To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

 and 

We can then solve these two equations to obtain .

To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

 and 

We can then solve these two equations to obtain

The equation of a circle is:  where  is the radius and  is the center. 

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of  that makes  equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 6.

The equation of a circle is:  where  is the radius and  is the center. 

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of  that makes  equal to , which is . We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 10.

* We have to change the time from minutes to hours, there are 60 minutes in one hour. 

Substitute into , and then substitute the answer into .

We will solve this system of equations by Elimination.  Multiply both sides of the first equation by 2, to get:

Then add this new equation, to the second original equation, to get:

or 

Plugging this value of  back into the first original equation, gives:

or 

Since the equation given in the question is based off of the parent function , we can write the general form for transformations like this:

determines the vertical stretch or compression factor.

• If is greater than 1, the function has been vertically stretched (expanded) by a factor of .
• If is between 0 and 1, the function has been vertically compressed by a factor of .

In this case, is 4, so the function has been vertically stretched by a factor of 4.

determines the horizontal stretch or compression factor.

• If  is greater than 1, the function has been horizontally compressed by a factor of .
• If  is between 0 and 1, the function has been horizontally stretched (expanded) by a factor of .

In this case,  is 6, so the function has been horizontally compressed by a factor of 6. (Remember that horizontal stretch and compression are opposite of vertical stretch and compression!)

determines the horizontal translation.

• If is positive, the function was translated units right.
• If is negative, the function was translated units left.

In this case, is 3, so the function was translated 3 units right.

 determines the vertical translation.

• If  is positive, the function was translated  units up.
• If  is negative, the function was translated  units down.

In this case,  is -7, so the function was translated 7 units down.

A normal distribution is one in which the values are evenly distributed both above and below the mean.  A population has a precisely normal distribution if the mean, mode, and median are all equal.  For the population of 3,4,5,5,5,6,7, the mean, mode, and median are all 5.

In a normal distribution, the number of values within one positive standard deviation of the mean is equal to the number of values within one negative standard deviation of the mean.  The reason for this is that the values below the population mean exactly parallel the values above the mean.