Question

Help me please asappp!!!!​

Answer 1

Answer:

For the function $y=2x^2$:

$y=0\\y=2\\y=8$

For the second function $y=2^x$:

 $y=1\\y=2\\y=4$

Step-by-step explanation:

You only need to make the following substitution:

For the first function $y=2x^2$:

When $x=0$

$y=2x^2\\ y=2(0)^2\\ y=0$

When $x=1$

$y=2(1)^2\\ y=2(1)^2\\ y=2$

When $x=2$

$y=2x^2\\ y=2(2)^2\\ y=2*4\\ y=8$

For the second function $y=2^x$:

When $x=0$

$y=2^x\\ y=2^0\\ y=1$

When $x=1$

$y=2^x\\ y=2^1\\ y=2$

When $x=2$

$y=2^x\\y=2^2\\ y=2^2\\ y=4$

Answer 2

ANSWER

When x=1, $y =2{x}^{2}$ is 2

and $y = {2}^{x}$ is 2

When x=2, $y =2{x}^{2}$ is 8

and $y = {2}^{x}$ is 4

EXPLANATION

The given functions are:

$y =2{x}^{2}$

and

$y = {2}^{x}$

We want to find the functional values for the given x-values.

When x=1,

$y = 2( {1})^{2} = 2$

and

$y = {2}^{1} = 2$

When x=2,

The first function gives us:

$y = 2 {(2)}^{2} = 8$

and the second function gives

$y = {2}^{2} = 4$