I guess that's f(x) = -3x^2+2
Well :
-3x^2+2=0
-3x^2=-2 / : (-3)
x^2= -2/-3
x^2= 2/3 / sqrt
x= +/ - sqrt(2)/sqrt(3)
We got the zero points.
x ~ +/- 0.81
f(0)= -3(0)^2 + 2
f(0)= 2
Mark three points:
(0.81;0)
(-0.81;0)
(0;2)
Connect with parabolic line and ya got a your graph
The question is the thing that needs help. What does that hanging "a leading 2" mean? And rational real coefficients? Obviously if they're rational they're gonna be real, but really.
I would say "a leading 2" means the highest power of x has a coefficient to two, which is none of the above.
The last two are of degree two, which is the lowest degree. They both have integer coefficients, so are necessarily real and rational as well.
Neither has a leading 2 as far as I can tell. The last one is monic (a leading coefficient of 1). I like monic polynomials so I'd pick that one, but that doesn't make it right.
None of the above
What do you have to find?
do you just need to solve the problem with those numbers?
Answer: The answer is A, 40.
2 x 2 = 4 + 4 x 4 = 16
4 + 16
= 20 + 2 x 40
20 + 80 = 100
(x = 40)
