All of Rational Functions in Under 1 Hour ultimate study guide jensenmath ca

rational function is function that can be expressed as the ratio of two polinomial functions so for example if the function ofx is rational function it would be the quotient of two polinomial functions let's say function ofx / function ofx and remember division by zero is always undefined so rational functions often have restrictions the function in the denominator ofx can't be zero and when looking at the graphs of different types of rational functions you'll notice that there's lots of different shapes they can take and you'll notice in these graphs they often have different types of discontinuities such as vertical asmp tootes or holes in the graph and those discontinuities happen at any values that make the denominator of the rational function zero so let me take you through the top 10 things you need to know about rational functions so that you fully understand the properties and graphs of all different types of rational functions must know number one simplifying and stating restrictions before we can start graphing rational functions you have to be confident working with them algebraically and being able to simplify them and state any restrictions on the variable and when we do this simplifying process we're going to follow Two Steps step one is to factor the Expressions that are in the numerator and denominator and then cancel any common factors and then step two is to State any restrictions on the variable and restriction on the variable would be any value of the variable that makes the rational function undefined which would be any value that makes the denominator of the rational function be zero so let's look at two different rational functions we'll simplify them and state any restrictions on the variables for both of these rational functions let's go ahead and try and simplify them and state any restrictions on the variable throughout the simplification process you should always be looking if you can find any restrictions on the VAR variable restriction would be value of that makes the function undefined and the function would be undefined if the denominator is zero well since the function in the denominator is quadratic and standard form it's tough to see what value of would make it be zero but if we Factor it first that'll make the restrictions more obvious so let's go ahead and start this example one by factoring that quadratic that's in the denominator I'll leave the numerator the same and then this quadratic that's in the denominator to factor it I'll be looking for numbers that have product of8 and sum of three the numbers that satisfy that product and sum are 6 and -3 so know that quadratic would factor to + 6 * - 3 and now that the denominator is in factored form it's easy to see what the restrictions on the variable would be the rational function is undefined if the denominator is zero well the denominator would be zero if either of these two factors factors were zero this factor of + 6 would be0 if was -6 so one of the restrictions on this rational function is that cannot be -6 and then this second factor in the denominator also can't be zero because that would make the whole denominator be zero and what would make - 3 be zero is if was 3 so can also not equal 3 if was either of those two values it would make the denominator be zero and Division by 0 is undefined but now that I've stated restrictions and factored this rational function as much as possible can look for any common factors that cancel out well notice have factor of - 3 in the numerator and in the denominator anything divided by itself is 1 so can cancel those factors out - 3 / - 3 is 1 which means this rational function simplifies to 1/ + 6 and then have to make sure State those restrictions on the variable again still can't equal -6 or three and stating these restrictions is not only important so that we know when this function is undefined but it's also important that we're actually able to state that this function is equal to this function because if you were to sub three into this function you would get 0 over 0 which is undefined but if you sub sub it into this function you get 1 over9 but since we've said cannot be equal to 3 subbing in any other value other than three into these two functions you get the exact same value so having this restriction here allows us to state that it's equivalent and as we go through this video we'll talk about graphically how these discontinuities show up on the graph of rational function but we'll get there so let's practice one more let's do example two where we simplify that and state its restrictions as well once again the first thing we should do would be to factor the functions that are in the numerator and denominator that way we'll be able to see if any common factors cancel and also we'll be able to see any restrictions on the variable well in the numerator there's difference of squares it's an x^2 minus 3^ 2 so that would factor to - 3 * + 3 and in the denominator there is quadratic trinomial so I'll find numbers that have product of 12 and sum of the numbers that satisfy that product and sum are 4 and 3 which means the quadratic in the denominator factors to + 4 multiplied + 3 and now the denominator is in factored form can easily see the restrictions on the variable what would make that denominator be zero well the denominator would be zero if either of these factors were zero and + 4 would be zero if was -4 so that's one restriction on the variable and + 3 would be if was -3 so that's another restriction on the variable and now should check if any common factors cancel out notice both in the numerator and denominator there are factors of + 3 that can cancel out because + 3 / + 3 is 1 which means as long as does not equal -4 -3 this rational function would be equal to the version where those factors of + 3 are canceled it's equal to just - 3 + 4 and then make sure to State the restrictions on the variable again cannot be -4 or -3 must know number two operations with rational expressions let's start with how to multiply rational expressions anytime there's an operation between rational expressions the goal is to find simplified answer and also State the restrictions on the variable so the first thing you want to do that's going to help with the simplification and with the stating of restrictions is to check and see if anything is factorable in either rational expression notice this quadratic trinomial can be factored it could be factored to + 3 * + 1 and now that the denominators are in factored form it would be easy to see the restrictions on the variable since division by 0 is undefined this product would be undefined if any of the factors in the denominators were zero well 2x would be zero if was 0 + 3 would be 0 if was -3 and + 1 would be 0 if was -1 so those are the restrictions on the variable and now we can work on further simplifying this product when multiplying rational expressions we are allowed to cross reduce factors meaning we could cancel factor that's in numerator of one rational expression with factor that's in the denominator of the other rational expression or we could combine these fractions first and then cancel factors think I'll do that because when multiplying fractions it's equal to the product of the numerators divid the product of the denominators and now I'll do my canceling of factors these factors of + 1 can cancel since something divided by itself is one we can cancel those out and also there's factor of in both the numerator and denominator so those can be canceled as well and on each line of my simplification should keep writing the restrictions and now can write my final simp simpied answer notice in the numerator I'm just left with 3 and in the denominator I'm left with 2 * + 3 and then it's important the restrictions are written again and now let me make some room and let's do some other operations next up is dividing rational expressions the rule for dividing fractions says to keep the first fraction exactly the same change the division operation to multiplication and flip the second fraction write its reciprocal and and now this is just like multiplication question before actually multiplying these rational expressions we should check if anything is factorable can common factor 2x^2 - 8X can common factor out to 2x this quadratic can factor to - 5 * + 2 and this quadratic can factor to - 5 * - 4 and now the next thing want to do is before canceling any factors want to State the restrictions on the variable so let me make bit of room so can write those restrictions the restrictions are any value of that would make the original expression undefined well the original expression would be undefined if either of the rational expressions were undefined and that would happen if either of the denominators were zero but also the original division question would be undefined if the value of this fraction was Zero because you can't divide by zero and that fraction would be zero if the 4x^2 was zero so to check for restrictions for division question you're going to have to check in these three spots meaning on the line below I'll be checking for restrictions in all three of these spots so any value of that makes any of these factors be zero is restriction on the variable which means cannot equal 5 -2 0 5 again or 4 and now that the restrictions are stated can start canceling factors any Factor that's in numerator in either of the fractions can be canceled if the same factor is in the denominator of either of the fractions so these factors of - 5 can be canel this factor of can be canceled with one of those two factors of and this 2 over 4 can be reduced to half and now can multiply the fractions have - 4 * another - 4 which could just write as - 4^ 2 and that's all divided by + 2 * 2 and then once again I'll have to restate those restrictions on the variable and now I'll make some rooms so we can do an example where we add or subtract rational expressions in this example we have two rational expressions being added when adding rational expressions you need common denominator if we're going to combine and simplify them but before trying to get common denominator good first step is to check if anything is factorable notice have this quadratic trinomial it would factor to - 8 * -1 now that the denominators are in factored form it's easy to see the restrictions on the variable can't be 8 or 1 or 8 again if was either 8 or one it would make the sum of these rational expressions undefined because division by 0 is undefined so I'll State those restrictions and now I'll go ahead and try and get common denominator notied this denominator is product of - 8 and -1 while this denominator just has factor of - 8 so if multiplied this fraction top and bottom by - one would then have common denominator and now that these denominators are the exact same can combine these into single fraction by keeping that common denominator and adding their numerators and then all that's left to do is simplify the numerator I'll distribute the two into the -1 and if collect my like terms + 2x is 3x and there's the final simplified answer with the restrictions must know number three vertical asmp tootes now now that we know how to simplify and state restrictions on rational expressions we can start talking about the properties of their graphs the first main property that lot of rational expressions have are vertical ASM tootes if function let's call it ofx has vertical ASM toote at xals three things would be true number one the value of the function at so at would equal some nonzero number divid zero which would make the original function undefined at that value of and number two the limit of the function as the value approaches would be either positive or negative Infinity as function approaches vertical ASM toote it always goes towards infinity or negative infinity and that's because number three the graph will always just approach the vertical ASM toote but it'll never touch it or cross it so let me make some room and let's look at an example of function that has two two different vertical ASM tootes looking at this rational function let's try and figure out where the vertical ASM tootes are and then also I'll show you its graph and we'll look at its one-sided limits to know where its vertical ASM tootes are what I'll start by doing is looking at the restrictions of this function so I'll try and find what values of make this denominator zero and would be able to see those more easily if put it into factored form the numbers that multiply to8 and add to -2 are -4 and 2 so can factor that quadratic in the denominator to - 4 * + 2 and now can more easily see what values of make this function undefined it'll be undefined if either of the factors in the denominator were zero so it'll be undefined if is either -2 or 4 now restrictions on the variable do always result in discontinuity of the graph but it's not always vertical ASM toote to verify that there are vertical ASM tootes at = -2 and = 4 I'll show you that when subbing those values into our function we get nonzero number over zero at -2 is 1/0 and at 4 is also 1/0 since both of those values result in nonzero number being divided by zero know that there's going to be vertical ASM toote at those values so there are vertical ASM tootes at both = -2 and = 4 and let me show you what the graph of this function roughly looks like notice this function is divided into three branches and that's because of the two vertical ASM tootes that the function has also notice the function has horizontal ASM toote at equals 0 but I'll explain to you about horizontal ASM tootes in the next section what want you to notice before moving on is that the function never touches or crosses these two vertical ASM tootes as it approaches those vertical asmt the function either goes up towards infinity or down towards negative Infinity so if were to record the one-sided limits of this function as it approaches the vertical ASM tootes the limit of the function as approaches -2 from the left well from the left of -2 as the function approaches -2 it goes up towards Infinity but as the function approaches -2 from the right of -2 it goes down towards negative infinity and then as it approaches this vertical ASM toote of four from the left of four it goes down towards negative Infinity but as approaches 4 from the right of four the function goes up towards positive Infinity notice the limits of the function as approaches vertical ASM toote are all either Infinity or negative Infinity that's always what happens as you approach vertical ASM toote and that's why these vertical ASM tootes are often called infinite discontinuities must know number four horizontal ASM tootes not only can rational functions have vertical ASM tootes but they also sometimes have horizontal asmp tootes the rational function ofx has horizontal ASM toote at the line = if the limit of ofx as approaches infinity and or negative Infinity is equal to so basically at the extremes if the values of the function Trend towards specific finite value there's horizontal ASM toote at equals that finite value and now let's look at two types of rational functions that have horizontal asmp tootes and I'll show you how to calculate where those horizontal asmp tootes are the first type of rational function that has horizontal ASM toote is the type where the degree of the function in the denominator is greater than the degree of the function in the numerator when that happens it's guaranteed that there's going to be horizontal ASM toote at the line equals 0 and let's take look at an example of function where that's the case and I'll show you why the horizontal ASM toote is at equals 0 so here have rational function where the function in the numerator is degree 1 and the function in the denominator is degree 2 since the function in the denominator is higher degree than the function in the numerator there's going to be horizontal ASM toote at equal 0 but let me actually show you why it's at equals 0 by finding the limit of the function as approaches Infinity now to see what happens to this function as approaches Infinity there's actually trick for rewriting it you find the highest power of that's in the denominator and you actually divide all of the terms by that highest power of or algebraically what's really happening is we are multiplying this expression by one over the highest power of top and bottom so really we're just multiplying the expression by one which doesn't change its value but it'll be written in different format which will be easier to find the limit of so I'll be finding the limit as goes to Infinity of this function with all of its terms being divided by x^2 and now when finding the limit of this rational function as goes to Infinity can just find the limit of each individual term as goes to Infinity well as goes to Infinity all of these four terms are going to approach zero leaving me with 0 over 1 so that limit will just be equal to 0 over 1 which is just equal to 0 which is why there's horizontal ASM toote at the line equal 0 but you don't have to go through this whole limit calculation every time you want to find the horizontal ASM toote just check if the function in the denominator is higher degree than the function in the numerator you know there's going to be horizontal asmo at the line equals 0 what about case 2 what if the degree of the numerator is equal to the degree of the denominator well when that happens the horizontal ASM toote is actually at the quotient of the leading coefficients so for example this function because the numerator is degree one polinomial and the denominator is also degree one polinomial since their degrees are equal to each other the horizontal ASM toote is going to be at the quotient of their leading coefficients the leading coefficient of the function in the numerator is three and the leading coefficient of the function in the denominator is 1 so the horizontal ASM toote would be at the line = 3 / 1 which is just 3 but let me show you the limit at Infinity so you can see why it's at = 3 when finding the limit of FX as approaches Infinity could once again rewrite this function by dividing all of the terms by the highest power of that's in the denominator there's an to the one in the denominator so I'll divide all the terms by and as these X's approach Infinity the value of these fractions are going to be approaching zero leaving me just with 3 / 1 which is equal to 3 which is why there's horizontal ASM toote at the line equal 3 and then let's actually take look at the graph of this function when looking at the graph of that function notice that as approaches Infinity so as you move forever to the right the values of the function approach three and in fact as approaches negative Infinity the values approach three as well so that's what happens when you have horizontal ASM toote at the extremes as approaches infinity and or negative Infinity you'll see the function approaching finite yvalue and in this case the yv values are approaching three now should also mention that function only approaches its horizontal ASM toote at the extremes so that means for some finite value of the function actually can touch or cross its horizontal ASM toote unlike vertical asmp tootes that functions can never touch or cross must know number five slant asmp tootes not all rational functions have horizontal asmp tootes some of them have what are called slant or oblique asmp tootes so for some rational function ofx that is quotient of two functions of if the degree of the function in the numerator of is one greater than the degree of the function in the denominator ofx then that means there is an oblique or slant ASM toote so let's take look at rational function where the degree of the function in the numerator is one degree higher than the function in the denominator notice the numerator is degree 2 well the denominator is 1 de less it's degree 1 when that's the case there's slant ASM toote now when finding the equation of horizontal ASM toote showed you that we can just find the limit as the function approaches Infinity but if we tried to do that for this function we would see that it doesn't have limit as goes to Infinity goes to Infinity as well but even though it doesn't go towards specific finite value it does actually grow along linear path and we can find the equation of that line that it grows more closely towards called the slant ASM toote by performing The polom Division and then analyzing its limit so let me go ahead and actually divide the numerator by the denominator and can do that using any division method I'll use synthetic division for synthetic division write the zero of the divisor up here and then the coefficients of the dividend 1 4 and 10 go across the top of my table then can bring the one down 2 * 1 is 2 add the column to get six and then just keep repeating the process and that answer tells me that when do x^2 + 4x + 10 / - 2 the quotient is 1 + 6 with remainder of 22 so can rewrite this in quotient form it's equal to 1x + 6 plus the remainder 22 over the divisor of - 2 and now you can see how this function grows as goes to Infinity as goes to Infinity this fraction is going to go towards zero so the values are going to go towards Infinity along this line + 6 which makes this quotient + 6 the equation of the slant ASM toote and now let me show you what the graph of this function looks like and also notice it's going to have vertical ASM toote at xal 2 because the denominator of the function is zero when is 2 so here's graph of the function notice it has vertical ASM toote at xal 2 but that diagonal line is the slant ASM toote yal + 6 and as goes to infinity or negative Infinity notice that the function keeps getting closer and closer and closer to that line = + 6 must know number six holes earlier showed you that division by zero causes discontinuity in the graph of rational function it can cause vertical ASM toote or in this section I'll show you how it can actually cause hole in the graph of rational function ofx has whole at xal if is the undefined expression 0 / 0 to show you how we can tell when rational function has vertical ASM toote versus when restriction is going to cause hole let's take look at this function to find the restrictions on this rational function we should start by factoring it notice the numerator is difference of squares so can Factor it - 2 * + 2 and the quadratic in the denominator the numbers that multiply to two and add to -3 are2 and -1 so can Factor it to - 2 * -1 and now can more easily see what the restrictions on are since division by 0 is undefined whatever values of make this denominator be zero would be restrictions on the variable causing discontinuities in the graph of the function well the denominator would be zero if either of these factors were so the restrictions on are that can't be 2 or 1 and these restrictions graphically will either result in vertical ASM toote called an infinite discontinuity or hole in the graph called removable discontinuity and how we could tell is by subbing both of these values into our rational function and analyzing the result let me start by seeing what undefined expression get when sub in one for the numerator becomes -3 and the denominator becomes zero now this expression is undefined but because the numerator is nonzero number and the denominator is zero know that that's going to cause vertical ASM toote at xal 1 and now let's check what happens when sub in two into the rational function if sub in two notice that it'll make both the numerator and the denominator be zero and when you get 0 over 0 that's going to cause hole in the graph at the point where is 2 now don't know what the y-coordinate of this hole is quite yet but after simplify this expression by canceling these factors of - 2 that create the whole in the graph get new function that is actually defined when is 2 so could sub two into the simplified version and when do that would get the cordinate of the hole now know here in our restrictions it says can't equal to that's why we're allowed to say that this function is actually equal to the one above by stating this restriction but if we ignore the restriction and actually sub two into the function it does give us the coordinate of the hole so I'll make note of that to find the location of the hole sub = 2 into the simplified equation and if sub two into this function would get 4 over 1 which is four so know the coordinate of the hole is going to be at four now let me quickly show you what the graph of this function looks like notice the function has vertical ASM toote an infinite discontinuity at xal 1 and it also has hole in the graph at the point xal 2 that's the removable discontinuity and the reason why it's called removable discontinuity is because if we look at the rational function that we have here the factor that creates the discontinuity at two this factor of - 2 in the denominator we are able to cancel out that factor so remove that factor that creates the whole to get simplified expression that tells us the location of the hole and also because the degree of the numerator and denominator are equal to each other they're both degree 2 there's horizontal osmote at equals the quotient of the leading coefficients 1 / 1 which is why there's horizontal ASM toote at = 1 and also want to point out the limits of this function as they approach the restrictions on as the function approaches = 1 from the left it goes down to negative infinity and from the right it goes up to positive Infinity that's always what happens as you approach vertical ASM toote but what happens as the function approaches an value of two is the function approaches the finite value of four which the whole is at and we could have actually calculated that finite value of four using limits to find the limit of this function as approaches 2 could find the limit of this simplified version of the function as approaches two those limits would be equal because the functions are equal to each other at all values as long as is not equal to two must know number seven the and intercepts of rational function let me show you the method for calculating both the and intercepts of rational function ofx has an intercept at = if equals so that just means the intercept of the function is just at whatever value makes the y-coordinate be zero and for the intercept of rational function ofx has intercept at equals of 0 so that just means whatever the value is when you set to Zer that's the intercept of the function so let's look at an example of rational function where we'll calculate both the and intercepts of the function for this rational function let me start by finding its intercept to find the intercept just have to find whatever value makes the function have y-coordinate of zero so I'll just set the function's yvalue to zero and then solve for and before solving should note the restriction on the function notice that cannot equal 2 that would make the denominator and division by is undefined what makes rational function zero is if the numerator is zero so can just set the numerator equal to zero and then solving that equation see that if was -4 this function would be 0 over -6 which is 0 so there's an intercept at the point-4 0 and now let me calculate the intercept the intercept is that whatever value you get when you set to 0 so I'll just evaluate at 0 subbing in for these X's get 4 over -2 which is -2 so my intercept is at the point 0 -2 and before showing you the graph of this function let's quickly make note about the asmp tootes of this function because the numerator and denominator are both degree one they're equal degree the horizontal ASM toote will be at the quotient of their leading coefficients 1 / 1 is 1 so there will be horizontal ASM toote at the line yal 1 and there'll be vertical ASM toote at equals whatever the restriction on is cannot equal 2 subbing two into this function gives me 6 over nonzero number over zero is vertical ASM toote so vertical ASM toote at the line = 2 and now looking at this graph we'll see those ASM tootes and we also see the intercept of 4 and the intercept of -2 must know number eight graphing reciprocal functions let me start by quickly summarizing some properties of reciprocal functions if we have the function ofx which is equal to the reciprocal of the function ofx so if it equals 1/ ofx there are few things we should know about how this function in the denominator relates to the reciprocal function ofx first of all whatever makes ofx is going to make of undefined because division by is undefined and whatever makes ofx be zero would be considered the intercept of ofx so the intercepts of ofx are equal to the vertical asmp tootes of ofx and since ofx is just the reciprocal of ofx the reciprocal of function doesn't change its sign so whenever ofx is positive 1/ ofx of will also be positive and same if ofx is negative ofx will also be negative and next whenever the ofx value is increasing whenever it's getting bigger the value of of is going to be getting smaller because when the denominator of fraction gets bigger the overall value of the fraction gets smaller so the intervals of when ofx and of are increasing or decreasing are actually opposites of each other because the numerator of of is degree our denominator is going to be higher than degree 0 creating horizontal ASM toote at equals 0 and the last thing want to mention whenever ofx is at local Max of will be at local Min and whenever ofx is at local Min of will be at local Max and now let me make bit of room so we can graph few reciprocal functions let me start with the reciprocal of linear function our function ofx is the reciprocal of linear function the linear function is - 3 when graphing this function think it's best if we start by analyzing the Key properties of the function that's in the denominator so let's just look at the function - 3 to start with one of the key things I'm interested in is what is the intercept of this function right what makes it be zero well this would be zero if was 3 so it has an intercept at the 3 0 so if were to make table of values for this linear function would put three in the middle and then choose couple points to left and couple points to the right and then subbing these values into this linear function could calculate the corresponding values and now using these properties and these points can try and figure out what the graph of the reciprocal of xus 3 looks like well remember reciprocal functions always have horizontal ASM tootes at equals 0 and there's going to be vertical ASM toote at whatever value of makes the function in the denominator be zero and recorded that over here at an value of three the function in the denominator is zero so there will be vertical ASM toote at = 3 and now let me get some points for this rational function I'll choose the same values that chose for the - 3 function but notice ofx is equal to 1 / - 3 so if these are the values for - 3 my reciprocal function would have to do one divided by each of those values and this would be enough information to sketch pretty accurate graph of the reciprocal function on that grid think I'll actually graph the function xus 3 as well so I'll do that to start with plotting these points get that linear function - 3 and then for the reciprocal function ofx could graph the ASM tootes and then could roughly plot these points and then could connect those points using the proper behavior know the function is going to exhibit as goes to infinity or negative Infinity know it's going to be approaching that horizontal ASM toote and as the values approach the vertical ASM toote the function is going to be going either down to negative infinity or up to positive infinity and then hopefully this graph allows you to see all of these properties we have over here ofx is increasing over its entire domain which is why 1 over ofx is decreasing over its entire domain and whenever ofx is negative the reciprocal is also negative and whenever ofx is positive the reciprocal is also positive and the intercept of the ofx function makes the reciprocal become 1 over which is undefined which creates this vertical ASM toote and now let me shrink this work and let me show you what some reciprocal quadratic functions could look like for reciprocal quadratic functions there are three General shapes they could make instead of giving you equations and coming up with tables of values and sketching them very accurately let's just use these properties to figure out what the graphs of reciprocal quadratics could look like well remember quadratic functions can have either 0 1 or 2 intercepts and that's what's going to cause the different shapes for the reciprocal functions let me start by sketching quadratic with 2 intercepts if zoom in on this quadratic that has 2 intercepts let's call this function function ofx and let's graph its reciprocal 1/ ofx Well know reciprocal functions have horizontal asmp tootes at equal 0 and whenever ofx has an intercept so whenever its value is zero the value of the reciprocal function will be 1 /0 which is undefined and creates vertical ASM toote and now to graph the branches of the reciprocal function have to keep some properties in mind know it's going to approach the horizontal ASM toote at the extremes and know as it approaches the vertical ASM toote it's either going to go up to infinity or down to negative Infinity notice to the left left of the first vertical ASM toote ofx is positive so the reciprocal function will also be positive meaning it will look something like this and same thing to the right of the second vertical ASM toote ofx is positive so it's reciprocal will also be positive and also notice in those two sections when ofx is decreasing the reciprocal is increasing and when ofx is increasing the reciprocal is decreasing and now let's look between the vertical asmp tootes this section is negative meaning the reciprocal will also be negative and since ofx has local Min right there the reciprocal will have local Max at that same value making this branch of the function look like this and now let's see what happens if the quadratic only has 1 intercept if this is our ofx function and we want to graph 1 over ofx once again reciprocal functions have horizontal ASM tootes at equal 0 but there is only one place where ofx is 0 meaning there is only one place where 1/ ofx is 1/0 which is undefined so there's only one vertical ASM toote and now to the left of the vertical ASM toote ofx is positive and decreasing so the reciprocal will still be positive but it will be increasing and to the right of the vertical ASM toote ofx is still positive but increasing so the reciprocal function 1 over ofx will be positive but decreasing and now let's do one more where the quadratic the third scenario has no intercepts if this is our ofx function notice it has no intercepts which means there are going to be no vertical asmp tootes but it still will have horizontal ASM toote at the line equals 0 and because this function is always positive the reciprocal will always be positive and when this function is decreasing the reciprocal will be increasing and then when this function switches to increasing the reciprocal will switch to decreasing and because see there's local Min here at that same value is where there's going to be local Max for the reciprocal making it look like this and there's our reciprocal of ofx so there you have it hopefully you understand the connection between function and the graph of its reciprocal must know number nine solving rational equations when solving rational equation you should always start by stating the restrictions that way you can make sure they're excluded from the solution set and then before trying to solve you can eliminate any fractions by multiplying both sides of the equation by the lowest common denominator and then after eliminating the fractions you can just go ahead and use your algebra skills to solve the equation so let's look at few different examples in this first example we need to solve for what value of makes this equation true can get rid of these fractions If multiply both sides of the equation by the lowest common denominator between these denominators here and since both the denominators are their lowest common denominator is of course so all would have to do is multiply both sides of this equation by can then distribute the to both terms on the left side of the equation and then all the fractions should simplify leaving me with on the left side of the equation just 3 + 2x and on the right side just 5 and then you can just rearrange this equation to solve for have 2X = 5 - 3 which is 2 which means is equal to 1 and then don't forget anytime you solve an equation you should check the final answer in the original equation to make sure it's true 3 1 + 2 is 5 now let's try second example in this equation have two fractions if look at their denominators they're -1 and the lowest common denominator between those two denominators would just be the product of them so * -1 will be my lowest common denominator that means should multiply both sides of this equation by * - 1 and on the left I'll distribute that product to all three terms that are on the left side of the equation and on the right the product of those three factors will just be zero and now all the fractions should simplify nicely looking at the first term can cancel these factors of - one and the third term can cancel these factors of so there are no more fractions left I've got * 3 + 5x * - 1 I'll distribute that that 5x + -1 * 2 I'll distribute that 2 to the -1 and now just have quadratic equation to solve it's 5x^2 and then 3x + 2x is 5x - 5x is 0x and then - 2 so have 5x^2 - 2 = 0 when you have single term with variable we can just rearrange it to isolate the variable would have x^2 = 2s which means = + or minus the < TK of 25 so those would be my answers to the original equation and for both of the equations we just solved forgot to mention the restrictions on the variable in the first equation the only restriction was that can't be zero and the solution was one so that's fine and in the second equation the restrictions would be can't be 1 and can't be zero must know number 10 solving rational inequalities here are the steps for Sol rational inequality step one move all terms to one side step two combine Expressions to make single rational expression step three find the intercepts and the restrictions of the function and then step four use those numbers to divide the function into intervals and then lastly test value within each interval to see if it satisfies the inequality and let me make some room and let's go ahead and solve rational inequality when solving this in equality the first thing should do is move all the terms to one side so think what I'll do is I'll take this term and bring it to the left by subtracting it and now need to combine these Expressions by getting common denominator and my common denominator will just be the product of these denominators so this fraction will be multiplied by - 3 top and bottom and this fraction will be multiplied by + 1 top and bottom and now notice these denominators are the same so can combine these two fractions into single fraction by keeping that common denominator and then in the numerator have - 3 * + 3 which is x^2 - 9us the product of - 2 and + 1 I'll make sure to put that product in Brackets and it's x^2 - - 2 and then in the numerator would have to subtract all three of the terms that are in Brackets and then collect my like terms when doing that the numerator will simplify to x^2 - x^2 is 0 I've got- that's and -9 - -2 that's -7 now to figure out when this function is greater than or equal to zero I'm basically trying to figure out when is it above the xaxis so I'm going to divide it into intervals based on where its intercept is and its vertical ASM tootes are because know those are the places where the function might switch from being above to being below so remember its intercept will be at an value that makes the numerator be zero so it'll be an intercept at 7 and the vertical ASM tootes will happen at the restrictions of the variable since division by 0 is undefined whatever makes the denominator to be zero is restriction so can't be three or1 and now have to divide the function into intervals based on these values so what I'll do is I'll actually make chart where I'm going to do testing of values in the intervals create call it sign chart or Factor table here's what it looks like notice I've arranged the intercepts and restrictions in ascending order at the top here and always start with negative infinity and end with positive infinity and now what want to do is choose test value within each of the intervals I've created so between negative infinity and -1 need test value any number in that interval will do so I'll choose -2 and then the next interval is from -1 to 3 I'll choose test value of 0o and then between 3 and 7 I'll I'll choose 5 and from 7 to Infinity I'll choose 10 and now what want to do is basically figure out is this function positive or negative in each of those intervals so I'll take these test values that chose and I'll sub them into this function and the way I've created this table allows you to do it in an organized way which makes the calculation easy I'm just going to sub into each of these three factors that make up the rational function and not even record the answer but just whether it's going to be positive or negative so when is -2 it makes all three of those factors be negative which means we would have negative / negative time negative that's negative over positive which means the overall sign of the function will be negative if you have an odd number of negative factors the overall sign will be negative and an even number of negative factors the overall sign will be positive and knowing the overall sign of the function is helpful in answering the inequality because we're is interested in when is it greater than or equal to zero so that means when is it positive well in the first interval it's negative but after we move across this vertical ASM toote looking within the next interval my test value is zero when sub that into these factors 0 - 7 is negative 0 - 3 is negative but 0 + 1 is positive because have an even number of negative signs that's going to be positive result which means this function is positive which is greater than or equal to so this interval from -1 to 3 will be part of my final answer and then for the test value of five get negative positive positive when sub them into those factors one negative sign will give negative result and then subbing in 10 get positive positive positive which is positive result so since want to know when is the function greater than or equal to zero the intervals where the function has an overall positive sign so between -1 and 3 and between 7 and infinity are going to be my answers to this inequality so I'll say my answer is that the inequality is true when is an element of numbers within this first interval between -1 and 3 or if is between 7 and infinity and notice because the original inequality wants to know when it's greater than or equal to0 do want to include the intercept in my answer which is why put square bracket at seven but made sure to put round brackets at my restrictions because the function doesn't even exist there so they can't be part of my final answer and it might be good to end the video by making quick sketch of the graph of the rational function we were working with so you can see how this is the answer to the inequality so I'll make bit of room and this is the function we're working with here we know it has an intercept of and these two vertical ASM tootes at1 and 3 and because the function in the denominator is higher degree than the function in the numerator there'll be horizontal ASM toote at equal 0 and also there's an intercept at 7 which means this function actually touches and goes through the horizontal ASM toote so what this chart showed me is that between negative Infinity And1 the function is negative which means it must be down here and then between -1 and 3 the function positive so so it must be up here and then from 3 to 7 the function is negative and then from 7 to Infinity the function is positive so it's above the x-axis but it does have to start approaching the horizontal ASM toote so there's rough sketch of the function and notice how it shows that this is the answer for when the function is greater than or equal to zero hopefully this video helped you understand rational functions let me know in the comments what you want top 10 of next