Given the function f and the point Q, find all points P on the...

Question

Given the function f and the point Q, find all points P on the graph of f such that the line tangent to f at P passes through Q. Check your work by graphing f and the tangent lines.

f(x) = 1/x; Q (-2, 4)

Accepted Answer

Welcome back, everyone. In this problem, the function G and the point A are given as G of X equals -3 divided by X and A is the 0.46. Determine all points P on the graph of G such that the line tangent to G at P passes through A. A says it's -2, 3 houses and 13. B 23 halves and 31 C 13 and 62 and D 3-1 and 65. Now, what are we trying to figure out here? How are we supposed to figure out the points P on the graph of G such that the tangent line to G passes through A? Well, let's try to make sense of this with a quick sketch, OK? So we know that we have the graph of -3 divided by X, so it's gonna be an asymptote that probably looks something like this, OK? And we're saying that we have some point A on our graph, which is uh 46. Now, we're trying to determine all of the points P on the graph of G, such that the line tangent to G at P passes through A. In other words, if we have a tangent line passing through A here and another tangent line passing through a here, OK, on our graph, then at those points, let's call those points P, we're trying to find out what the values of X and Y are so that we can figure out the point on our graph G. But now before we figure out what those points are, we'll need the slope to our tangent line because once we have the slope to our tangent line, we may be able to use them to figure out the value of X at our slope and thus plug it into our function G to find the value of Y, OK? So let's break it down. Let's first start by finding the tangent, sorry, by finding the slope of G of X. And we can find the slope using the limit definition, OK? Now, given Gaics. Being equal to -3 divided by X. Then by the limit definition, by the limit definition. The derivative of G of X, OK. Is going to be equal to the limit as it approaches 0 of G X plus H minus G of X all divided by H. Let's find G X plus H, substitute G of X, and try to simplify to get the derivative of G of X. When we do that, OK, then what we're really saying here. Is that we're gonna have uh GX -3 divided by X to be -3 divided by X plus H, OK? We're gonna be subtracting uh negative 3X. OK. And then we're dividing all of this by H. and we're finding the limit of this as H approaches 0, OK? No, we can combine our, our fractions in our numerator here. So no, this is gonna be the limit as H approaches 0 of -3 X plus 3 multiplied by X plus H, all divided by X multiplied by X + H. we've combined both fractions and then we're dividing this by H. Now, we can simplify this a bit further, OK? So we put back our limit as H approaches 0. By saying in our numerator we'll have -3 X plus 3 multiplied by X3X + 3 H. we've we've distributed 3 across X plus H. and now we're dividing that by X sorry, by H multiplied by X. H multiplied by X multiplied by X plus H, OK? No negative 3X + 3 X equals 0, so you simplify that to the limit as H approaches 0 of 3 H divided by HX multiplied by X plus H, H cancers H and no here we are with the limit as H approaches 0 of 3 divided by X multiplied by X plus H. No we can directly substitute H equals 0 into our expression. So no, that's going to be equal to 3 divided by X multiplied by X + 0. X + 0 is X and X multiplied by X equals X squared. So the derivative of G of X is equal to 3 divided by X2. That's the slope of our tangent line. Now that we have our slope, OK, then the next question we're asking ourselves is how can we use this slope to figure out the X values of our points. Well, note that each point on the graph is represented by X-3 of X, OK? So, each point on G. Each point on G, and let me write that properly here. is represented by X by X coordinate and -3 are divided by X, the Y value. OK. So for the points, OK. 4 X -3 divided by X. And another point on our tangent line for 6, OK? We can use the slope formula to figure out the value of X because we know that the slope M. Will be equal to the difference between the Y values Y2 minus Y1 divided by the difference between the X values X2 minus X1. So, here, if we call XX1 and -3 divided by XY1 and call 4 X2 and 6 Y2, OK, then this implies M is going to be equal, OK. To 6 minus -3 divided by X divided by 4 minus X. And remember earlier we found that our slope M, the derivative of G of X, is 3 divided by X2. So 3 divided by X2 is equal to this entire expression. Let's try to simplify it to see if we can solve for X. Now if we cross multiply here, then we're going to have 3 multiplied by 4 minus X equal to X squared multiplied by 6 + 3 divided by X. replied by 4 is 12, 3 multiplied by negative X is negative 3 X. And on the right-hand side, we're gonna have 6 X squared plus 3 X. OK? Now if we group our like terms here, OK, and right, this is in, in the quadratic form. We're gonna have 6 X squared plus 6 X minus 12 equal to 0. If we factor out 6, that's gonna be 6 multiplied by X2 + X minus 2 equals 0. And then if we divide through our entire equation by 6, let me make some space on the right here. We're gonna have X2 + X minus 2 being equal to 0. If we factor here quadratically x minus 1 multiplied by X + 2 will be equal to 0, which implies then. That either X minus 1 equals 0, which means X equals 1, or X + 2 equals 0, which makes X equals -2. So we have the values of X that would fit into our function, that would work on our slope meeting the point at A, OK? Know that we have our values of X, then we can find our Y values because no. When X equals 1. G 1 is going to be -3 divided by 1, which is -3. So that means the 0.13 is going to be on our tangent line. And then when X equals -2, G of -2 will be -3 divided by -2, which equals 3 divided by 2 or 3 houses. So that means -2 3 halves will also be on our tangent line. Therefore, the points on our tangent line are -2, 3 halves and 1-3. If we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Given the function f and the point Q, find all points P on the graph of f such that the line tangent to f at P passes through Q. Check your work by graphing f and the tangent lines.
f(x) = 1/x; Q (-2, 4)

Key Concepts

Tangent Line

Finding Points on a Graph

Graphing Functions and Tangents

Watch next