Consider the quadratic function shown in the table below. x y 0 0 1 3 2 12 3 27 which exponential function grows at a faster rate than the quadratic function for 0

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.