Derivatives and tangent lines
b. Determine an equation o...

Question

Derivatives and tangent lines

b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.

f(x) = 1/3x-1; a= 2

Accepted Answer

Welcome back, everyone. In this problem, we want to find the equation of the tangent line to the curve H at the point A H of A, where H X equals 3 divided by 5 X minus 3 and A equals 3. A says the equation is Y equals -54/8 of X plus 9/16. B says it's -548 of X plus 27/16. C 54/8 of X plus 9/16 and D-548 of X minus 9/16. Now what do we know about the equation of a line? Well, recall that the equation of a line is Y equals MX + C, where M represents the slope and C represents the Y intercept. But if we don't have the Y intercept or the slope, we can also write the equation of a line in the point slope form, OK? We can also write it as Y minus Y1 equals M multiplied by X minus X1, OK, where Y1 and X1 are the coordinates of a point and M is the slope. So from our point slope form, if we have a coordinate and the slope of our line, then we should be able to write it in the general form for the line of, for the equation of the line. No, how can we do that? Well, here we know that if we find the derivative of H of X, that will be our slope if we substitute the point where it's at, OK, and then we can substitute an X value into our formula 4 H X to get a Y value, OK, to get both of those points that we need. So let me shade that in blue here actually, because now, let's see if we can find the slope and then the points substitute them to get the equation of the line. So let's start with the slope. Let's start by differentiating H of X. Now here We know H X equals 3 divided by 5 X minus 3. And we could rewrite that, OK. As 3 multiplied by 5 x minus 3 raise to the power of -1. No, to differentiate, we could use the, the, the chain rule, sorry, and the chain rule tells us that the derivative of H of X. To get it first, we'll bring down the power, so that's -1. We'll multiply it by our expression, OK. Sorry, we multiply it by our expression where we've subtracted 1 from the power. So that's 5 X minus 3 raises to the power of -1 minus 1, which is -2, OK. And then we'll multiply that by the derivative of what's inside our expression. And the derivative of 5 X minus 3 is 5. And then don't forget that we had the constant 3 already in our problem. Now when we simplify this expression, it's going to be equal to -15 multiplied by 5 X minus 3 divided by -2, which we could rewrite as -15 divided by 5 X minus 3 squared. This is the derivative of H of X. Now that we have the derivative, that means we can find the tangent, sorry, the, the slope of the tangent line at A or X equals 3, OK. So at X equals 3, OK. Then our tangent of 3. Is going to be equal to -15 divided by 5 X minus 3 squared. So that's gonna be um 5 multiplied by 3 minus 3 all squared. 5 multiplied by 3 is 15 and 15 minus 3 is 12, so that's negative 15 divided by 12 squared. This -15 divided by 144, which if we cancel out the common factor of 3, we can simplify this to -5 divided by 48. This is the slope of the tangent line at X equals 3. So I could also write here that this equals M just just for clarity. Next, we can find a point on the graph using the equation for H of X, OK. So next, OK. At X equals 3, then H of 3. Is going to be equal to 3 divided by 5 multiplied by 3 minus 3. That's 3 divided by 15 minus 3, which is 12, and 3/12 equals a quarter, OK? Therefore, What this means Is that 3. Quarter Is a point on the tangent line. OK. We're here now, we can refer to 3 as X1 and a quarter as Y1. Let's do a quick check. Now, we have our slope, -5408. We have our coordinates X1, which is 3, and Y1, which is a quarter. So we can use the point slope form. And we can substitute these values in the point slope form to get the equation off the line. So now by substituting these values, OK, that means. Y minus Y125 equals M -548 multiplied by X minus X1, which is 3. Now, if we simplify that here, we'll get Y to be equal to -548 X, OK, plus 1548. OK. And oops, sorry, Y minus quarter here. My apologies, should have Y minus quarter here. So we'll get Y minus quarter to be equal to -548 plus 154/8. If we expand negative 548s across our bracket, if we distribute it, OK? And no, if we were to solve for Y then to get the general form, Y equals -548 X plus 154848. Plus a quarter we add a quarter to both sides, and when we simplify that we get Y to be equal to -548 X plus 9/16. So this Is going to be the equation of the tangent line to the curve at A equals 3. If we look back at our answer choices, that tells us that A is the correct answer. Thanks a lot for watching everyone. I hope this video helps.

Derivatives and tangent lines
b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.
f(x) = 1/3x-1; a= 2

Key Concepts

Derivatives

Tangent Line

Point-Slope Form

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