Question

create a table of values to show that the exponential expression 3(2)^x eventually overtakes the quadratic expression 3x^2 + 2x.

Explanation:

Step1: Choose values for \( x \)

We'll choose integer values for \( x \) starting from a small number and increasing. Let's pick \( x = 0, 1, 2, 3, 4, 5, 6 \).

Step2: Calculate \( 3x^2 + 2x \) for each \( x \)

• For \( x = 0 \): \( 3(0)^2 + 2(0) = 0 \)
• For \( x = 1 \): \( 3(1)^2 + 2(1) = 3 + 2 = 5 \)
• For \( x = 2 \): \( 3(2)^2 + 2(2) = 12 + 4 = 16 \)
• For \( x = 3 \): \( 3(3)^2 + 2(3) = 27 + 6 = 33 \)
• For \( x = 4 \): \( 3(4)^2 + 2(4) = 48 + 8 = 56 \)
• For \( x = 5 \): \( 3(5)^2 + 2(5) = 75 + 10 = 85 \)
• For \( x = 6 \): \( 3(6)^2 + 2(6) = 108 + 12 = 120 \)

Step3: Calculate \( 3(2)^x \) for each \( x \)

• For \( x = 0 \): \( 3(2)^0 = 3(1) = 3 \)
• For \( x = 1 \): \( 3(2)^1 = 3(2) = 6 \)
• For \( x = 2 \): \( 3(2)^2 = 3(4) = 12 \)
• For \( x = 3 \): \( 3(2)^3 = 3(8) = 24 \)
• For \( x = 4 \): \( 3(2)^4 = 3(16) = 48 \)
• For \( x = 5 \): \( 3(2)^5 = 3(32) = 96 \)
• For \( x = 6 \): \( 3(2)^6 = 3(64) = 192 \)

Step4: Create the table

Now we'll organize the values into a table:

\( x \)	\( 3x^2 + 2x \)	\( 3(2)^x \)
1	5	6
2	16	12
3	33	24
4	56	48
5	85	96
6	120	192

From the table, we can see that when \( x = 5 \), \( 3(2)^5 = 96 \) which is greater than \( 3(5)^2 + 2(5) = 85 \), and as \( x \) increases further, the exponential function \( 3(2)^x \) will grow much faster than the quadratic function \( 3x^2 + 2x \), so it "overtakes" the quadratic expression.

Answer:

The table of values is:

\( x \)	\( 3x^2 + 2x \)	\( 3(2)^x \)
1	5	6
2	16	12
3	33	24
4	56	48
5	85	96
6	120	192

We can observe that \( 3(2)^x \) overtakes \( 3x^2 + 2x \) when \( x = 5 \) (and continues to be larger for \( x > 5 \)).