Questions: Question 6 Choose the equation representing the graph below. f(x)=1+√x f(x)=√(1-x) f(x)=√(x-1) None of these choices f(x)=-√x+1

Solution Steps

The graph starts at the point (1, 0) and passes through the point (0, 1). It is a decreasing function and looks like a square root graph reflected over the y-axis and shifted one unit to the right.

$f(x) = 1 + \sqrt{x}$. If $x=0$, $f(0) = 1 + \sqrt{0} = 1$. If $x=1$, $f(1) = 1 + \sqrt{1} = 2$. So this equation doesn't work.

$f(x) = \sqrt{1-x}$. If $x=0$, $f(0) = \sqrt{1-0} = 1$. If $x=1$, $f(1) = \sqrt{1-1} = 0$. This looks promising so far. If $x=-3$, $f(-3) = \sqrt{1-(-3)} = \sqrt{4} = 2$, which also appears to lie on the graph.

$f(x) = \sqrt{x-1}$. If $x=0$, $f(0) = \sqrt{0-1} = \sqrt{-1}$, which is not a real number, so this equation doesn't work.

$f(x) = -\sqrt{x} + 1$. If $x=0$, $f(0) = -\sqrt{0} + 1 = 1$. If $x=1$, $f(1) = -\sqrt{1} + 1 = 0$. If $x=4$, $f(4) = -\sqrt{4} + 1 = -1$. The point (4, -1) is not on the graph.

Final Answer