My Math Forum Nature of Roots for Quadratic and Cubic Functions

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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

Thanks
Attached Images
 Screenshot 2016-10-22 15.59.29.jpg (16.3 KB, 11 views) Screenshot 2016-10-22 15.59.57.jpg (9.4 KB, 7 views) Screenshot 2016-10-22 16.09.30.jpg (10.4 KB, 4 views)

Last edited by GardeeZak; October 22nd, 2016 at 06:13 AM.

 October 22nd, 2016, 06:57 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 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.
October 22nd, 2016, 07:45 AM   #3
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Quote:
 Originally Posted by 123qwerty 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

 October 22nd, 2016, 09:57 AM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 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.
 October 23rd, 2016, 11:18 AM #5 Newbie   Joined: Oct 2016 From: South Africa Posts: 3 Thanks: 0 Hi - I'm so sorry about the unclear pics; here they are in clearer form: http://bit.ly/2eyoBdz I have done calculus with derivatives - I have not done integration at school level. I am not sure how your explanation helps me to understand the concepts I don't get to answer 8.1.4, 7.4 and 4.5 Please assist me, and thanks. Last edited by skipjack; January 30th, 2017 at 11:21 AM.
 January 30th, 2017, 11:22 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2201 Your link no longer has relevant content.

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