Calculus Problem ***REALLY IMPORTANT*** Please Help!

Test Preparation
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<p>Sorry, I didn’t know where I should post my question. I’m in HL Math (IB) and we’re working on an IA assignment and I have no idea how to get started.</p>

<li>Consider the cubic function f(x)= 2x^3 + 6x^2 - 4.5x - 13.5. Graph this function and find an appropriate window so that the graph will look like this: (shows graph)</li>
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<p>Record your window values. State the roots of this cubic and confirm using the Remainder Theorem. Then, taking the roots two at a time, find the equations of the tanget lines to the average of two of the three roots. Find where the tangent lines at the average of the two roots intersect the curve again. Does this observation hold regardless of which two roots you averae? State a conjecture concerning the rots of the cubic and the tangent lines at the average value of these roots.</p>

<li>Test your conjecture in other similar cubic functions.</li>
<li>Prove your conjecture</li>
<li>Investigate the above properties with cubic functions which have:
(a.) one root
(b.) two roots
(c.) one real and two complex roots.</li>
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<p>Please PM me or respond. If someone really helps me out I will give them all QAS’s, Grammatix, Online Courses, Xiggi’s Method…so on and so forth. PLEASE HELP…I’m DYING!!! lol</p>

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<p>you might wanna post that in cramster.com</p>

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<p>1) The roots of f(x): x = -3, -1.5, and 1.5</p>

<p>Remainder Theorem Formula:
f(x)= q(x)g(x) + r(x) where q, g, and r are the the quotient, divisor, and remainder.
You divide f(x) by x+3, and you are left with a quadratic 2x^2-4.5.</p>

<p>f(x)= (2x^2-4.5)(x+3) + 0</p>

<p>When f(3), it is 0; r(3) is also 0.</p>

<p>Finding the equations of the tangent lines to the average of two of the three roots:
-3 and -1.5: average is x=-2.25 and the tangent line is y= -1.125x+1.6785.
-1.5 and 1.5: average is x=0 and the tangent line is y= -4.5x-13.5.
-3 and 1.5: average is x=-0.75 and the tangent line is y= -10.125x-15.1875.</p>

<p>That’s as far as I got…I can’t really find a conjecture. Maybe I’m reading the problem wrong or something.</p>

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<p>Oh! I see! When you see where the tangent lines and the curve intersect, they intersect at the maximum, minium, and at the two outer x-intercepts.</p>

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<p>The tangent lines don’t intersect at the local maximum and minimum of the graphs, only the outer roots (x-intercepts) of f(x).</p>

<p>Example:
f(x) and tangent line y= -10.125x-15.1875 intersect at (-2.22, 4.17), but the local maximum of f(x) is (-2.33, 4.26).</p>

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<p>right, right, right. I found that out after doing a bit more work.</p>

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<p>I’m still incredibly confused.</p>

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<p>That’s a tough calc problem :slight_smile:
I got a 5 on the AB exam and have no clue how to do that problem :)</p>

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<p>Omg, I Need Help With This Problem!!! Please Help!!!</p>

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<p>I could help out, but it’s too late at night to do so, I’ll probably make a mistake, plus it’ll probably take an hour to type up. I’ll see if I can do something about it in about 18 hours.</p>

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<p>You could kind of cheat by just pressing Zoom->Zoom Fit. I know it isn’t correct but maybe your teacher has a sense of humor.</p>

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<p>jask925, do you still need help? I’ll try to figure out the solution today and then post it in about 5 hours</p>

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<p>ewww…I remember doing this last year and am SOOOOO happy that I never have to take another math class EVER AGAIN!!! good luck finishing IB…the exams are much easier than the IAs…well, actually I cried after the IB math exam (i somehow got a 5, though…), but the rest of the exams were pretty easy. yay for 7s in English and Spanish! :slight_smile: </p>

<p>o and sorry about being so off topic…this problem just brought back horrible IB memories…good luck!!!</p>