### All High School Math Resources

## Example Questions

### Example Question #1 : Pre Calculus

What is the center and radius of the circle indicated by the equation?

**Possible Answers:**

**Correct answer:**

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 .

### Example Question #2 : Conic Sections

What is the shape of the graph indicated by the equation?

**Possible Answers:**

Hyperbola

Circle

Ellipse

Parabola

**Correct answer:**

Ellipse

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

### Example Question #1 : Pre Calculus

A conic section is represented by the following equation:

What type of conic section does this equation represent?

**Possible Answers:**

Ellipse

Parabola

Hyperbola

Circle

**Correct answer:**

Hyperbola

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.

### Example Question #1 : Pre Calculus

A conic section is represented by the following equation:

Which of the following best describes this equation?

**Possible Answers:**

horizontal hyperbola with center and asymptotes with slopes of and

vertical ellipse with center and a major axis length of

vertical hyperbola with center and asymptotes with slopes of and

horizontal hyperbola with center of and asymptotes with slopes of and

vertical parabola with vertex and a vertical stretch factor of

**Correct answer:**

horizontal hyperbola with center of and asymptotes with slopes of and

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 .

### Example Question #3 : Pre Calculus

Find the vertex for a parabola with equation

**Possible Answers:**

**Correct answer:**

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 :

=

### Example Question #4 : Pre Calculus

What is the minimal value of

over all real numbers?

**Possible Answers:**

No minimum value.

**Correct answer:**

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 .

### Example Question #5 : Pre Calculus

The function is such that

When you take the second derivative of the function , you obtain

What can you conclude about the function at ?

**Possible Answers:**

The point is an inflection point.

The point is a local maximum.

The point is a local minimum.

The point is an absolute maximum.

The point is an absolute minimum.

**Correct answer:**

The point is an inflection point.

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.

### Example Question #6 : Pre Calculus

Consider the function

Find the maximum of the function on the interval .

**Possible Answers:**

**Correct answer:**

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.

### Example Question #7 : Pre Calculus

What is the domain of the function below:

**Possible Answers:**

**Correct answer:**

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).

### Example Question #8 : Pre Calculus

What is the domain of the function below?

**Possible Answers:**

**Correct answer:**

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 .