#### MethMath11

##### Junior Member

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- Thread starter MethMath11
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Nice question! What have you tried?

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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?

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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 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?

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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?

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

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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

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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 , CMIIWRPlease define y1 and x1 and m in the context of the problem given.

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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

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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?

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

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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?

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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)

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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?

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

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