Answer:
y=3x^2 is the quadratic that presents the table
x | 0 1 2 3
y | 0 3 12 27
The exponential that includes (1,4) takes off a bit faster than the others near x=0.
Step-by-step explanation:
If the table is:
x | 0 1 2 3
y | 0 3 12 27
It says it is a quadratic and we know the y-intercept is 0 since we have the point (0,0).
y=ax^2+bx
Now we need to use 2 more mores and we should wind up with a system to solve for a and b.
(1,3) (2,12)
3=a(1)^2+b(1) 12=a(2)^2+b(2)
3=a+b 12=4a+2b
6=2a+b (divided both sides by 2)
So we are solving the system
3=a+b
6=2a+b
-------------- I'm going to choose elmination because it is already setup that way. I'm going to subtract the equations. In doing so, the b's will cancel.
3-6=a-2a
-3=-a
a=3
so since 3=a+b and a=3 then b=0
So the quadratic is y=3x^2
And perhaps we could have looked for an easier way to solve this since we winded up with such a simple quadratic. I'm saying it might have been just easier looking for a pattern but something it won't be that easy.
