Question

Which could be the function graphed below? On a coordinate plane, a curve opens down and to the right in quadrants 1 and 4. The curve starts on the y-axis in quadrant 4 and goes through the x-axis into quadrant 1. f (x) = StartRoot x EndRoot minus 2 f (x) = StartRoot x minus 3 EndRoot + 1 f (x) = StartRoot 2x + 4 EndRoot f (x) = StartRoot x + 1 EndRoot + 8

Answer 1

Given:

1. f (x) = StartRoot x EndRoot minus 2 f (x) = StartRoot x minus 3 EndRoot + 1 f (x) = StartRoot 2x + 4 EndRoot f (x) = StartRoot x + 1 EndRoot + 8

To find:

Which could be the function graphed below? On a coordinate plane, a curve opens down and to the right in quadrants 1 and 4. The curve starts on the y-axis in quadrant 4 and goes through the x-axis into quadrant

Solution:

Consider the attached figure while going through the following steps.

From given, we have,

1. f (x) = StartRoot x EndRoot minus 2

⇒ f(x) = √x - 2

2. f (x) = StartRoot x minus 3 EndRoot + 1

⇒ f(x) = √(x - 3) + 1

3. f (x) = StartRoot 2x + 4 EndRoot

⇒ f(x) = √(2x + 4)

4. f (x) = StartRoot x + 1 EndRoot + 8

⇒ f(x) = √(x + 1) + 8

All the above functions are graphed in the image attached. Only the first option f(x) = √x - 2 satisfies the given condition.