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Fird &1 ccuation Of the tangent line to the graph of the functian ut the given point: Ax) sntz (2, 13)Usd graphing utilty araph thc function and thc tangent lin...

Question

Fird &1 ccuation Of the tangent line to the graph of the functian ut the given point: Ax) sntz (2, 13)Usd graphing utilty araph thc function and thc tangent line in the same viewing window.

Fird &1 ccuation Of the tangent line to the graph of the functian ut the given point: Ax) sntz (2, 13) Usd graphing utilty araph thc function and thc tangent line in the same viewing window.



Answers

Find an equation of the tangent line to the graph of the function at the given point. Then use a graphing utility to graph the function and the tangent line in the same viewing window. $$ \begin{array}{ll}{\text { Function }} & {\text { Point }} \\ {f(x)=\frac{x-2}{x+1}} & {\left(1,-\frac{1}{2}\right)} \end{array}$$

Okay. In this question, we will be talking about quotient and product rules in the question, we want to find the tangent line to this curve at this point. And obviously to do that, we're going to have to use product rule because this is a product of two functions. So, first of all, since our goal is to find the tangent line, we need to find a slope of this tension line which will be the derivative of this function. Mhm. And that's where this product rule, as I said, comes into play. The derivative of a product of two things is the derivative of the first times the second. So the derivative of this times X -2 plus the derivative of The second, which in this case is one times the first. Yeah. And so We can simplify that by factoring out two X -1 and then being left with two x -4 Here at X -1 here. So three X -5. Now, since we have this derivative, we can plug in X equals zero. To find what it will be at this specific point, So -1 times -5 gives us positive five. Uh huh. Okay. So now we have all the information. We need to find the equation of the tangent. So to write the equation of any line, we know we need the y intercept. And this look We just said that the slope of our line is five. So that's taken care of. We just need the one intercept. The only information other than the slope that were given or that we have is that it passes through this point because it is tangent at that point. And well what do you know this point is exactly the one intercept Which tells us that the lines except is -2. Yeah. Yeah. And so there is our equation Y equals five x -2. Let's take a moment to graph that along with this function. So here's the function to which the line is tangent, here is the tangent point, And if we have y equals five X -2. Yeah, we see that that line nicely fits the description of being tangent to the curve.

Mm. The topic of this question is product and quotient rules. And this question asks us to find uh the equation of the line tangent to the curve given by this equation at this point. So to find this equation of the tension line. First, we have to uh well, first, let's consider the equation we know since it's a line that it can be written in the form Y equals M times X plus B. Where M is the slope and be as the Y intercept. Okay, now, uh by our knowledge of handed lines we know. So say we have a curve and a tangent line to the curve. We know the tangent line after the point of tendency must have the same direction in a sense as the curve. So it must have the same slope at that point and slope of curves we describe as derivatives. So if this curve has equation H of X, it's derivative which will will be H. Prime of X. And depending on what this X value is, uh the slope that derivative or the pounds at that point, the derivative will depend on the value of X at that point. So if it were the slope at two, uh then that would be a prime of two. So since we want to find slope of the tangent line, we want to find the derivative of this and X equals negative two. So let's differentiate using the product rule. Since that's the topic of this question. We can write this square as a product so that we can use product rule, although it may be easier to use chain rule if you know how to use that. Yeah. Yeah. So we to do the product rule, we have the two terms which I will call F entity two expressions. So the derivative of age, which is F times G is the derivative of us times three. Yeah. Plus the derivative of G. Which is the same as if time's up, which is the same as geek. So we really have just double of this. And since we want to find a prime at negative two, we substitute in X equals negative two. And so our slope of the tangent at that point is negative 24. Mhm. Yeah. Oh we want to find the right intercept of this tangent line. We know it's slope is negative 24. And so it can be written like this but we need to find A. B. And so if you know the formula for tangent lines have a special form, then you should use that because it is easier. All you have to do is put X. Subtract the X coordinate of the point of tendency from X directly. And then that gets multiplied by the slope. And then to all of that all together you add the y coordinate of the point of tangent C. And so that is our tangent line. Right? And so we had minus 48 plus nine which is minus 39. Yeah, except there is our tangent line. Now if you don't know this formula yet, I would encourage you to get to be familiar with it because it really helps. But uh get comfortable with the idea or if you don't want to use it yet, then you can also substitute this point into this equation to find B. And you would also find the answer to be negative 39. Now we want to graph this and so that's simply done. That can be simply done. Using days most this online graphing software we have our curve x squared minus one squared and the line is minus 24 x minus 39 because the slope is minus 24 and the Y intercept as negative 39. And so there we go. The point of tangent c minus 29 is right here somewhere.

Okay. Um So here I'll give it a given function is well one over the square root of X squared minus three X plus four. And I've given point is the 40.0.3 comma one half. Now to find the equation of the tangent line to the graph at a given point, we need to find the slope of the tangent line at that point. The slope of the tangent line is just given by the derivative. So the derivative of our function um is just going to be equal to while we have a um well negative X squared minus three X plus four to the negative three half over two and then times two x minus three. So the derivative um here is going to be while we get a negative two x minus three. Um Times X squared minus three X plus four to the negative three halves. Train role here. Um All over to um And then the slope of the tangent at the given 0.3 comma one half. Well that's going to be equal to a negative uh See two x minus three times X squared minus six was four, 2 -3 half. All over two. So we get a um more negative two times three minus streets becomes a negative three. Um and then just times four To the -3/2ves and four to the negative three halves all over two which gives us a negative three times to two negative three all over to which becomes negative three. Um Over two times eight which is negative 3/16. So the slope here of the tangent line at the given point of three, common one half is going to be negative 3/16. We can then go ahead and just plug that into um our point slope form because the point slope form, we can write the equation of any line if we ever have any point on the line and the slope. So if we had the slope of the tangent there's there are slope. And then you know that our point slope form is just why minus Y one is equal to the slope m times x minus x one. So here we have well y minus one half is equal to negative 3/16. Um X plus 9/16. So we end up here with our Cash in line is going why is equal to negative 3/16 X-plus 1716. So there is our um our tangent line at a given point and then you can go ahead we can grab the function and the tangent line. So here we have our function in blue that's one over the square root of X squared minus three X plus four. And our stupid tangent line while our tangent line is gonna be, Y is equal to negative 3 16 X plus 7 16. And we see here at the point Um right here at the .3 common 1/2 we are tangent to our function.

In this question we will understand the concept of how to find the derivative and the equation of tangent to the given girl like. So we have a function equation of Nico which is Doing to access where -1. To leave Power three. We have to find the derivative and also the engine of the line. So the question of the tangent at two f. too late. So the approach to solve this question is to find the equation of detention will be needing the slope and the point they so to find the slope, we have to find a derivative. So aftershocks would be equal to two multiplied by to the Access with -1 Square multiplied by two. works. Right? So this is it was too dread legs X squared minus one squared. So the slope would be At F-2 Which is equal to 12 multiplied by two two square minus one square light, which is equal to 216 To find the equation of the attendant will be needing the point. So after would be equal to two and two, two sq -1 to the Power three, which is equal to 54. So the equation of that engine would be why minus have to Is equals to have-2 x minus two. We can substitute the values here, So Y -216. Sorry, by -54 is equal to 216, Multiplied by X -2. Hence Why is equal to 216, x minus two plus 54. So this is the equation of the danger and we can check the same with graphing utility has been like.


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