### All College Algebra Resources

## Example Questions

### Example Question #1 : Transformations Of Parabolic Functions

If the function is depicted here, which answer choice graphs ?

### Example Question #1 : Graphing Parabolic Inequalities

Which of the following graphs correctly represents the quadratic inequality below (solutions to the inequalities are shaded in blue)?

**Possible Answers:**

**Correct answer:**

To begin, we analyze the equation given: the base equation, is shifted *left* one unit and *vertically stretched* by a factor of 2. The graph of the equation is:

To solve the inequality, we need to take a test point and plug it in to see if it matches the inequality. The only points that cannot be used are those directly on our parabola, so let's use the origin . If plugging this point in makes the inequality *true*, then we shade the area containing that point (in this case, *outside *the parabola); if it makes the inequality *untrue*, then the opposite side is shaded (in this case, the *inside* of the parabola). Plugging the numbers in shows:

Simplified as:

Which is *not* true, so the area inside of the parabola should be shaded, resulting in the following graph:

### Example Question #61 : Polynomial Functions

How many zeroes does the following polynomial have?

**Possible Answers:**

**Correct answer:**

is a degree 3 polynomial, so we don't have any easy formulas for calculating possible roots--we just have to check individual values to see if they work. We can use the **rational root test** to narrow the options down. Remember, if we have a polynomial of the form then any rational root will be of the form p/q where p is a factor of and q is a factor of . Fortunately in this case, so we only need to check the factors of , which is -15. Let's start with the easiest one: 1.

It doesn't work.

If we try the next number up, 3, we get this:

It worked! So we know that a factor of our polynomial is . We can divide this factor out:

and now we need to see if has any roots. We can actually solve quadratics so this is easier.

There aren't any real numbers that square to get -5 so this has no roots. Thus, only has one root.

### Example Question #63 : Polynomial Functions

is a polynomial function. , .

True or false: By the Intermediate Value Theorem, cannot have a zero on the interval .

**Possible Answers:**

False

True

**Correct answer:**

False

As a polynomial function, the graph of is continuous. By the Intermediate Value Theorem, if or , then there must exist a value such that .

Set and . It is not true that , so the Intermediate Value Theorem does not prove that there exists such that . However, it does not *disprove* that such a value exists either. For example, observe the graphs below:

Both are polynomial graphs fitting the given conditions, but the only the equation graphed at right has a zero on .

### Example Question #1 : Graphing Polynomials

True or false:

The polynomial has as a factor.

**Possible Answers:**

False

True

**Correct answer:**

True

One way to answer this question is as follows:

Let . By a corollary of the Factor Theorem, is divisible by if and only if the sum of its coefficients (accounting for minus symbols) is 0. has

as its coefficient sum, so is indeed divisible by .

### Example Question #1 : Graphing Polynomials

True or false:

The polynomial has as a factor.

**Possible Answers:**

True

False

**Correct answer:**

False

Let . By a corollary of the Factor Theorem, is divisible by if and only if the alternating sum of its coefficients (accounting for minus symbols) is 0.

To find this alternating sum, it is necessary to reverse the symbol before all terms of odd degree. In , there is one such terms, the term, so the alternating coefficient sum is

,

so is not divisible by .

### Example Question #1 : Graphing Polynomials

is a polynomial function. and .

True or false: By the Intermediate Value Theorem, must have a zero on the interval .

**Possible Answers:**

True

False

**Correct answer:**

True

As a polynomial function, the graph of is continuous. By the Intermediate Value Theorem, if or , then there must exist a value such that .

Setting , and looking at the second condition alone, this becomes: If , then there must exist a value such that - that is, must have a zero on . The conditions of this statement are met , since - and - so does have a zero on this interval.

### Example Question #67 : Polynomial Functions

Let be an even polynomial function with as a factor.

True or false: It follows that is also a factor of .

**Possible Answers:**

True

False

**Correct answer:**

True

By the Factor Theorem, is a factor of a polynomial if and only if . It is given that is a factor of , so it follows that .

is an even function, so, by definition, for all in its domain, . Setting , ; by substitution, . It follows that is also a factor of , making the statement true.

### Example Question #1 : Graphing Parabolas

Which of the following graphs matches the function ?

**Possible Answers:**

**Correct answer:**

Start by visualizing the graph associated with the function :

Terms within the parentheses associated with the squared x-variable will shift the parabola horizontally, while terms outside of the parentheses will shift the parabola vertically. In the provided equation, 2 is located outside of the parentheses and is subtracted from the terms located within the parentheses; therefore, the parabola in the graph will shift down by 2 units. A simplified graph of looks like this:

Remember that there is also a term within the parentheses. Within the parentheses, 1 is subtracted from the x-variable; thus, the parabola in the graph will shift to the right by 1 unit. As a result, the following graph matches the given function :

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