Functions and Graphs - High School Math

This question is asking for the equation's slant asymptote. To find the slant asymptote, divide the numerator by the denominator. Long division gives us the following:

However, because we are considering as it approaches infinity, the effect that the last term has on the overall linear equation quickly becomes negligible (tends to zero). Thus, it can be ignored.

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 .

The domain is defined as the set of all possible values of the independent variable . First, must be greater than or equal to zero, because if , thenwill be undefined. In addition, the total expression under the radical, i.e. must be greater than or equal to zero:

That means that the expression under the radical is always positive and therefore is defined. Hence, the domain of the function is .

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.

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.

A function's symmetry is related to its classification as even, odd, or neither.

Even functions obey the following rule:

Because of this, even functions are symmetric about the y-axis.

Odd functions obey the following rule:

Because of this, odd functions are symmetric about the origin.

If a function does not obey either rule, it is neither odd nor even. (A graph that is symmetric about the x-axis is not a function, because it does not pass the vertical line test.)

To test for symmetry, simply substitute into the original equation.

Thus, this equation is even and therefore symmetric about the y-axis.

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for , we're hoping that the equation ends up equaling zero. Let's see if is a zero:

Since the function equals zero when is , one of the factors of the polynomial is . This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

Now we can factor the function this way:

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try :

When we factor using synthetic substitution for , we get the following result:

Using our quadratic factoring rules, we can factor completely:

Thus, the zeroes of are

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

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 .

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 .

FOIL:

These are our possible solutions. However, we need to test them.

:

The equation becomes . This is true, so is a solution.

:

However, negative numbers do not have logarithms, so this equation is meaningless. is not a solution, and is the one and only solution. Since this is not one of our choices, the correct response is "The correct solution set is not included among the other choices."

First, multiply the outside term with each term within the parentheses:

Rearranging the polynomial into fractional form, we get:

To simplify the polynomial, begin by combining like terms:

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 .

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.

Since and , it follows that and