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October 22nd, 2016, 06:10 AM  #1 
Newbie Joined: Oct 2016 From: South Africa Posts: 2 Thanks: 0  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 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: 763 Thanks: 278 Math Focus: Linear algebra, linear statistical models 
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  
Newbie Joined: Oct 2016 From: South Africa Posts: 2 Thanks: 0  Quote:
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: 421 Thanks: 169 
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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cubic, functions, nature, quadratic, roots 
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