Let $$ (x+4)^{2}+(y-5)^{2}=40 $$. Find the exact $$ y $$-intercepts of the graph (if any). O There are no $$ y $$-intercepts. $$ \begin{array}{l} ext { O(0,5 } \pm 2 \sqrt{6}) \ O(0,5 \pm 2 \sqrt{10}) \ O(0,0.101) ext { and }(0,9.899) .\end{array} $$

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Let $$ (x+4)^{2}+(y-5)^{2}=40 $$. Find the exact $$ y $$-intercepts of the graph (if any). O There are no $$ y $$-intercepts. $$ \begin{array}{l}	ext { O(0,5 } \pm 2 \sqrt{6}) \ O(0,5 \pm 2 \sqrt{10}) \ O(0,0.101) 	ext { and }(0,9.899) .\end{array} $$

UpStudy AI · Accepted Answer

To find the $$ y $$-intercepts of the graph given by the equation $$ (x+4)^{2}+(y-5)^{2}=40 $$, we need to determine the points where the graph intersects the $$ y $$-axis. The $$ y $$-intercepts occur when $$ x = 0 $$.

1. Substitute $$ x = 0 $$ into the equation:
   $$
   (0 + 4)^{2} + (y - 5)^{2} = 40
   $$
   This simplifies to:
   $$
   4^{2} + (y - 5)^{2} = 40
   $$
   $$
   16 + (y - 5)^{2} = 40
   $$

2. Next, isolate $$ (y - 5)^{2} $$:
   $$
   (y - 5)^{2} = 40 - 16
   $$
   $$
   (y - 5)^{2} = 24
   $$

3. Now, take the square root of both sides:
   $$
   y - 5 = \pm \sqrt{24}
   $$
   Since $$ \sqrt{24} = 2\sqrt{6} $$, we have:
   $$
   y - 5 = \pm 2\sqrt{6}
   $$

4. Solve for $$ y $$:
   $$
   y = 5 \pm 2\sqrt{6}
   $$

Thus, the $$ y $$-intercepts are:
$$
(0, 5 + 2\sqrt{6}) \quad \text{and} \quad (0, 5 - 2\sqrt{6})
$$

Now, let's summarize the results:
- The exact $$ y $$-intercepts of the graph are $$ (0, 5 + 2\sqrt{6}) $$ and $$ (0, 5 - 2\sqrt{6}) $$.

From the options provided, the correct answer is:
$$
O(0,5 \pm 2 \sqrt{6})
$$
The $$ y $$-intercepts are $$ (0, 5 + 2\sqrt{6}) $$ and $$ (0, 5 - 2\sqrt{6}) $$.

Let . Find the exact -intercepts of the graph (if any).
O There are no -intercepts.

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Let . Find the exact -intercepts of the graph (if any). O There are no -intercepts.