Functions and Graphs - Math
Card 0 of 144
List the transformations that have been enacted upon the following equation:
List the transformations that have been enacted upon the following equation:
Since the equation given in the question is based off of the parent function 
, we can write the general form for transformations like this:
determines the vertical stretch or compression factor.
- If
is greater than 1, the function has been vertically stretched (expanded) by a factor of 
.
- If
is between 0 and 1, the function has been vertically compressed by a factor of 
.
In this case, 
is 4, so the function has been vertically stretched by a factor of 4.
determines the horizontal stretch or compression factor.
- If
is greater than 1, the function has been horizontally compressed by a factor of 
.
- If
is between 0 and 1, the function has been horizontally stretched (expanded) by a factor of 
.
In this case, 
is 6, so the function has been horizontally compressed by a factor of 6. (Remember that horizontal stretch and compression are opposite of vertical stretch and compression!)
determines the horizontal translation.
- If
is positive, the function was translated 
units right.
- If
is negative, the function was translated 
units left.
In this case, 
is 3, so the function was translated 3 units right.
determines the vertical translation.
- If
is positive, the function was translated 
units up.
- If
is negative, the function was translated 
units down.
In this case, 
is -7, so the function was translated 7 units down.
Since the equation given in the question is based off of the parent function
- If
- If
In this case,
- If
- If
In this case,
- If
- If
In this case,
- If
- If
In this case,
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Based on the figure below, which line depicts a quadratic function?
Based on the figure below, which line depicts a quadratic function?
A parabola is one example of a quadratic function, regardless of whether it points upwards or downwards.
The red line represents a quadratic function and will have a formula similar to 
.
The blue line represents a linear function and will have a formula similar to 
.
The green line represents an exponential function and will have a formula similar to 
.
The purple line represents an absolute value function and will have a formula similar to 
.
A parabola is one example of a quadratic function, regardless of whether it points upwards or downwards.
The red line represents a quadratic function and will have a formula similar to
The blue line represents a linear function and will have a formula similar to
The green line represents an exponential function and will have a formula similar to
The purple line represents an absolute value function and will have a formula similar to
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Which of the following functions represents a parabola?
Which of the following functions represents a parabola?
A parabola is a curve that can be represented by a quadratic equation. The only quadratic here is represented by the function 
, while the others represent straight lines, circles, and other curves.
A parabola is a curve that can be represented by a quadratic equation. The only quadratic here is represented by the function
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Write 
in slope-intercept form.
Write
Slope-intercept form is 
.
Slope-intercept form is
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Find the 
-intercepts for the circle given by the equation:
Find the
To find the 
-intercepts (where the graph crosses the 
-axis), we must set 
. This gives us the equation:
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
and 
We can then solve these two equations to obtain 
.
To find the
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
We can then solve these two equations to obtain
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Find the 
-intercepts for the circle given by the equation:
Find the
To find the 
-intercepts (where the graph crosses the 
-axis), we must set 
. This gives us the equation:
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
and 
We can then solve these two equations to obtain
To find the
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
We can then solve these two equations to obtain
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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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Let 
. What is 
?
Let
We are asked to find 
, which is the inverse of a function.
In order to find the inverse, the first thing we want to do is replace f(x) with y. (This usually makes it easier to separate x from its function.).
Next, we will swap x and y.
Then, we will solve for y. The expression that we determine will be equal to 
.
Subtract 5 from both sides.
Multiply both sides by -1.
We need to raise both sides of the equation to the 1/3 power in order to remove the exponent on the right side.
We will apply the general property of exponents which states that 
.
Laslty, we will subtract one from both sides.
The expression equal to y is equal to the inverse of the original function f(x). Thus, we can replace y with 
.
The answer is 
.
We are asked to find
In order to find the inverse, the first thing we want to do is replace f(x) with y. (This usually makes it easier to separate x from its function.).
Next, we will swap x and y.
Then, we will solve for y. The expression that we determine will be equal to
Subtract 5 from both sides.
Multiply both sides by -1.
We need to raise both sides of the equation to the 1/3 power in order to remove the exponent on the right side.
We will apply the general property of exponents which states that
Laslty, we will subtract one from both sides.
The expression equal to y is equal to the inverse of the original function f(x). Thus, we can replace y with
The answer is
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What is the inverse of 
?
What is the inverse of
The inverse of 
requires us to interchange 
and 
and then solve for 
.
Then solve for 
:
The inverse of
Then solve for
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If 
, what is 
?
If
To find the inverse of a function, exchange the 
and 
variables and then solve for 
.
To find the inverse of a function, exchange the
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Find the -intercepts for the circle given by the equation:
Find the
To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
and
We can then solve these two equations to obtain .
To find the
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
We can then solve these two equations to obtain
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Find the -intercepts for the circle given by the equation:
Find the
To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
and
We can then solve these two equations to obtain
To find the
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
We can then solve these two equations to obtain
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Solve:
Solve:
Use substution to solve this problem:
becomes 
and then is substituted into the second equation. Then solve for 
:
, so 
and 
to give the solution 
.
Use substution to solve this problem:
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Solve for the 
- and 
- intercepts:
Solve for the
To solve for the 
-intercept, set 
to zero and solve for 
:
To solve for the 
-intercept, set 
to zero and solve for 
:
To solve for the
To solve for the
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Solve:
Solve:
Use substution to solve this problem:
becomes and then is substituted into the second equation. Then solve for :
, so and to give the solution .
Use substution to solve this problem:
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Solve for the - and - intercepts:
Solve for the
To solve for the -intercept, set to zero and solve for :
To solve for the -intercept, set to zero and solve for :
To solve for the
To solve for the
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Find the center and radius of the circle defined by the equation:
Find the center and radius of the circle defined by the equation:
The equation of a circle is: 
where 
is the radius and 
is the center.
In this problem, the equation is already in the format required to determine center and radius. To find the 
-coordinate of the center, we must find the value of 
that makes 
equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that 
. To find the radius we take the square root of the constant on the right side of the equation which is 6.
The equation of a circle is:
In this problem, the equation is already in the format required to determine center and radius. To find the
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Find the center and radius of the circle defined by the equation:
Find the center and radius of the circle defined by the equation:
The equation of a circle is: 
where 
is the radius and 
is the center.
In this problem, the equation is already in the format required to determine center and radius. To find the 
-coordinate of the center, we must find the value of 
that makes 
equal to 
, which is 
. We do the same to find the y-coordinate of the center and find that 
. To find the radius we take the square root of the constant on the right side of the equation which is 10.
The equation of a circle is:
In this problem, the equation is already in the format required to determine center and radius. To find the
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Find the radius of the circle given by the equation:
Find the radius of the circle given by the equation:
To find the center or the radius of a circle, first put the equation in the standard form for a circle: 
, where 
is the radius and 
is the center.
From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula 
.
, so 
.
and 
, so 
and 
.
Therefore, 
.
Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:
Now we do the same for 
:
We can now find 
:
To find the center or the radius of a circle, first put the equation in the standard form for a circle:
From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula
Therefore,
Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:
Now we do the same for
We can now find
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Find the center of the circle given by the equation:
Find the center of the circle given by the equation:
To find the center or the radius of a circle, first put the equation in standard form: 
, where 
is the radius and 
is the center.
From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula 
.
, so 
.
and 
, so 
and 
.
This gives 
.
Because the constant, in this case 9, was not in the original equation, we must add it to both sides:
Now we do the same for 
:
We can now find the center: (3, -9)
To find the center or the radius of a circle, first put the equation in standard form:
From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula
This gives
Because the constant, in this case 9, was not in the original equation, we must add it to both sides:
Now we do the same for
We can now find the center: (3, -9)
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