General cubic: find condition given slope of tangent at inflection

MethMath11 said:


Yes and I'm stuck
Click to expand...
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?
 
Subhotosh Khan said:
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?
Click to expand...
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
 
Subhotosh Khan said:
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?
Click to expand...
Is it the y-y1 = m(x-x1) ?
 
Tell us the x value at the point of inflection.
 
What is the definition of point of inflection of a curve?
 
MethMath11 said:
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
Click to expand...
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.
 
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 said:
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?
Click to expand...
Is that -b/2a as a formula or what? Sorry
 
Harry_the_cat said:
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?
Click to expand...
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?
 
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