what is going on here? (find a function of the form ax^2 +bx+c given certain information relating...)

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allegansveritatem

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This is the solution to a problem that asked the student to find a function of the form ax^2 +bx+c given certain information relating to the slopes at certain values of x. Also a set of coordinates was given . Anyway, I worked out the problem and managed with the help of the solutions manual (I already had most of the solution on my own) to get the function, namely:solution.PNG
I worked out the derivative and found the function for a tangent line, as follows:
tangent.PNG
This was not asked for in the problem but I wanted to see what a graph of the tangent would look like. Here is what it looked like:
graph solution.PNG
Obviously this is not the picture I was expecting. But I can't for the life of me see what is wrong. 6x-2 should yield a tangent line here and it doesn't. Can someone tell me why? I apologize for not including a photo of the actual problem involved but I didn't feel it was necessary in this case and so I neglected to make one. If anyone feels they need to see the problem I will post it tomorrow. My whole dilemma is this: Given the function -2/3 x^2 + 14/3 x why did I not get a tangent line with the function 6x-2? By the way, the point of tangency is (1,4)
 
d/dx [-(2/3)x^2 + (14/3)x] = -(4/3)x + 14/3 Please put does parenthesis. Without them, it completely confused me for a good 30 seconds.

I don't see why f'(-1) = 6.
Fix that and post back.
 
allegansveritatem said:
This is the solution to a problem that asked the student to find a function of the form ax^2 +bx+c given certain information relating to the slopes at certain values of x. Also a set of coordinates was given . Anyway, I worked out the problem and managed with the help of the solutions manual (I already had most of the solution on my own) to get the function, namely:View attachment 36159
I worked out the derivative and found the function for a tangent line, as follows:
View attachment 36160
This was not asked for in the problem but I wanted to see what a graph of the tangent would look like. Here is what it looked like:
View attachment 36161
Obviously this is not the picture I was expecting. But I can't for the life of me see what is wrong. 6x-2 should yield a tangent line here and it doesn't. Can someone tell me why? I apologize for not including a photo of the actual problem involved but I didn't feel it was necessary in this case and so I neglected to make one. If anyone feels they need to see the problem I will post it tomorrow. My whole dilemma is this: Given the function -2/3 x^2 + 14/3 x why did I not get a tangent line with the function 6x-2? By the way, the point of tangency is (1,4)
Click to expand...
Your work is full of big and little errors; here are corrections:

1690167888679.png​

The big error, which I couldn't correct, is in yellow: You used 1 to find y, but -1 to find y'. So your line is not going to be tangent at your point.

Lesson: when your check fails, check your work carefully, before you ask someone else to do it. And writing what you mean makes that easier.
 
Steven G said:
d/dx [-(2/3)x^2 + (14/3)x] = -(4/3)x + 14/3 Please put does parenthesis. Without them, it completely confused me for a good 30 seconds.

I don't see why f'(-1) = 6.
Fix that and post back.
Click to expand...
Thanks for your reply. I am going to work on this again today to make sure I used the right value. I will also take a photo of the problem and post it.
 
Dr.Peterson said:
Your work is full of big and little errors; here are corrections:


The big error, which I couldn't correct, is in yellow: You used 1 to find y, but -1 to find y'. So your line is not going to be tangent at your point.

Lesson: when your check fails, check your work carefully, before you ask someone else to do it. And writing what you mean makes that easier.
Click to expand...
Sorry, the work was sloppy. I will go at it again today and also take a photo of the problem. In the problem it was stated what the slope was at two values of x. I may have mixed things up somehow. I will work this all out again today and post results. Thanks very much for pointing out the inconsistencies in the above.
 
Here is the problem I am working with:
problem solution.PNG
In the process of solving or trying to solve this I looked at the solutions manual and understood that they way to go was with a system of equations. I then found the the function asked for. Then, I wanted to find the linear functions of the tangents so I could see how they looked on a graph and I proceeded thus:

solution tangents.PNG
I first dealt only with the 6x-2 tangent but today I found the -2x +6 and graphed them both together thus:
graph two tangents.PNG
And so again I ask: what is going on that neither of these lines is tangential (to coin a phrase?). Where am I going wrong?
 
allegansveritatem said:
Here is the problem I am working with:
View attachment 36166
In the process of solving or trying to solve this I looked at the solutions manual and understood that they way to go was with a system of equations. I then found the the function asked for. Then, I wanted to find the linear functions of the tangents so I could see how they looked on a graph and I proceeded thus:

View attachment 36167
I first dealt only with the 6x-2 tangent but today I found the -2x +6 and graphed them both together thus:
View attachment 36168
And so again I ask: what is going on that neither of these lines is tangential (to coin a phrase?). Where am I going wrong?
Click to expand...
First, your answer is correct: the quadratic function you found has the right tangents at the indicated points.

What you did wrong is in the part they didn't ask for, finding the equations of the tangent lines as a check of the solution. You made lines through (1, 4) with the two given slopes, rather than a line through (-1, ?) with slope 6, and a line through (5, ?) with slope -2. If you find those two lines, they will be tangent to your curve.
 
There are three unknowns, a, b and c, and you are given three pieces of information. YES, you need to solve a system of equations.
 
Dr.Peterson said:
First, your answer is correct: the quadratic function you found has the right tangents at the indicated points.

What you did wrong is in the part they didn't ask for, finding the equations of the tangent lines as a check of the solution. You made lines through (1, 4) with the two given slopes, rather than a line through (-1, ?) with slope 6, and a line through (5, ?) with slope -2. If you find those two lines, they will be tangent to your curve.
Click to expand...
Thanks for the reply. I will have to ponder what you are saying here. I don't see clearly what you are getting at but I have a vague sense of it. I will work on this again tomorrow keeping in mind this post and get back here with my results.
 
Steven G said:
There are three unknowns, a, b and c, and you are given three pieces of information. YES, you need to solve a system of equations.
Click to expand...
Right. After I found a and b I found c by plugging 1 into the equation of the parabola and setting y to 4. Working this out I got 0 for c.
 
allegansveritatem said:
Thanks for the reply. I will have to ponder what you are saying here. I don't see clearly what you are getting at but I have a vague sense of it. I will work on this again tomorrow keeping in mind this post and get back here with my results.
Click to expand...
Read the problem again!

1690320957216.png

The tangent lines are at x = -1 and x = 5; so find the points (-1, ???) and (5, ???) on the curve (that is, evaluate f(-1) and f(5), and use those points (not (1, 4)) to find the equations of the lines.
 
Dr.Peterson said:
Read the problem again!

View attachment 36174

The tangent lines are at x = -1 and x = 5; so find the points (-1, ???) and (5, ???) on the curve (that is, evaluate f(-1) and f(5), and use those points (not (1, 4)) to find the equations of the lines.
Click to expand...
Yes. I started thinking about your post at three in the morning--it doesn't cure insomnia but puts it to some use anyway--and in a flash saw the light! What I had been doing was using the right coordinates with the wrong slope. Exactly what I was thinking about when I did that I'm not sure...probably just didn't digest the problem thoroughly. I knew I was doing something maximally dumb but...Anyway here is an image of finding the correct slope for the points (1,4).:
correcttangent equation.PNG
and here is the graph:
correct tangent.PNG
Thanks for the help.
 
allegansveritatem said:
Yes. I started thinking about your post at three in the morning--it doesn't cure insomnia but puts it to some use anyway--and in a flash saw the light! What I had been doing was using the right coordinates with the wrong slope. Exactly what I was thinking about when I did that I'm not sure...probably just didn't digest the problem thoroughly. I knew I was doing something maximally dumb but...Anyway here is an image of finding the correct slope for the points (1,4).:
View attachment 36177
and here is the graph:
View attachment 36178
Thanks for the help.
Click to expand...
No, you were using the wrong coordinates with the right slopes, as I told you at least twice.

1690411408435.png​

The problem asks you to make a function that will have specified slopes at x=-1 and x=5, not at x=1. So if your goal was to see that your function fits that description, then you should not be finding the tangent line at (1,4), but the tangent lines shown here, which have the required slopes:

1690411777687.png​

But congratulations on successfully doing what you chose to do instead, which is unrelated to the problem.
 
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