Question 1193021: State two different ways to determine the number of zeros of the function
f(x)=2(x+1)^2 -6
Found 3 solutions by ikleyn, greenestamps, Edwin McCravy:
Answer by ikleyn(52887)   (Show Source):
You can put this solution on YOUR website! .
Use the Gauss theorem (another name: Fundamental Theorem of Algebra), which states that the number of (complex) roots
of any polynomial is equal to the degree of this polynomial.
See this Wikipedia article
https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
Or derive your own formula for this simple quadratic equation, giving its roots explicitly (as Sir Edwin did in his post).
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
Technically, it is a polynomial of degree 2, so it has 2 zeros.
If you mean real zeros, then....
(1) Convert the equation to standard form  and look at the discriminant  . There are 2 zeros if the discriminant is positive, 1 zero if it is 0, and no (real) zeros if it is negative.
(2) The equation as given is in vertex form. The vertex is (-1,-6) and the parabola opens upward, so there are 2 real zeros.
Answer by Edwin McCravy(20064)  (Show Source):
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