### All Algebra II Resources

## Example Questions

### Example Question #1 : Parabolic Functions

Write a quadratic equation having as the vertex (vertex form of a quadratic equation).

**Possible Answers:**

**Correct answer:**

The vertex form of a quadratic equation is given by

Where the vertex is located at

giving us .

### Example Question #28 : Exponents

What are the -intercepts of the equation?

**Possible Answers:**

There are no -intercepts.

**Correct answer:**

To find the x-intercepts of the equation, we set the numerator equal to zero.

### Example Question #2 : Parabolic Functions

What is the minimum possible value of the expression below?

**Possible Answers:**

The expression has no minimum value.

**Correct answer:**

We can determine the lowest possible value of the expression by finding the -coordinate of the vertex of the parabola graphed from the equation . This is done by rewriting the equation in vertex form.

The vertex of the parabola is the point .

The parabola is concave upward (its quadratic coefficient is positive), so represents the minimum value of . This is our answer.

### Example Question #3 : Parabolic Functions

**Find the coordinates of the vertex of this quadratic function**:

**Possible Answers:**

**Correct answer:**

Vertex of quadratic equation is given by .

For ,

,

so the coordinate of vertex is .

### Example Question #4 : Parabolic Functions

What are the x-intercepts of the graph of ?

**Possible Answers:**

**Correct answer:**

Assume y=0,

,

### Example Question #5 : Parabolic Functions

Find the vertex of the parabola given by the following equation:

**Possible Answers:**

**Correct answer:**

In order to find the vertex of a parabola, our first step is to find the x-coordinate of its center. If the equation of a parabola has the following form:

Then the x-coordinate of its center is given by the following formula:

For the parabola described in the problem, a=-2 and b=-12, so our center is at:

Now that we know the x-coordinate of the parabola's center, we can simply plug this value into the function to find the y-coordinate of the vertex:

So the vertex of the parabola given in the problem is at the point

### Example Question #3 : How To Use The Quadratic Function

Give the minimum value of the function .

**Possible Answers:**

This function does not have a minimum.

**Correct answer:**

This is a quadratic function. The -coordinate of the vertex of the parabola can be determined using the formula , setting :

Now evaluate the function at :

### Example Question #6 : Parabolic Functions

What is the equation of a parabola with vertex and -intercept ?

**Possible Answers:**

**Correct answer:**

From the vertex, we know that the equation of the parabola will take the form for some .

To calculate that , we plug in the values from the other point we are given, , and solve for :

Now the equation is . This is not an answer choice, so we need to rewrite it in some way.

Expand the squared term:

Distribute the fraction through the parentheses:

Combine like terms:

### Example Question #7 : Parabolic Functions

Determine the maximum or minimum of .

**Possible Answers:**

**Correct answer:**

To find the max or min of , use the vertex formula and substitute the appropriate coefficients.

Since the leading coefficient of is negative, the parabola opens down, which indicates that there will be a maximum.

The answer is:

### Example Question #8 : Parabolic Functions

Factorize:

**Possible Answers:**

**Correct answer:**

To simplify , determine the factors of the first and last term.

The factor possibilities of :

The factor possibilities of :

Determine the signs. Since there is a positive ending term and a negative middle term, the signs of the binomials must be both negative. Write the pair of parenthesis.

These factors must be manipulated by trial and error to determine the middle term.

The correct selection is .

The answer is .

### All Algebra II Resources

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