IBKen-Z Posted May 18, 2011 Report Share Posted May 18, 2011 Can anybody tell me how to find the zeros of a cubic function on a graphing calculator?Example: x^3+3x^2+2x+8I thought I knew how to do it but the problem I'm having is that my graph of this function does not look like a regular cubic function. It only passes through the x-axis once! Can anybody help me?! Please please please?!?!?! Reply Link to post Share on other sites More sharing options...
Rigel Posted May 18, 2011 Report Share Posted May 18, 2011 (edited) I don't do Maths HL, but i'll do my best try: The graph just passes through the x-axis once, don't worry. (Real numbers). If you solve the equation, you will see that the only real root is something like this: -3.1663. Or, graphically, you can press F5 when you see the graph, then F6, then X-Cal (F2), and type in 0. You will see when does it pass through the x-axis. Hope this was helpful! In functions with more than 1 real root, do the same process, and you can find the intercepts using the arrow pad. Edited May 18, 2011 by Ipos Manger Reply Link to post Share on other sites More sharing options...
IBKen-Z Posted May 18, 2011 Author Report Share Posted May 18, 2011 I don't do Maths HL, but i'll do my best try: The graph just passes through the x-axis once, don't worry. (Real numbers). If you solve the equation, you will see that the only real root is something like this: -3.1663. Or, graphically, you can press F5 when you see the graph, then F6, then X-Cal (F2), and type in 0. You will see when does it pass through the x-axis. Hope this was helpful! In functions with more than 1 real root, do the same process, and you can find the intercepts using the arrow pad. That is the answer that I have been getting. However, my text book says the answer is -1, 2 or -4. Everything seems to say that a cubic function passes through the x-axis three times. It's causing me a lot of confusion haha Reply Link to post Share on other sites More sharing options...
Rigel Posted May 18, 2011 Report Share Posted May 18, 2011 Your book is probably mistaken. The other 2 roots are complex roots. Also, that's not always the case in a cubic function. Reply Link to post Share on other sites More sharing options...
Rigel Posted May 18, 2011 Report Share Posted May 18, 2011 Also, it depends on the constant term. If it's positive, it will only pass once. If it's negative, it will pass three times. (My deductions using the calculator). Reply Link to post Share on other sites More sharing options...
dessskris Posted May 18, 2011 Report Share Posted May 18, 2011 This is your question x³+3x²+2x+8This is the answer in your textbook x=-1, 2, -4; meaning (x+1), (x-2) and (x+4) are factors of the cubic equation.Let's work backwards.(x+1)(x-2)(x+4)=0(x+1)(x²+2x-8)=0x³+2x²-8x+x²+2x-8=0x³+3x²-6x-8=0Therefore there must have been a misprint. Which textbook do you use? If you use the Oxford textbook in fact it has soooooooo many mistakes so you should not be surprised if your answer doesn't match with the answer key.Anyway to answer your question:Factorise x³+3x²+2x+8=0 to be a product of a linear function and a quadratic function. You get the real root by graphing, get the quadratic expression algebraically. Then solve the quadratic one using x=(-b±√(b²-4ac))/2a and you'll get your complex roots.Have fun! Reply Link to post Share on other sites More sharing options...
Jchh44 Posted May 18, 2011 Report Share Posted May 18, 2011 misprints and Math HL textbooks go together like peanut butter and jelly Reply Link to post Share on other sites More sharing options...
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