General cubic: find condition given slope of tangent at inflection

MethMath11

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Harry_the_cat

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Nice question! What have you tried?
 

Subhotosh Khan

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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?
 

MethMath11

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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

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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

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Tell us the x value at the point of inflection.
 

Subhotosh Khan

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What is the definition of point of inflection of a curve?
 

Subhotosh Khan

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JeffM

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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

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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

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We got the X from Y'' = 0 and the y after you substitute the X with the one you get from Y'' = 0 , CMIIWR
Please show your work mathematically - instead of using words.
 

Harry_the_cat

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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

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MethMath11

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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

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MethMath11

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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

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