Question

sketch the graph of the first function. then, on the same coordinate plane, use a transformation to sketch the second graph. y = \sqrt{49 - x^{2}}, y = -\sqrt{49 - x^{2}}
which graph shows y = \sqrt{49 - x^{2}} in red and y = -\sqrt{49 - x^{2}} in blue? a. b. c. d.

Explanation:

Step1: Analyze the first - function

The function $y = \sqrt{49 - x^{2}}$ can be rewritten as $y^{2}=49 - x^{2}$ or $x^{2}+y^{2}=49(y\geq0)$. It represents the upper - half of a circle centered at the origin $(0,0)$ with radius $r = 7$.

Step2: Analyze the second - function

The function $y=-\sqrt{49 - x^{2}}$ can be rewritten as $y^{2}=49 - x^{2}$ or $x^{2}+y^{2}=49(y\leq0)$. It represents the lower - half of a circle centered at the origin $(0,0)$ with radius $r = 7$.

Step3: Determine the correct graph

The graph of $y = \sqrt{49 - x^{2}}$ (red) should be the upper - half of the circle and $y=-\sqrt{49 - x^{2}}$ (blue) should be the lower - half of the circle.

Answer:

The graph that shows the upper - half of the circle in red and the lower - half of the circle in blue is the correct one. Without seeing the actual options in detail, the correct graph will have the red curve above the x - axis and the blue curve below the x - axis for the circle $x^{2}+y^{2}=49$. If we assume the standard orientation of the coordinate axes, the correct graph is the one where the red part is the positive y - value part of the circle and the blue part is the negative y - value part of the circle.