#### MethMath11

##### Junior Member

- Joined
- Mar 29, 2019

- Messages
- 73

- Thread starter MethMath11
- Start date

- Joined
- Mar 29, 2019

- Messages
- 73

- Joined
- Mar 16, 2016

- Messages
- 1,764

Nice question! What have you tried?

- Joined
- Mar 29, 2019

- Messages
- 73

- Joined
- Jun 18, 2007

- Messages
- 19,980

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

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

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

- Joined
- Mar 29, 2019

- Messages
- 73

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 partYou 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?

- Joined
- Mar 29, 2019

- Messages
- 73

Is it the y-y1 = m(x-x1) ?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?

- Joined
- Jun 18, 2007

- Messages
- 19,980

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

- Joined
- Jun 18, 2007

- Messages
- 19,980

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

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.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

- Joined
- Mar 29, 2019

- Messages
- 73

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

- Joined
- Jun 18, 2007

- Messages
- 19,980

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

- Joined
- Mar 29, 2019

- Messages
- 73

- Joined
- Mar 16, 2016

- Messages
- 1,764

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?

- Joined
- Mar 16, 2016

- Messages
- 1,764

You could use this to find theIs it the y-y1 = m(x-x1) ?

- Joined
- Mar 29, 2019

- Messages
- 73

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

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?

- Joined
- Mar 16, 2016

- Messages
- 1,764

Sorry that last line should have been x = -b/3a not 2aIs that -b/2a as a formula or what? Sorry

(I've edited my previous post)

- Joined
- Mar 29, 2019

- Messages
- 73

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

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?

f'(-b/3a) =( b^2)/3a -( 2b^2)/3a + c

Somehow I think m1 = m2, is it true?

- Joined
- Mar 16, 2016

- Messages
- 1,764

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

- Joined
- Mar 29, 2019

- Messages
- 73

Already got the answer needed, thank youAnd you also know the gradient at the POI is 2c (you are told that), so ....