A circle is defined by an equation in the format .

The center is indicated by the point  and the radius .

In the equation , the center is  and the radius is .

An ellipse has an equation that can be written in the format. The center is indicated by , or in this case .

The simplest way to know what kind of conic section an equation represents is by checking the coefficients in front of each variable. The equation must be in general form while you do this check. Luckily, this equation is already in general form, so it's easy to see. The general equation for a conic section is the following:

Assuming the term  is 0 (which it usually is):

• If A equals C, the equation is a circle.
• If A and C have the same sign (but are not equal to each other), the equation is an ellipse.
• If either A or C equals 0, the equation is a parabola.
• If A and C are different signs (i.e. one is negative and one is positive), the equation is a hyperbola.

First, we need to make sure the conic section equation is in a form we recognize. Luckily, this equation is already in standard form:

The first step is to determine the type of conic section this equation represents. Because there are two squared variables ( and ), this equation cannot be a parabola.  Because the coefficients in front of the squared variables are different signs (i.e. one is negative and the other is positive), this equation must be a hyperbola, not an ellipse.

In a hyperbola, the squared term with a positive coefficient represents the direction in which the hyperbola opens. In other words, if the term is positive, the hyperbola opens horizontally. If the term is positive, the hyperbola opens vertically. Therefore, this is a horizontal hyperbola.

The center is always found at , which in this case is .

That leaves only the asymptotes. For a hyperbola, the slopes of the asymptotes can be found by dividing  by  (remember to always put the vertical value, , above the horizontal value, ). Remember that these slopes always come in pairs, with one being positive and the other being negative.

In this case,  is 3 and  is 2, so we get slopes of and .

For any parabola of the form   ,  the -coordinate of its vertex is 

So here, we have

= 

We plug this back into the original equation to find :

= 

Since this is an upwards-opening parabola, its minimum value will occur at the vertex.  The -coordinate for the vertex of any parabola of the form

is at

So here, 

We plug this value back into the equation of the parabola, to find the value of the function at this . 

Thus the minimal value of the expression is . 

We have a point at which . We know from the second derivative test that if the second derivative is negative, the function has a maximum at that point. If the second derivative is positive, the function has a minimum at that point. If the second derivative is zero, the function has an inflection point at that point.

Plug in 0 into the second derivative to obtain 

So the point is an inflection point.

Notice that on the interval , the term  is always less than or equal to . So the function is largest at the points when . This occurs at  and .

Plugging in either 1 or 0 into the original function  yields the correct answer of 0.

The domain is defined as the set of possible values for the x variable. In order to find the impossible values of x, we should:

a) Set the equation under the radical equal to zero and look for probable x values that make the expression inside the radical negative:

There is no real value for x that will fit this equation, because any real value square is a positive number i.e. cannot be a negative number.

b) Set the denominator of the fractional function equal to zero and look for probable x values:

Now we can solve the equation for x:

There is no real value for x that will fit this equation.

The radical is always positive and denominator is never equal to zero, so the f(x) is defined for all real values of x. That means the set of all real numbers is the domain of the f(x) and the correct answer is .

Alternative solution for the second part of the solution:

After figuring out that the expression under the radical is always positive (part a), we can solve the radical and therefore denominator for the least possible value (minimum value). Setting the x value equal to zero will give the minimum possible value for the denominator.

That means the denominator will always be a positive value greater than 1/2; thus it cannot be equal to zero by setting any real value for x. Therefore the set of all real numbers is the domain of the f(x).

The domain is defined as the set of all values of x for which the function is defined i.e. has a real result. The square root of a negative number isn't defined, so we should find the intervals where that occurs:

The square of any number is positive, so we can't eliminate any x-values yet.

If the denominator is zero, the expression will also be undefined.

Find the x-values which would make the denominator 0:

Therefore, the domain is .