Question

At what integer value will the values of y=2(3)x
surpass the values of y=8x2+2
?(1 point)
Responses

x=5
x is equal to 5

x=4
x is equal to 4

x=2
x is equal to 2

x=3

Answer 1

To determine at what integer value \( x \) the function \( y = 2(3)^x \) surpasses the function \( y = 8x^2 + 2 \), we can evaluate both functions for the integer values given in the options.

1. **For \( x = 2 \)**:
- \( y = 2(3)^2 = 2 \times 9 = 18 \)
- \( y = 8(2)^2 + 2 = 8 \times 4 + 2 = 32 + 2 = 34 \)
- \( 18 < 34 \)

2. **For \( x = 3 \)**:
- \( y = 2(3)^3 = 2 \times 27 = 54 \)
- \( y = 8(3)^2 + 2 = 8 \times 9 + 2 = 72 + 2 = 74 \)
- \( 54 < 74 \)

3. **For \( x = 4 \)**:
- \( y = 2(3)^4 = 2 \times 81 = 162 \)
- \( y = 8(4)^2 + 2 = 8 \times 16 + 2 = 128 + 2 = 130 \)
- \( 162 > 130 \)

4. **For \( x = 5 \)**:
- \( y = 2(3)^5 = 2 \times 243 = 486 \)
- \( y = 8(5)^2 + 2 = 8 \times 25 + 2 = 200 + 2 = 202 \)
- \( 486 > 202 \)

From the calculations, we can see that:

- At \( x = 2 \), \( y = 2(3)^x \) is less than \( y = 8x^2 + 2 \).
- At \( x = 3 \), \( y = 2(3)^x \) is still less.
- At \( x = 4 \), \( y = 2(3)^x \) surpasses \( y = 8x^2 + 2 \).

Thus, the value of \( x \) at which \( 2(3)^x \) first surpasses \( 8x^2 + 2 \) is:

**x is equal to 4.**