Question

sketch the graph of the first function. then, on the same coordinate plane, use a transformation to sketch the second graph
( y = sqrt { 1 6 - x ^ { 2 } } cdot y = - sqrt { 1 6 - x ^ { 2 } } )
which graph shows ( y = sqrt { 1 6 - x ^ { 2 } } ) in red and ( y = - sqrt { 1 6 - x ^ { 2 } } ) in blue?

Explanation:

Step1: Analyze the first function \(y = \sqrt{16 - x^{2}}\)

Square both sides: \(y^{2}=16 - x^{2}\), then \(x^{2}+y^{2}=16\) (\(y\geq0\)). This is the upper - half of a circle with center \((0,0)\) and radius \(r = 4\).

Step2: Analyze the second function \(y=-\sqrt{16 - x^{2}}\)

Square both sides: \(y^{2}=16 - x^{2}\), then \(x^{2}+y^{2}=16\) (\(y\leq0\)). This is the lower - half of a circle with center \((0,0)\) and radius \(r = 4\).

Since the problem is about graphing functions which are parts of a circle (a geometric shape), and we are dealing with function transformations (reflection about the \(x\) - axis as \(y = f(x)\) to \(y=-f(x)\)), we can conclude the following:

The graph of \(y = \sqrt{16 - x^{2}}\) is the upper - half of the circle \(x^{2}+y^{2}=16\) and the graph of \(y=-\sqrt{16 - x^{2}}\) is the lower - half of the circle \(x^{2}+y^{2}=16\).

If we assume the options:

• Option A: If it shows the upper - half (red) and lower - half (blue) of the circle \(x^{2}+y^{2}=16\) (center at the origin, radius 4), then it is the correct option.
• Option B: Incorrect if it does not match the upper - lower half description.
• Option C: Incorrect if it does not match the upper - lower half description.
• Option D: Incorrect if it does not match the upper - lower half description.

Answer:

Assuming Option A shows the upper - half (red) \(y = \sqrt{16 - x^{2}}\) and lower - half (blue) \(y=-\sqrt{16 - x^{2}}\) of the circle \(x^{2}+y^{2}=16\), then A. Option A.