In Exercises 89–94, the equation for f is given by the simplif...

Question

In Exercises 89–94, the equation for f is given by the simplified expression that results after performing the indicated operation. Write the equation for f and then graph the function. 5x²/(x²−4) ⋅ (x²+4x+4)/(10x³)

Accepted Answer

Welcome back. I am so glad you're here. We are asked to graph the given expression. After performing the indicated operation, our expression is in the numerator of the first fraction X squared minus six X plus nine that's divided by in the denominator X squared minus nine. Now that fraction is multiplied by the second fraction. The numerator of that is six X cubed and that's divided by 10 X to the fourth. And we have for our answer choices four graphs, we will evaluate these graphs after simplifying the expression. So now let's move this expression to where we've got a little bit more room on the screen. So the indicated operation here is multiplication. But before we begin multiplying, let's be sure to factor all of these numerators and denominators. And if you need a refresher on factoring, please go check out some of the previous lesson videos. But because this is a complex problem, I am going to factor them quickly. So for the numerator of the first fraction that will factor out to B we'll put in our numerator two open sets of parentheses that will be X minus three in the first set of parentheses multiplied by the quantity of X minus three. And looking at our denominator, we recognize a difference of squares in X squared minus nine. So that will factor to be the quantity of X plus three multiplied by the quantity of X minus three. Now that's still multiplied by our second fraction which is six X cubed divided by 10 X raised to the fourth power. Now that everything is factored, we can see if we have any common factors in the numerator and the denominator. And yeah, we've got an X minus three factor in both the numerator and the denominator and we can cancel those out, those become ones being multiplied. So now we can rewrite this. And let's take our six X cubed and put that in front of the quantity of X minus three that's divided by 10 X to the fourth multiplied by the quantity of X plus three. We can still do some simplification with the six X cubed and the X to the fourth six and 10 are both divisible by two. So dividing those both by 26, divided by two is 35 di or excuse me, 10, divided by two is five and, and X to the fourth are both divisible by X cubed, X cubed divided by X cubed is just one and X raised to the fourth power divided by X cubed is just one single X left. So this will be in the numerator three multiplied by the quantity of X minus three. That's all divided by, in the denominator, five X multiplied by the quantity of X plus three. Now, that's as simplified as this one gets, we could distribute the three in the numerator and the five X in the denominator. Both of these are useful at different points of what we'll be doing um in the next few steps. So distributing the three in the numerator will have three X minus nine. And that's divided by distributing the five X in the denominator will have five X squared plus 15 X. So there are the two fractions that we will work with in order to graph this. So for graphing this, we recall from previous lessons that we're going to start off by checking our domain and we are going to identify any domain issues where X cannot exist. And we want to look at the original problem as well as what we've got here in order to identify domain issues. So let's look at the denominators for our original multiplication problem. We have this X squared minus nine. If we set that equal to zero and sulfur X, we would be adding nine to both sides. We'd have X squared equals a positive nine. We take the square root of both sides and we get X is equal to plus or minus three, meaning that X cannot equal plus or minus, it can't be a positive or a negative three because of that domain restriction. And then looking at the denominator for this 10 multiplied by X to the fourth. Well, in this case, X just cannot equal zero. If we set the 10 X to the fourth equal to zero. And we solved for X, that would just be zero. So looking at our final denominator, the five X squared plus 15 X, it's a little bit easier to identify our zeros actually from the factored version of that five X multiplied by the quantity of X plus three setting that equal to zero, setting each factor equal to +05 X equal to zero and X plus three equal to zero. Well, for the first one, if we divide both sides by five, we see X cannot equal zero, which is what we have. And for the second one, X plus three equals zero. If we subtract three from both sides, we see that X cannot equal negative three. So we have got the domain issues squared away. And now we can begin the steps from previous lessons to graph this. I'll draw a rough sketch of a coordinate plane. We'll have a vertical Y axis and a horizontal X axis. They come together at the origin in the middle. First, let's determine our vertical asom totes, our vertical as some totes are found by setting the denominator equal to zero. And we've already done that. We set five X multiplied by the quantity of X plus three equal to zero. And we found vertical asom totes where X equals zero and where X equals negative three. So we can draw that on our coordinate plane, we can add X equals negative three. So from the origin going to the left three units, we can draw a vertical ASOM tote where X equals negative three. And then along the Y axis, we can draw another vertical ASOM tote where X equals zero. Next, let's work on our horizontal asom totes. That almost looks like a two step two, horizontal as and totes, we find these by looking at the degrees of our numerators and our denominators. So the degree of the numerator is one, we just have X to the first power. There's an unwritten one and the degree of the denominator is two. That's the highest degree in the denominator. So which one's greater? The denominator when the denominator's degree is greater than the numerator, we have a horizontal asom tote at Y equals zero. So along the X axis, we can draw a horizontal asom tote at Y equals zero. All right. Step three. Let's ask ourselves what our Y intercepts are. Well, this one, normally, what we would do is we would plug zero in for X and solve. But we know that if we plug zero in for X, we'll have a zero in our denominator, we'd have five multiplied by zero, squared plus 15, multiplied by zero, which would just be zero. And that's undefined, we cannot have that. So there are no Y intercepts here. There are none. Alright. How about X intercepts? Step four X intercepts? We find the X intercepts by finding the zeros of the numerator will essentially we set the numerator which is three X minus nine equal to zero in solving for X. All right, we can add nine to both sides and we've got three X equals nine dividing both sides by three and we have X equals a positive three. So the X intercept is at a positive three. However, recalling what we did at the beginning, establishing our domain, we know that X cannot equal a positive or a negative three, there's actually going to be a hole at our X intercept. So because X cannot equal three, there will be a hole at 30. All right. Next, we want to see if our graph intersects the horizontal Asymptote. So the way we solve for that, seeing if the graph intersects the horizontal asom tote is we set our expression. So we set our three X minus nine divided by five X squared plus 15 X equal to our horizontal Asymptote, which is zero. So solving for X, well, what we're doing is multiplying this denominator by both sides, that'll just be zero. So we'll have three X minus nine equals zero. And solving for zero. We've already done that earlier when we were finding the X intercepts, we know that that is where X equals three. So it makes sense that we would have an intersection with the horizontal aps asim tote at X equals three. However, there will be a hole there. All right. And the last step here is we're going to plot some points to see what's going on with our graph. So plotting some points, we want to choose some points for each of the regions between the Asom totes. So we'll look between the regions of the vertical asom totes. So we want somewhere that where X is less than negative three, somewhere where X is between negative three and zero and somewhere where X is greater than zero. So for these points, I am going to choose for X value that is less than negative three. I'm going to choose negative four for an X value that is between negative three and zero. I'm going to choose negative two for an X value greater than zero. I am going to choose one and now we plug those values in to our original well, to our simplified expression which is three X minus nine divided by five X squared plus 15 X. I'm just going to give the final simplified answers here, but you're welcome to pause the video and see how to do the math on that. So where X equals negative four, Y equals a negative 21 20th. So that's very close to negative one. We can graph that. So X equals negative four Y is approximately negative one. So we can put that about here where X equals negative two, Y equals 15 10th, which is about 1.5. So at X equals negative two, we go from the origin left two units and then go up about 1.5 units on the Y axis approximately here and four X equals one, Y equals negative 6/20. And so from the origin, we'll go to the right one unit for our X value. And then down, that's like a half a unit for our Y value approximately here. All right. So now we know that these are going to hug up against the Asom totes as much as possible. And we have an ASOM tote along the X axis. So from negative infinity for X, it's going to come close to the X axis intercept with our plotted point and then start heading down along the ASOM tote where X equals negative three for toward negative infinity for the Y values. Now plotting the second line we're looking in between our asom totes of negative three and zero where X equals negative three and X equals zero, that's going to be up in the second quadrant. So it's coming, it's heading up toward negative infinity along the ASOM tote where X equals negative three that comes down hits our point curves and then it's going to come back up and start heading up toward positive infinity along the Y or along the Y axis where X is equal to zero. And then for our final line, we are in the fourth quadrant, it's going to come up from negative infinity along the Y axis. It's going to intersect our point that we plotted and then it's going to, we can draw it for now going through the 0.0.30 where we had that X intercept and then it's just kind of heading toward positive infinity for the X value on the right. But we know that we do have that whole at the 0.30. So there is our graph and now what we can do is bring this graph back up. And let's take a closer look at our, I'm going to shrink it a little bit at our other answer. Choices, we'll compare those and see which one is the closest match. Mm Alright. So we want to look for asymptotes along, we want a horizontal Asymptote along Y equals zero answer choices A and B do not have horizontal asom totes along Y equals zero. A has one along Y equals negative 3/5 and B has one at Y equals a positive 3/5. So we can eliminate answer choices A and B. Now we need vertical asom totes at X equals negative three and at X equals zero answer choice C has vertical asom totes at X equals negative three and X equals zero as does answer choice D. So both of those are have potential and they have the correct vertical asom totes or excuse me, horizontal asom tote at Y equals zero. So now we want to look for one that has the lines for our graph, the line for the section where X is less than negative three is going to be down in the fourth quadrant. And for answer choice C it's up in the, excuse me, it's going to be down in the third quadrant. And for answer choice C it's up in the second quadrant. Answer choice D it is down in the third quadrant. So answer choice D is oriented correctly. We can eliminate answer choice C and we can choose answer choice D well done. We'll catch you on the next one.

In Exercises 89–94, the equation for f is given by the simplified expression that results after performing the indicated operation. Write the equation for f and then graph the function. 5x²/(x²−4) ⋅ (x²+4x+4)/(10x³)

Key Concepts

Simplifying Rational Expressions

Multiplication of Rational Expressions

Graphing Rational Functions

Watch next

In Exercises 89–94, the equation for f is given by the simplified expression that results after performing the indicated operation. Write the equation for f and then graph the function. 5x2/(x2−4) ⋅ (x2+4x+4)/(10x3)

Key Concepts

Simplifying Rational Expressions

Multiplication of Rational Expressions

Graphing Rational Functions

Watch next

In Exercises 89–94, the equation for f is given by the simplified expression that results after performing the indicated operation. Write the equation for f and then graph the function. 5x²/(x²−4) ⋅ (x²+4x+4)/(10x³)