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tdddd:
for x^3 + 2x^2 + ax + b how do I:
Find the value of a so that there is stationary point(s)
and find the values of a and b so that the points of inflection are the x intercepts

Thank you!

keltingmeith:

--- Quote from: tdddd on June 15, 2020, 11:17:25 am ---for x^3 + 2x^2 + a + b how do I:
Find the value of a so that there is stationary point(s)
and find the values of a and b so that the points of inflection are the x intercepts

Thank you!

--- End quote ---

Interesting question, which I think you might have copied wrong. What are you struggling with specifically?

tdddd:

--- Quote from: keltingmeith on June 15, 2020, 11:28:12 am ---Interesting question, which I think you might have copied wrong. What are you struggling with specifically?

--- End quote ---

Nope thats the question, struggling with the entire thing

tdddd:

--- Quote from: keltingmeith on June 15, 2020, 11:28:12 am ---Interesting question, which I think you might have copied wrong. What are you struggling with specifically?

--- End quote ---

edit: ax

keltingmeith:
Well then, how about I give you some tips to help you out:

Remember that a curve will have stationary points when its derivative equals 0. So, if you differentiate the curve they've given to you, when will it equal 0? What do you require for it to equal 0?

For the next part, remember that a point of inflection occurs when the SECOND derivative is 0 (you can think of it as the maximum/minimum gradient. When does a function have a maximum? When its derivative is 0. So when is the gradient a maximum? When the gradient function's [that is, the derivative of the function's] derivative is 0). So, when is the second derivative 0?

Finally, remember that b is just a translation factor - all it will do is move the graph up and down. It's not going to affect the gradient of the graph at all (question to think about: how do you know this?)

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