# General cubic: find condition given slope of tangent at inflection

#### Harry_the_cat

##### Senior Member
Nice question! What have you tried?

• topsquark

#### Subhotosh Khan

##### Super Moderator
Staff member

Yes and I'm stuck
You must have "started" to get stuck. Where exactly are you stuck? Do you understand the terms:

cubic function, slope of the tangent and point of inflection?

If yes - how the"slope of a tangent" and "point of inflection" are related?

• JeffM

#### MethMath11

##### Junior Member
You must have "started" to get stuck. Where exactly are you stuck? Do you understand the terms:

cubic function, slope of the tangent and point of inflection?

If yes - how the"slope of a tangent" and "point of inflection" are related?
The relation between gradient and point of inflection, if there's any link for me to learn, please do ask. So far, my teacher only told me how to get the (x,y) of a point of inflection, but never told me that part

#### MethMath11

##### Junior Member
You must have "started" to get stuck. Where exactly are you stuck? Do you understand the terms:

cubic function, slope of the tangent and point of inflection?

If yes - how the"slope of a tangent" and "point of inflection" are related?
Is it the y-y1 = m(x-x1) ?

#### pka

##### Elite Member
Tell us the x value at the point of inflection.

#### Subhotosh Khan

##### Super Moderator
Staff member
What is the definition of point of inflection of a curve?

#### Subhotosh Khan

##### Super Moderator
Staff member
Is it the y-y1 = m(x-x1) ?
Please define y1 and x1 and m in the context of the problem given.

#### JeffM

##### Elite Member
The relation between gradient and point of inflection, if there's any link for me to learn, please do ask. So far, my teacher only told me how to get the (x,y) of a point of inflection, but never told me that part
You have not answered the questions posed by Subhotosh. He is trying to get you to think about applying what you have learned to a new situation.

#### MethMath11

##### Junior Member
Please define y1 and x1 and m in the context of the problem given.
We got the X from Y'' = 0 and the y after you substitute the X with the one you get from Y'' = 0 , CMIIWR

#### Subhotosh Khan

##### Super Moderator
Staff member
We got the X from Y'' = 0 and the y after you substitute the X with the one you get from Y'' = 0 , CMIIWR

#### Harry_the_cat

##### Senior Member
OK so far so good.
Although you don't need to find the y-value of the point of inflection so your last line is unnecessary to answer the question.

You have correctly found the x-coordinate of the POI ie x = -b/3a.

You are told that the slope of the tangent at the POI is 2c.

The slope of the tangent at any point is found by substituting the x-value into the expression for y'.

So forgetting about the 2c at the moment, what is the slope of the tangent when x = -b/2a ?

Then what do you do?

#### Harry_the_cat

##### Senior Member
Is it the y-y1 = m(x-x1) ?
You could use this to find the equation of the tangent but you're not asked for that.

#### MethMath11

##### Junior Member
OK so far so good.
Although you don't need to find the y-value of the point of inflection so your last line is unnecessary to answer the question.

You have correctly found the x-coordinate of the POI ie x = -b/3a.

You are told that the slope of the tangent at the POI is 2c.

The slope of the tangent at any point is found by substituting the x-value into the expression for y'.

So forgetting about the 2c at the moment, what is the slope of the tangent when x = -b/2a ?

Then what do you do?
Is that -b/2a as a formula or what? Sorry

#### Harry_the_cat

##### Senior Member
Is that -b/2a as a formula or what? Sorry
Sorry that last line should have been x = -b/3a not 2a
(I've edited my previous post)

#### MethMath11

##### Junior Member
OK so far so good.
Although you don't need to find the y-value of the point of inflection so your last line is unnecessary to answer the question.

You have correctly found the x-coordinate of the POI ie x = -b/3a.

You are told that the slope of the tangent at the POI is 2c.

The slope of the tangent at any point is found by substituting the x-value into the expression for y'.

So forgetting about the 2c at the moment, what is the slope of the tangent when x = -b/2a ?

Then what do you do?
The slope of you substitute f'(x)= 3ax^2 + 2bx + c. X = -b/3a
f'(-b/3a) =( b^2)/3a -( 2b^2)/3a + c

Somehow I think m1 = m2, is it true?

#### Harry_the_cat

##### Senior Member
And you also know the gradient at the POI is 2c (you are told that), so ....

#### MethMath11

##### Junior Member
And you also know the gradient at the POI is 2c (you are told that), so ....
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