Graphing Polynomials - College Algebra
Card 0 of 9
Which of the following graphs matches the function 
?
Which of the following graphs matches the function
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 
:
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
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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If the function 
is depicted here, which answer choice graphs 
?
If the function
The function 
shifts a function f(x) 
units to the left. Conversely, 
shifts a function f(x) 
units to the right. In this question, we are translating the graph two units to the left.
To translate along the y-axis, we use the function 
or 
.
The function
To translate along the y-axis, we use the function
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Which of the following graphs correctly represents the quadratic inequality below (solutions to the inequalities are shaded in blue)?
Which of the following graphs correctly represents the quadratic inequality below (solutions to the inequalities are shaded in blue)?
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:
To begin, we analyze the equation given: the base equation,
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
Simplified as:
Which is not true, so the area inside of the parabola should be shaded, resulting in the following graph:
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How many zeroes does the following polynomial have?
How many zeroes does the following polynomial have?
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.
If we try the next number up, 3, we get this:
and now we need to see if
There aren't any real numbers that square to get -5 so this has no roots. Thus,
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is a polynomial function. 
, 
.
True or false: By the Intermediate Value Theorem, 
cannot have a zero on the interval 
.
True or false: By the Intermediate Value Theorem,
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 
.
As a polynomial function, the graph of
Set
Both are polynomial graphs fitting the given conditions, but the only the equation graphed at right has a zero on
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True or false:
The polynomial 
has 
as a factor.
True or false:
The polynomial
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 
.
One way to answer this question is as follows:
Let
as its coefficient sum, so
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is a polynomial function. 
and 
.
True or false: By the Intermediate Value Theorem, must have a zero on the interval .
True or false: By the Intermediate Value Theorem,
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.
As a polynomial function, the graph of
Setting
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True or false:
The polynomial 
has 
as a factor.
True or false:
The polynomial
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 
.
Let
To find this alternating sum, it is necessary to reverse the symbol before all terms of odd degree. In
so
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Let 
be an even polynomial function with 
as a factor.
True or false: It follows that 
is also a factor of 
.
Let
True or false: It follows that
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.
By the Factor Theorem,
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