Tangent line question

[Bree]

Okay, so this is the question:

But I'm not sure how to go about solving it.
What's my first step?

 

[stapel]

Your image is way too tiny (at least on my display) to be viewed. In general, it is better to simply type out the text:

\(\displaystyle \mbox{For the fun}\)\(\displaystyle \mbox{tion }\, y\, =\, ax^3\, +\, bx^2\, +\, 2,\, \mbox{ where }\)

\(\displaystyle y'(1)\, =\, 9\, \mbox{ and }\, y"(1)\, =\, 12,\, \mbox{ the equa}\)\(\displaystyle \mbox{tion of the line}\)

\(\displaystyle \mbox{that is tangent to the func}\)\(\displaystyle \mbox{tion at its inflection point is:}\)

As for solving, it would probably be wise to start by finding the inflection point. ;-)

 

[srmichael]

Okay, so this is the question:
View attachment 3433
But I'm not sure how to go about solving it.
What's my first step?

Bree, sometimes when you don't know where to start, simply do what you know you have to eventually do. In this case, you will need to at least find the first and second derivative. Then you will notice that you will have to find the values of "a" and "b" so how can we do this? Well, they gave you some information that may help you, right?

Take a stab and then come back to us if you hit a roadblock.

 

[Bree]

Bree, sometimes when you don't know where to start, simply do what you know you have to eventually do. In this case, you will need to at least find the first and second derivative. Then you will notice that you will have to find the values of "a" and "b" so how can we do this? Well, they gave you some information that may help you, right?

Take a stab and then come back to us if you hit a roadblock.

Okay, so I did like you suggested and took the first and second dreivative:

y'=3ax²+2bx
y''=6ax+2b

And you said I'd have to find the values of a and b, but I'm not sure how to do this with the information they gave me.

 

[srmichael]

Okay, so I did like you suggested and took the first and second dreivative:

y'=3ax²+2bx
y''=6ax+2b

And you said I'd have to find the values of a and b, but I'm not sure how to do this with the information they gave me.

Correct on your derivative calculations!

Well, they say y'(1) = 9 and y''(1) = 12. This can also be represented as coordinates, namely (1, 9) for the first derivative and (1, 12) for the second derivative. If you plug these values into the approriate derivatives you will have a system of equations in terms of "a" and "b" and can thus solve for "a" and "b".

 

[pka]

Okay, so I did like you suggested and took the first and second dreivative:
y'=3ax²+2bx
y''=6ax+2b
And you said I'd have to find the values of a and b, but I'm not sure how to do this with the information they gave me.

From the qiven
\(\displaystyle \begin{align*} y'(1)=9&=3a+2b\\y''(1)=12 &=6a+2b\end{align*}\)

Solve for \(\displaystyle a~\&~b~.\)

 

[Bree]

Correct on your derivative calculations!

Well, they say y'(1) = 9 and y''(1) = 12. This can also be represented as coordinates, namely (1, 9) for the first derivative and (1, 12) for the second derivative. If you plug these values into the approriate derivatives you will have a system of equations in terms of "a" and "b" and can thus solve for "a" and "b".

Here are my calculations for a and b by using a system of equations:

(9)=3a(1)²+2b(1)
9=3a+2b
b=-3/2(a-3)

(12)=6a(1)+2b
12=6a+2b
12=6a+2(-3/2(a-3))
12=6a-3(a-3)
12=6a+9_3a
12=3a+9
3a=12-9
a=1

What do I have to do next?

 

[srmichael]

Here are my calculations for a and b by using a system of equations:

(9)=3a(1)²+2b(1)
9=3a+2b
b=-3/2(a-3)

(12)=6a(1)+2b
12=6a+2b
12=6a+2(-3/2(a-3))
12=6a-3(a-3)
12=6a+9_3a
12=3a+9
3a=12-9
a=1

What do I have to do next?

I was hoping that as you started doing this problem the light bulb would go on and you would see how to eventually do this problem, but it does not seem that way. It looks like we would have to eventually tell you how to to do the entire problem which would not do you any good because you would not be learning anything. Has your teacher not explained this concept very well? You may want to get some classroom help so you can grasp the concept and then be able to tackle problems like this one on, oh I don't know, your test.

 

[Bree]

I was hoping that as you started doing this problem the light bulb would go on and you would see how to eventually do this problem, but it does not seem that way. It looks like we would have to eventually tell you how to to do the entire problem which would not do you any good because you would not be learning anything. Has your teacher not explained this concept very well? You may want to get some classroom help so you can grasp the concept and then be able to tackle problems like this one on, oh I don't know, your test.

I've tried getting help from my teacher, but he said he was too busy.

I'm guessing that the next step I have to take in this question is find the inflection point and, once I have found that, plug the values for a and b and the inflection point into the original equation and that will give me the equation of the tangent line. Is that even close?

 

[DrPhil]

I've tried getting help from my teacher, but he said he was too busy.

I'm guessing that the next step I have to take in this question is find the inflection point and, once I have found that, plug the values for a and b and the inflection point into the original equation and that will give me the equation of the tangent line. Is that even close?

Reviewing what I think you know so far:

\(\displaystyle y = a\ x^3 + b\ x^2 + 3\)

From values of y'(1) and y''(1), you found \(\displaystyle a=1\).
Did you plug that back into one of the equations to get \(\displaystyle b=3\)? Then

\(\displaystyle y = x^3 + 3\ x^2 + 3\)

\(\displaystyle y' = 3\ x^2 + 6\ x\)

\(\displaystyle y'' = 6\ x + 6\)

That takes care of plugging in a and b. Now find the inflection point. Also find the slope at the inflection point. Use that point and slope to get the equation of the tangent line.

 

[Bree]

Reviewing what I think you know so far:

\(\displaystyle y = a\ x^3 + b\ x^2 + 3\)

From values of y'(1) and y''(1), you found \(\displaystyle a=1\).
Did you plug that back into one of the equations to get \(\displaystyle b=3\)? Then

\(\displaystyle y = x^3 + 3\ x^2 + 3\)

\(\displaystyle y' = 3\ x^2 + 6\ x\)

\(\displaystyle y'' = 6\ x + 6\)

That takes care of plugging in a and b. Now find the inflection point. Also find the slope at the inflection point. Use that point and slope to get the equation of the tangent line.

Thank you! That all made sense and I was able to figure out the equation.