Question

which function could represent the values shown in the table?
a. $y = 2x - 1$
b. $y = 3x - 2$
c. $y = 4x - 3$
d. $y = 5x + 3$

Explanation:

Step1: Analyze Option A ($y = 2x - 1$)

For \(x = 1\): \(y = 2(1)-1 = 1\), but the table's \(y\) when \(x = 1\) should be checked. Wait, the table has \(x\) values (let's assume the first column is \(x\): 1,2,3,4,5? Wait, the table shows \(x\) maybe 1,2,3,4,5 and \(y\) as 1,4,7,10,13? Wait, no, the table: first column (x?): 1,2,3,4,5; second column (y?): 1,4,7,10,13? Wait, no, the options: let's take \(x = 1\), check each option.

Wait, let's list the table properly. From the image, the table has two columns. Let's assume \(x\) values: 1, 2, 3, 4, 5 (first column) and \(y\) values: 1, 4, 7, 10, 13 (second column)? Wait, no, the options:

Option B: \(y = 3x - 2\)

Let's test \(x = 1\): \(y = 3(1)-2 = 1\) (matches if \(y=1\) when \(x=1\))

\(x = 2\): \(y = 3(2)-2 = 4\) (matches if \(y=4\) when \(x=2\))

\(x = 3\): \(y = 3(3)-2 = 7\) (matches \(y=7\) when \(x=3\))

\(x = 4\): \(y = 3(4)-2 = 10\) (matches \(y=10\) when \(x=4\))

\(x = 5\): \(y = 3(5)-2 = 13\) (matches \(y=13\) when \(x=5\))

Now check Option A: \(y = 2x - 1\)

\(x=1\): \(y=2(1)-1=1\) (matches), \(x=2\): \(y=2(2)-1=3\) (but table's \(y\) for \(x=2\) is 4? Wait, maybe I misread the table. Wait the table: first column (x) 1,2,3,4,5; second column (y) 1,4,7,10,13? Wait no, the second column: first row 1, second 4, third 7, fourth 10, fifth 13.

Wait Option B: \(y=3x - 2\)

\(x=1\): 3(1)-2=1 ✔️

\(x=2\): 3(2)-2=4 ✔️

\(x=3\): 3(3)-2=7 ✔️

\(x=4\): 3(4)-2=10 ✔️

\(x=5\): 3(5)-2=13 ✔️

Option A: \(x=2\): 2(2)-1=3 ≠4 ❌

So Option B works.

Step2: Confirm with other options (if needed)

But since Option B works for all \(x\) values in the table, it's the correct function.

Answer:

B. \(y = 3x - 2\)