Asked by Joey
Determine the equation of a cubic function with zeros 2 √3 and -1 whose graph passes through the point (2, -18)
Answers
Answered by
oobleck
y = a(x-2)(x-√3)(x+1)
since y(2) = -18, this is impossible. So maybe you meant the zeroes are
2±√3 and -1. In that case,
y = a(x-(2-√3))(x-(2+√3))(x+1) = a(x^3 - 3x^2 - 3x + 1)
since y(2) = -18, that means that
a(-9) = -18
so a = 2 and y = 2(x^3 - 3x^2 - 3x + 1)
since y(2) = -18, this is impossible. So maybe you meant the zeroes are
2±√3 and -1. In that case,
y = a(x-(2-√3))(x-(2+√3))(x+1) = a(x^3 - 3x^2 - 3x + 1)
since y(2) = -18, that means that
a(-9) = -18
so a = 2 and y = 2(x^3 - 3x^2 - 3x + 1)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.