# Slopes of tangents of a cubic function

• fawk3s
In summary, the conversation was about a problem involving a cubic function with only one tangent with a slope of 4, and the point of tangency being at x=-1/3. The problem raised a question about whether a cubic function should have two tangents with the same slope in every two corresponding points on the graph, except for the extremum point. The conversation then shifted to solving the problem and finding the values of a, b, and c. It was eventually discovered that the function must have a discriminant of 0 to have only one solution for y'=4, and this was the missing piece to correctly solving the problem.
fawk3s
Recently I came across this problem where it was stated:

"A cubic function y=ax3+bx2+cx+1 has only one tangent which has the slope of 4, and which touches the graph in x=-1/3"

I did the best I could to translate it into English. The problem went on ofcourse, but this is the part which raised the following question in my head:
shouldnt a cubic function like this have 2 tangents with the same slope in every 2 according points on the graph, EXCEPT for the extremum point?
Where do I go wrong?

When you solved the problem, you got that y=3x3+9x2+9x+1
You can easily draw it out on http://rechneronline.de/function-graphs/

hi fawk3s!
fawk3s said:
"A cubic function y=ax3+bx2+cx+1 has only one tangent which has the slope of 4, and which touches the graph in x=-1/3"

When you solved the problem, you got that y=3x3+9x2+9x+1

but that doesn't go through x = 1/3

I think you missed the minus there

fawk3s said:
I think you missed the minus there

ooh, yes

that doesn't go through x = -1/3

Now I am confused. What doesn't go through -1/3?

y=3x3+9x2+9x+1

If x=-1/3, then y=3*(-1/3)3+9*(-1/3)2+9*(-1/3)+1=-10/9
So the tangent is in the point (-1/3; -10/9)

It does, why shouldn't it?

yes, x = -1/3 would have to be an inflexion point, and have a slope of 4 …

how did you get that result?​

I did mention above that I didnt write out all of the problem. So you are missing some info here. The problem goes:

"There's a cubic function y=ax3+bx2+cx+1, which has only one tangent with a slope of 4, and that tangent touches the given function in x=-1/3. It is also known that the cubic function has extremum in x=-1. Find a,b and c."

Im not really familiar with all the terms in English so I don't know if my translation makes sense.

Anyways, I solved it by constructing a "function system" (dont know how its called in English ), which contains
y'(-1/3)=4
y'(-1)=0

And I had trouble getting that last third function to complete that "function system", so I used
y''(-1)=0

because I assumed that the funcion has only one extremum point and that its neither maximum nor minimum point. This actually confuses me, because say you take a function like
y=3x3+10x2+9x+1

and you get 2 extremums, of which one is maximum, and the other minimum.

I think I got lucky, because the given answer is a=3, b=9, c=9. But I am rather confused if I solved it correctly, because I made that assumption.But now to my real question:
doesnt a function like y=3x3+9x2+9x+1 have 2 tangents with identical slopes, except for the one in the extremum (assuming there's only one extremum)?

Im rather confused here acutally.

fawk3s said:
… And I had trouble getting that last third function to complete that "function system", so I used
y''(-1)=0

I think I got lucky, because the given answer is a=3, b=9, c=9.

shouldn't it be y''(-1/3)=0 ?

(and i get a = b = -c = 6 )

No, by my logic, the extremum, which is in x=-1, is neither a minimum nor a maximum, and therefore
y''(-1)=0

Im not sure if this is correct though since it involves an assumption that the given extremum is the only extremum the function has. I'll ask my teacher tomorrow for confirmation.
As I can see you have taken interest in the problem like I have, so I'll be sure to post what my teacher teacher will say tomorrow.

But what do you think about my original question, the one I created this thread for?

fawk3s

fawk3s said:
No, by my logic, the extremum, which is in x=-1, is neither a minimum nor a maximum, and therefore
y''(-1)=0

at an extremum, y' is 0, not y''

tiny-tim said:

at an extremum, y' is 0, not y''

Maybe my logic does fail here. I'll try and explain:

Say we have an extremum in x. So
y'(x)=0

If y''(x)>0, then the extremum is a minimum.
If y''(x)<0, then the extremum is a maximum.

Say for a function like y=x3, there is only one point where y'(x)=0. But this point is neither a minimum or maxiumum (take a look at the graph). So I figured that if
y''(x) cannot be bigger or smaller than 0, then it must be 0.
Therefore y''(x)=0

Is this a fail in my logic?

It did give me the right answer, so I figured it must be true, but I am not really satisfied with my solution, because it does involve an assumption about the extremum. I think I am missing something easier and more logical here which I could use as the third function.

fawk3s said:
if y''(x) cannot be bigger or smaller than 0, then it must be 0.

but i don't see where your premise comes from

I indeed have to admin I was wrong. My version holds true for the function I brought out, but you really can't make an assumption like that based on nothing - if it solves true, its just blind luck. Plus, my function did not hold true for the given fact that the function has only one tangent with the slope of 4. So I failed to solve it correctly.

It was easy to solve actually, but I missed something fairly simple. Since the problem states that the function has one and only one tangent with a slope of 4, then y'=4 has only one solution, which means the discriminant D=0.
So when y'=3ax2+2bx+c, then
3ax2+2bx+c=4
3ax2+2bx+c-4=0

Solving for x,
x=-2b+-sqrt[4b2-4*3a*(c-4)]/2*3a
from which D=0 is
4b2-4*3a*(c-4)=0

which would be the last function needed.

I also missed the obvious fact that you were indeed correct.
y''(-1/3)=0 also holds beautifully true.

You keep saying "the extremum, which is in x=-1, is neither a minimum nor a maximum" but an 'extremum', by definition, must be either a minimum or a maximum. A point where both y'= 0 and y''= 0 may be an inflection point but NOT an "extremum".

HallsofIvy said:
You keep saying "the extremum, which is in x=-1, is neither a minimum nor a maximum" but an 'extremum', by definition, must be either a minimum or a maximum. A point where both y'= 0 and y''= 0 may be an inflection point but NOT an "extremum".

Im sorry, that's what I meant actually, but I didnt know the term in English.

## What is a cubic function?

A cubic function is a polynomial function of degree 3, meaning it has the form f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants and x is the variable.

## What is the slope of a tangent line?

The slope of a tangent line is the instantaneous rate of change of a function at a specific point. It represents the steepness of the curve at that point.

## How do you find the slope of a tangent line for a cubic function?

To find the slope of a tangent line for a cubic function, you can use the derivative of the function. The derivative of a cubic function is a quadratic function, which can be used to find the slope at any point on the original cubic function.

## What does the slope of a tangent line tell us about a cubic function?

The slope of a tangent line tells us about the rate of change of the cubic function at a specific point. It can also indicate the direction in which the function is increasing or decreasing at that point.

## Can the slope of a tangent line be negative for a cubic function?

Yes, the slope of a tangent line can be negative for a cubic function. This would indicate that the function is decreasing at that point. It is also possible for the slope to be zero, indicating a point of inflection on the cubic function.

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