Graphing equations

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joannamartinez

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Feb 19, 2010
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I need your help please. I need to create an equation for the solution of x=2, -3. I have been trying for quite some time now and I need help. I believe the beginning of the equation begins with x^2 and the end =0 but I cannot get the remainder. Could you please advise? Thank you.
 
joannamartinez said:
I need your help please. I need to create an equation for the solution of x=2, -3. I have been trying for quite some time now and I need help. I believe the beginning of the equation begins with x^2 and the end =0 but I cannot get the remainder. Could you please advise? Thank you.
Click to expand...

If you want the roots of the function to be 'a' and 'b' - then the function is:

f(x) = A * (x - a) * (x - b) where 'A' is any arbitrary constant.
 
The equation would look something like this but this is not the equation: 6x^2 +11x-35=0. I do not know how to get the equation for the solutions of x=2,-3.
 
joannamartinez said:
The equation would look something like this but this is not the equation: 6x^2 +11x-35=0. I do not know how to get the equation for the solutions of x=2,-3.
Click to expand...

I understand - but can you factorize f(x) = 6x^2 + 11x - 35, and then tell me the roots of that function f(x).
 
Yes, I can. It would be 6x^2 +11x-35 = (3x-5)(2x+7) = 3x-5=0 and 2x+7=0 so 3x-5=0 would be x=5/3 and 2x+7=0 would be x= -7/2.
 
\(\displaystyle x \ = \ 2 \ and \ x \ = \ -3 \ \implies \ x-2 \ = \ 0 \ and \ x+3 \ = \ 0\)

\(\displaystyle Hence, \ (x-2)(x+3) \ = \ 0 \ \implies \ x^2+x-6 \ = \ 0\)
 
Thank you for your help. This math is not easy and your help and patience is very appreciated. Thank you for breaking it down for me. That makes it much easier to understand. Thank you for your time.
 
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