In Exercises 11–18, find the slope of the function’s graph at ...

Question

In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.

h(t) = t³ + 3t, (1, 4)

Accepted Answer

Hello. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. For the given function, determine the slope at the specified point and write the equation of the tangent line at that point. F of X is equal to 5 x 3 minus 18 x 2.4. Awesome. So it appears for this particular problem we're asked to take our given function of F of X and we're asked to solve for two separate answers. We're asked to first determine the slope at the specified point, and our second answer is we're asked to write the equation of the tangent line at this specific point. So now that we know that we're ultimately trying to use our function of F of X to help us find the slope and the tangent line of this particular function, our first step is to rewrite our function, since we need to know what function we're dealing with. So we're told that our function F of X is equal to 5x the power of 3 minus 18 X. And let us also note that we're told that the point, which we'll denote as capital P, our point is parentheses 24. And once again we need to note that we're ultimately trying to determine the slope, which is our first answer. And in order to determine the slope, we need to recall and use the difference quotient method. So as we should recall a note, the difference quotient method states that the limit as H approaches 0 of F of. X0 plus H minus F of X0 all divided by H. Fantastic, and let us know in this particular case that X0 is going to be equal to 2. Because we're told that our point, because as we should note, dealing with coordinate planes in case you've forgotten, the 2 represents the X value and the 4 represents the Y value. So moving right along here. So now our first step towards solving for the slope, we need to compute the values of F of X0 plus H, and we also need to determine F of X0. So, Now that we know we need to compute these values first, our first step is we need to find F of 2. So we're plugging in a value of 2 in for X0. So with that said, we need to take F of 2 is equal to 5 of 2 or actually, so it's gonna be FA 2, so FA 2 is equal to 5 multiplied because now we know that it's 5 multiplied by X to the 3, so when you take 5 multiplied by 2 to the power 3 minus 18 multiplied by 2. Which when we plug that into our calculator, we'll get an answer of 4. So now if we solve for F of 2 plus H, we will find that F of 2 + H is going to be equal to 5 multiplied by 2 + H to the power of 3 minus 18 multiplied by 2 + H. Fantastic. So if we were to expand, so moving right along here, if we were to expand the value of 2 + H to the power of 3. So once again we're trying to expand 2 + H to the power of 3. We will find that it's equal to the following, so it will be equal to 8 + 12 H + 6 H squared plus H to the power of 3. Fantastic. So, If we substitute this value back into our equation, we will find, or rather back into our function equation, we will get F of 2 plus H, it's going to be equal to 5 multiplied by 8 + 12 H plus 6 H quad plus H to the power of 3 minus 18. And we need to take 18, and we need to multiply 18 by 2 plus H. Fantastic. So when we simplify this statement out, which I'm going to skip a few steps, but when we skip it well skipping a few steps, just noting that you're using a little bit of algebra to simplify, or we will find that it's most simple form that 2 of F of 2 plus H, so F of 2 + H, when we write it in our most simple form will be equal to 4. Plus 42 H plus 30 H squared, and then we need to add Plus 5 H to the power of 3. Awesome. So now our next step is we now need to compute the difference quotient, so we need to take the limit as H approaches 0 of F of 2 plus H minus F of 2, all divided by H. So, when we plug in our known variables, we'll find that this is equal to the limit, as H approaches 0 of. 4 + 42 H plus 30H2 + 5 H 3 minus 4. All divided by H. Fantastic. So, when we do a little bit of simplifying, And we cancel out H, which I'm gonna skip a few steps, we will find that when we simplify and we cancel H, we will find that this will be equal to the limit, as H approaches 0 of 42. Plus 30 H plus 5 H squared, and if we take the limit as H approaches 0, we will find that 42 + 30 multiplied by 0. 0 + 5 multiplied by 0 is going to be simply equal to 42, and 42 is our value for the slope. Hooray, we did it. We found our first answer. So once again, our slope has to be 42. And once again, our slope at X equals 2 will be 42. Fantastic. So now, moving right along here to find our tangent line value now. So that's our third step. We need to find the tangent line equation, and to do that, we need to recall and use the point slope form formula. So we need to remember that the point slope forms equation states that Y minus Y1. is equal to the slope M multiplied by X minus X1. And let us also recall once again that M or slope is equal to 42. Let us also note that X1 is equal to 2 and Y1 is going to be equal to 4. So when we plug in our known variables, you will get Y minus 4 is equal to 42 multiplied by X minus 2, and skipping a few steps and rearranging our equation to isolate and solve for Y, you will find that Y is equal to 42. X minus 80, and that is our final answer for the tangent line. Hooray, we did it. So once again, Y is equal to 42 X minus 80. So quickly recapping what we learned together, our final two answers must be Once again, our slope is M is equal to 42, and Y are tangent line, so our tangent line Y is equal to 42X minus 80. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.

h(t) = t³ + 3t, (1, 4)

Key Concepts

Derivative

Tangent Line

Point-Slope Form

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