 3, HORIZONTAL AT Y = 2.