Question

match each function on the left to all points on the right that would be located on the graph of the function.
$f(x) = 2x + 2$
$f(x) = 2x^2 - 2$
$f(x) = 2\sqrt{x + 1}$
$(0, 2)$
$(2, 0)$
$(-1, 0)$
$(2, 6)$

Explanation:

Step1: Check \(f(x)=2x+2\)

Substitute points:

• \(x=0\): \(2(0)+2=2\) → \((0,2)\) matches.
• \(x=2\): \(2(2)+2=6\) → \((2,6)\) matches.

Step2: Check \(f(x)=2x^2-2\)

Substitute \(x=1\): \(2(1)^2-2=0\) → \((1,0)\)? No, but \(x=-1\): \(2(-1)^2-2=0\) → \((-1,0)\) matches.

Step3: Check \(f(x)=2\sqrt{x+1}\)

Substitute \(x=0\): \(2\sqrt{1}=2\)? No; \(x=2\): \(2\sqrt{3}\approx3.46≠6\); \(x=-1\): \(2\sqrt{0}=0\) → \((-1,0)\)? No. Wait correction: \(f(x)=2\sqrt{x+1}\) at \(x=8\) is 6 but no point. Wait original points: only \((0,2),(2,0),(-1,0),(2,6)\). Recheck:

• \(f(x)=2x+2\): \((0,2),(2,6)\)
• \(f(x)=2x^2-2\): \((1,0)\) not listed, \((-1,0)\) yes, \((2,6)\)? No \(2(4)-2=6\) → \((2,6)\) also matches? Wait no: \(2x^2-2=6→x²=4→x=±2\). So \(f(x)=2x²-2\) has \((2,6),(-1,0)\); \(f(x)=2x+2\) has \((0,2),(2,6)\); \(f(x)=2\sqrt{x+1}\) has no listed points except if \(x=0→2\), yes \((0,2)\). Wait the problem says "all points on the right that would be located on the graph". Assuming standard matches:
• \(f(x)=2x+2\): \((0,2),(2,6)\)
• \(f(x)=2x²-2\): \((2,0)\)? No \(2(4)-2=6≠0\); \((-1,0)\) yes
• \(f(x)=2\sqrt{x+1}\): \((0,2)\) yes

Answer:

• \(f(x)=2x+2\): \((0,2),(2,6)\)
• \(f(x)=2x^2-2\): \((-1,0)\)
• \(f(x)=2\sqrt{x+1}\): \((0,2)\)

(Note: If the problem expects one-to-one but points overlap, this is the correct membership.)