Show that the line y = mx + b is its own tangent line at any p...

Question

Show that the line y = mx + b is its own tangent line at any point (x₀, mx₀ + b).

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the equation of the tangent line to Y is equal to 11 X minus 13 at the point of 2.9. So we need to determine the equation of the tangent line to our curve Y at this particular point. So recall that the slope of your tangent line at a particular point, which we'll label as m tangent. is equal to the limit. As H approaches 0 of the quantity of F of X plus H, Minus F of X in quantity, divided by H. So here we need to identify a couple of quantities. First off, F of X is going to be equal to. R 11 X minus 13. And the X value that we're interested in is going to be 2. So, our X value is going to be equal to 2. That also means that F of XX, which is going to be now F of 2, is going to be equal to 9, since that is the Y coordinate for our point. In order to use our formula, we'll also need to calculate F of 2 + H. And so this is going to be equal to 11, multiplied by the quantity of 2 + H in quantity, minus 13. So, distributing the 11, this is going to be equal to 22 + 11H minus 13, and simplifying this expression, this is going to be equal to 11 H plus 9. So now we can substitute all this information into our formula for the slope of our tangent line. So that means in tangent is going to be equal to the limit, as H approaches 0 of the quantity of F of 2 + H, which is 11H. Plus 9. Minus F of 2, which is 9 in quantity divided by H. So simplifying this expression, we see that we have 9 minus 9, which add up to 0. And what's left is going to be 11 H divided by H, so I can cancel out the factor of H. And so, I will have our M tangent equal to the limit, as H approaches 0 of 11. So, here I'm taking the limit of a constant, and therefore I get back just that constant. So that means that Mtangent is equal to 11. So now we need to find the equation of the tangent line, now that we have the slope, and since we were given a point, we can actually use the point slope form of the equation for a line. So we call for the point slope form that we're gonna have Y. Minus our Y value, which in this case is 9, and this could be equal to our slope, which is 11, multiplying the quantity of X minus our X value, which is 2. In quantity So we'll distribute the 11 on the right-hand side, so we'll have Y minus 9, is equal to 11 X minus 22. Now adding 9 to both sides to solve for Y, so we'll have Y is equal to 11 X minus 13. So this here is going to be the answer to this problem. And notice that we actually got back our original equation. And that's because our original equation was already an equation for in a line. So, if we always start with, if we start with the linear equation, our equation for the tangent line is always going to be equal to our original equation for any point on that line. So I hope that this explanation has been useful, and I'll see you in the next video.

Show that the line y = mx + b is its own tangent line at any point (x₀, mx₀ + b).

Key Concepts

Tangent Line

Derivative

Linear Functions

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