Use a graphing utility to graph the curve and the tangent lin...

Question

Use a graphing utility to graph the curve and the tangent line on the same set of axes.

y = 2x² / (3x - 1); a = 1

Accepted Answer

Welcome back, everyone. In this problem, we want to use a graphing utility to graph the function Y equals X + 2X divided by X minus 1 and the tangent line to the curve at X equals 3. Now, if we're going to use a graphing utility, we know that we can simply input the equation of our function for it to do the graph, but it also means that if we're going to graph the tangent line to the curve, we need the equation of the tangent line to the curve at X equals 3. So if we can find the equation of that tangent line, then we should be able to input it into our utility to get our graph. So let's focus on doing that. What do we know about the equation of a line? Well, recall, OK, that we can write the equation of a line in the point slope form which says that it will be Y minus Y1 equals M multiplied by X minus X1, OK? Where X1 and Y1 are coordinates of a point on our graph and M is the slope of Y graph. Now, let's first try to find the slope of our graph M, and then we'll go ahead and try to find the coordinates X1 and Y1. What do we know about our slope? Well, we know that for the slope of the tangent line to the curve at X equals 3, if we can find the derivative of our function, OK, and then we can substitute X equals 3, then we'll get the, the slope for the tangent line to the curve. So let's differentiate Y and substitute X equals 3 into the derivative, OK? Now, given Y. Which is equal to x2 + 2 x divided by X minus 1. Notice that Y is a caution and we can differentiate Y then by using the quotient rule of differentiation. What does it say? Well, recall the quotient rule tells us that if we have Y equal to some quotient U divided by V where U and V are functions of X, then the derivative of Y. Is going to be equal to VU minus UV all divided by V2. So that means we'll need to differentiate U and V and substitute them into this formula to help us figure out the derivative. So let's go ahead and do that. As we said, our numerator is U, our denominator is V. When we let U be equal to X2 + 2 X, OK? Then the derivative of uu is going to be equal to 2 X. The derivative of X is 2 x plus 2, the derivative of 2 X is 2. Likewise, for our denominator X minus 1, its derivative would be equal to 1. Differentiate X we get 1, the derivative of a constant negative 1 is 0. Know that we have our terms and their derivatives, we can apply the quotient rule. There's no that tells us then that the derivative of Y will be equal to V X minus 1 multiplied by U2 X + 2. Minus u x2 + 2 X multiplied by V 1. And all of this is being divided by V2X minus 1 squared. Now we could simplify this a bit further by expanding our brackets. And for our first term when we expand, it's going to be 2 X2 + 2X minus 2X minus 2. And I subtracting this term that's gonna be minus X squad, OK, minus 2 X. That's all of what is gonna be in our numerator. We have the same denominator. Now if we start to cancel some terms here, uh, 2 X minus 2 X equals 02 X squared minus X2 is X2, OK, and we're left with minus 2 X and minus 2. So this means the derivative of Y with respect to X then is going to be X2. OK. X2 minus 2X minus 2 all divided by X minus 1 all squared. Now that's the derivative of Y. then we can evaluate the derivative at X equals 3. That's, that's the whole purpose of why we want the equation here. So now at X equals 3, then our derivative is going to be equal to 32 minus 2 multiplied by 3 minus 2. All divided by 3 minus 1 squared. Which is going to be 9 minus 6 minus 2, which is 1, divided by 3 minus 12 squared, which is 4. Thus, at X E equals 3, our derivative, which is gonna be our slope, M is 1/4. Now that we have our slope, can we figure out X1 Y1? Well, if we think about it, our curve meets the tangent line at the point where X equals 3. That means we can use our equation of the curve to find the Y value when X equals 3, OK? So in other words, we can also say that at X equals 3, let me do this as a little box over here. At the point where X equals 3, then Y is gonna be equal to 32, OK, plus 2 multiplied by 3, X2 + 2X divided by X minus +13 minus 1. This is gonna be 9 + 6, which is 15. Divided by 3 minus 1, which is 2. So this tells us that the 3152, sorry, 15 divided by 2 is on our graph. Know that we know that, OK, then we can substitute the values for our points on our slope into the equation of the line. Because now, this tells us then that Y minus Y1 is going to be Y minus 15 divided by 2. And that is gonna be equal to M14 multiplied by X minus X1, which is 3. If we simplify this some more here, then we add 15 halves to both sides and expand our bracket. Y will be equal to a quarter of X minus 3/4 plus 15 divided by 2. And know this if we simplify here, we're going to get Y to be equal to a 4th of X plus 27/4s. So this is going to be the equation of our tangent line to the curve at X equals 3. Now that we have that equation and we have the equation of our graph, we can go ahead and use the graph in utility to plot. And when you do that, we input both of those equations in the graphing utility. You should find that it looks like this, OK? Where our blue line represents the tangent line to the curve and the line in black represents the curve. Thanks a lot for watching everyone. I hope this video helped.

Use a graphing utility to graph the curve and the tangent line on the same set of axes.
y = 2x² / (3x - 1); a = 1

Key Concepts

Graphing Functions

Tangent Lines

Derivatives

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