what............there’s no pic
First, the graph is a parbole, this is a square function, so it can not be the option a) because it is a linear function.
Second, the vertix of the parabole is the point (0,3), so use the general vertex form of the parabole:
y = A(x - h)^2 + k, where h and k represent the vertex (h,k).
So, the form of the funtion is y = A (x-0)^2 - 3 = Ax^2 - 3
Now, you also know that the roots of the parabole are close to -1 and 1 but they are not those exact value.
So, find the roots for any of the possible options:
b) f(x) = 2x^2 - 3 = 0 => 2x^2 = 3 => x^2 = 3/2 => x = +/- √(3/2) ≈ +/- 1.22
That is a very plausible answer
c) f(x) = (1/2)x^2 - 3 = 0 => x^2 = 2*3 = 6 => x = +/- √6 ≈ +/- 2.45, which is very far from the roots shown on the graph.
3) f(x) = x^2 - 3 = 0 => x^2 = 3 => x = +/- √3 ≈ +/- 1.73, which is not so close to +/- 1.
So, the answer is the option b) f(x) = 2x^2 - 3
Thirty two and one hundred six thousandths.
First, find any zero of the polynomial. Since you didn't ask for work, I'll assume it's okay if I use my calculator. Your given polynomial has only one real root which is x=-4.
Now we use the rule that x-a divides the polynomial where a is a zero of said polynomial.
So x+4 divides 2x^3+2x^2-19x+20.
<span>(2x^3+2x^2-19x+20) / (x+4 equals 2x^2-6x+5).
If we factor out a two, we can use the quadratic formula.
2(x^2-3x+2.5) so we have x = (-(-3)+/-(9-4*1*2.5)^(1/2))/2*1)=(3+i)... or (3-i)/2 Where i is the square root of negative one. final answer:
2x^3+2x^2-19x+20=0
then x=-4, (3+i)/2, or (3-i)/2
</span>we have two imaginary number.
I hope it helped you
1850 x 0.22 (decimal form of 22%) = 407. 407 plastic beads are on clearance.
