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October 22nd, 2016, 06:10 AM   #1
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Nature of Roots for Quadratic and Cubic Functions

Hi

I am writing my final Mathematics exams for Grade 12 in South Africa in 5 days. I am well prepared with an aim of getting 100%, but one concept in functions might prevent that - the concept of how the nature of roots are affected by vertical/horizontal shifts in a function, and how to determine the values of the shift to obtain the required roots. I attach 3 example questions...
They are 8.1.4, 7.4 and 4.5

Please help me find Youtube Videos/Websites or any resource that might help me understand how to approach these questions.

Thanks
Attached Images
File Type: jpg Screenshot 2016-10-22 15.59.29.jpg (16.3 KB, 9 views)
File Type: jpg Screenshot 2016-10-22 15.59.57.jpg (9.4 KB, 7 views)
File Type: jpg Screenshot 2016-10-22 16.09.30.jpg (10.4 KB, 4 views)

Last edited by GardeeZak; October 22nd, 2016 at 06:13 AM.
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October 22nd, 2016, 06:57 AM   #2
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You could use the discriminant. For example, for the leftmost question, you can find the discriminant of f(x) - g(x) - k, set it as > 0 and solve the inequality.
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October 22nd, 2016, 07:45 AM   #3
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Quote:
Originally Posted by 123qwerty View Post
You could use the discriminant. For example, for the leftmost question, you can find the discriminant of f(x) - g(x) - k, set it as > 0 and solve the inequality.

At our level we have only been taught how to use the discriminant for quadratic graphs - the discriminant for a cubic graph is not examinable
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October 22nd, 2016, 09:57 AM   #4
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We have no idea what math you already know. Do you know derivatives?

Also, these thumbnails are very difficult to read. I have no idea what two of them say at all and must guess for some parts of the first problem.

So I think the first problem starts with $f(x) = x^3 + px^2 + qx - 12.$

Is that correct?

And it has a local maximum at (- 4, 36). Is that correct?

If all that is correct $f'(x) = 3x^2 + 2px + q.$ Furthermore,

$f(-\ 4) = 36 = (-\ 4)^3 + p(-\ 4)^2 + q(-\ 4) - 12 \implies 36 = 16p - 4q - 64 - 12 \implies$

$16p = 112 + 4q \implies 4p = 28 + q.$

And finally $f'(-\ 4) = 0 = 3(-\ 4)^2 + 2p(-\ 4) + q \implies 0 = 48 - 8p + q \implies 8p = 48 + q.$

$\therefore 8p - 4p = 48 + q - (28 + q) \implies 4p = 20 \implies p = 5.$

$\therefore q = 8 * 5 - 48 = -\ 8.$

$Or\ f(x) = x^3 + 5x^2 - 8x - 12.$

This involves nothing more than using the function, its derivative, and the value of both at x = - 4 to solve for the unknown parameters.

The next part of the problem involves nothing more than the fundamental theorem of algebra, the zero product property, and the quadratic formula.

$f(-\ 1) = -\ 1 + 5 * 1 - 8(-\ 1) - 12 = -\ 1 + 5 + 8 - 12 = 13 - 13 = 0.$

That - 1 is a zero of the function was given to you.

$\therefore x^3 + 5x^2 - 8x - 12 = \{x - (-\ 1)\}(x^2 + mx + n) \implies$

$\dfrac{x^3 + 5x^2 - 8x - 12}{x + 1} = x^2 + mx + n \implies x^2 + 4x - 12 = x^2 + mx + n.$

So the other zeroes are:

$\dfrac{-\ 4 \pm \sqrt{16 - 4(1)(-\ 12}}{2} = \dfrac{-\ 4 \pm \sqrt{16 + 48}}{2} = \dfrac{-\ 4 \pm \sqrt{64}}{2} = -\ 6\ or\ 2.$

Nothing after this is legible to me, but so far, none of this involves doing much more than using what you know about polynomials and their derivatives.
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