Question

the graph of a function f(x) passes through the following points: (0,2),(1,0),( - 1,0) which of the following could be f(x)? f(x)=2x + 2 f(x)= - 2x + 2 f(x)= - 2√x + 2 f(x)= - 2x² + 2

Explanation:

Step1: Test point (0, 2)

Substitute \(x = 0\) into each function.
For \(f(x)=2x + 2\), \(f(0)=2\times0 + 2=2\).
For \(f(x)=-2x + 2\), \(f(0)=-2\times0+2 = 2\).
For \(f(x)=-2\sqrt{x}+2\), when \(x = 0\), \(f(0)=-2\sqrt{0}+2=2\).
For \(f(x)=-2x^{2}+2\), \(f(0)=-2\times0^{2}+2 = 2\). All pass this - test.

Step2: Test point (1, 0)

Substitute \(x = 1\) into each function.
For \(f(x)=2x + 2\), \(f(1)=2\times1+2=4
eq0\).
For \(f(x)=-2x + 2\), \(f(1)=-2\times1 + 2=0\).
For \(f(x)=-2\sqrt{x}+2\), \(f(1)=-2\sqrt{1}+2=0\).
For \(f(x)=-2x^{2}+2\), \(f(1)=-2\times1^{2}+2=0\).

Step3: Test point (-1, 0)

Substitute \(x=-1\) into each function.
For \(f(x)=-2x + 2\), \(f(-1)=-2\times(-1)+2=4
eq0\).
For \(f(x)=-2\sqrt{x}+2\), the square - root of a negative number is not a real number when \(x=-1\), so this function is not valid.
For \(f(x)=-2x^{2}+2\), \(f(-1)=-2\times(-1)^{2}+2=-2 + 2=0\).

Answer:

\(f(x)=-2x^{2}+2\)