Findng Inverses of Formulas and Compositions of Inverse Functions

In this video math lesson, working with inverse functions is discussed. It is good to remember, when finding the the inverse of a formula, DO NOT SWITCH THE VARIABLES. The first example models finding the inverse of an existing formula:
d = 16t^2
16t^2 = d Symmetric Property of Equality
t^2 = \frac{d}{16} Division Property of Equality
\sqrt {t^2} = \sqrt{ \frac{d}{16}} Inverse of Square is Square Root
t^ = \frac{ \sqrt{d}}{4} Simplify Fraction to Simplest Radical Form

I know some readers may not be able to follow the above problem. The same problem, but with the example in the video, the auditory learner can benefit too.

The second example models finding the composition of a function and its inverse. This can be written as f^{-1}(f(x)) or f(f^{-1}(x)). In both cases are equal to the value of x. Performing the composition of a function and its inverse gives the value you started with. You will see in the video, how simple this process is.

If you have a question or this video helped you, leave a comment.

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Find the Domain of a Function and Its Inverse

In this post, I have embedded an Algebra 2 Video Math Lesson about the function f(x) = \sqrt {2x+2}. Before considering the function, the video defines f^{-1} as the inverse of f or as f inverse.

The video models how to:

  1. Find the domain and range of f(x) = \sqrt {2x+2}
  2. Find f^{-1}
  3. Find the domain and range of f^{-1}
  4. Determine if f^{-1} is a function

If you have any questions regarding the video lesson or other questions regarding finding the inverse of a function, use the comments section.

How To Find and Graph the Inverse of a Function

In this video Algebra 2 Math Lesson, I model how to graph a quadratic function and its inverse. The process starts with graphing a parabola in the form y=a^2+c. The vertex is given by (0,c). After graphing the parabola with four additional points, I create the inverse’s graph by moving the points about y = x line. Finally, I find the inverse of the original function.

If this video math lesson on finding the inverse of a function and then graphing it helped you, leave a comment. Heck, if it didn’t help leave a comment and let me know how to make it better.

How to Find the Inverse of a Relation and an Equation

This video lesson covers the definition of the inverse of a relation and reviews the concept of how to find the inverse of a relation.

Example 1 covers how to find the inverse of a relation from a table of values and provides a good visual of a relation that is a function and its inverse is not a function.

Before example 2, a discussion about how switching the x and the y in the equation is the best method for finding the inverse of a function. Example 2 models how to use the x and y switcheroo to find the inverse of a given function.

Using Properties of Parabolas to Graph a Parabola Problem 2

As I discussed in an earlier post on graphing a parabola, the following information allows to easily graph a parabola in standard form f(x)=ax^2+bx+c.

Properties of a Parabola or a Quadratic Function in Standard Form

Given that f(x)=ax^2+bx+c is a parabola when a\neq0.

  1. If a > 0, then the graph opens upward. If a < 0, then the graph opens downward.
  2. The axis of symmetry is given by the line -\frac{b}{2a}.
  3. The vertex is given by the ordered pair (-\frac{b}{2a},f(-\frac{b}{2a}))
  4. Finding 2 other points is easy. Choose any x-value and find the matching y-value. The second point is found by using the axis of symmetry and the location of the 1st points corresponding point.

In my first post the equation did not have a linear term and was in the form f(x)=ax^2+ c. The vertex could be found using (0,c) and the axis of symmetry (AOS) of x = 0. In terms of the coordinate plane, the y-axis is the AOS.

In this post, we change gears and will be graphing a parabola that is given in standard form f(x)=ax^2+bx+c. The point of this work is to sketch a graph using the core concepts behind the parabolic relation. Remember, because a parabola is symmetric every point except the vertex has a mirror image. Once you find the AOS and vertex, you will have to use the same process to find another point and then its mirror image. Let’s take this to the example.

Example Graphing a Quadratic Equation in Standard Form

Graph f(x)=2x^2+4x-3

Looking at the properties above in order, first we know this graph will open upwards, this is due to the fact the coefficient on the squared term is positive or a>0. The AOS is found by using the ratio -\frac{b}{2a}=-\frac{4}{2(2)}=-1 and can be written as x = -1. If you are trying to create the graph at the bottom of this post, you should draw a vertical dashed line at x = -1, because this represent the AOS.

Next we find the vertex. Since the equation or function is given as f(x)=2x^2+4x-3 and x = -1, it is natural to substitute -1 to get f(-1)=2(-1)x^2+4(-1)-3. Using the order of operations to simplify the right side, we arrive at f(-1)=-5. If you have no idea what is going on, here is an explanation. The AOS serves as the x-coordinate of the vertex. Since we know the x-value, we can find the y-value (***f(x) is the same as using the variable y because of ‘function notation’***). By plugging in the known x-value, we can find the related y-value and we now have an ordered pair. For this example the vertex is the ordered pair is (-1,-5). If you are trying to graph this problem, you should plot the point (-1,-5)

Lastly, we need to find two more points. To do this, pick any x-value to the left or right of the AOS. For this example I will use x = -3. Using the given function, f(x)=2x^2+4x-3, substituting -3 in for x gives f(-3)=2(-3)^2+4(-3)-3 applying the order of operations f(x)=3 and the ordered pair (-3,3). Again, because parabolas are symmetric, (-3,3) has a mirror image at (1,3). If you are graphing along with the post, plot the points and draw a curve to fit those points. Even though this is a sketch, you should try to make it neat.

Graphing a Parabola in Standard Form

Graphing a Parabola in Standard Form

Find the Quadratic Function That Models 3 Ordered Pairs

Find the Quadratic Function That Models 3 Ordered Pairs

Figure 1 is the solution to the second example on a post about finding the equation to a parabola when given three points.  The directions from the previous post read, “Find the quadratic function that models the given ordered pairs: (-2,-17), (-1,10), (5,-10).”

I plan on writing an explanation, but for now I wanted to get the solution up!

Cheers.



Find the Quadratic Function That Models 3 Ordered Pairs

Figure 1

Modeling Data With Quadratic Functions

Find the Quadratic Function That Models 3 Ordered Pairs

In a previous post about quadratic functions and the graphs, I discussed the standard form of a quadratic equation, f(x)=ax^2+bx+c. Because a parabola is symmetric, its equation can be found by solving a system of equations with three variables. I have only discussed solving equations with 2 variables previously.  If you know how to solve a system with 2 variables, then a system with 3 variables should be no problem. The process is rather simple. I will explain how to solve the system of three equations as well as a system of two equations in the examples.

Example 1 Find the Quadratic Function That Models 3 Ordered Pairs

Before discussing how to find the solution, there should be a discussion about what is to be found. Ultimately, the solution should be an equation in the standard form of a quadratic, f(x)=ax^2+bx+c. To find this equation or function involves multiple parts and each part consists of several steps. Many times we get so caught up in the details, we forget the overall picture.

The answer is: f(x)=x^2-6x+7. Yes, I just gave you the answer. Now, the question is:

Find the quadratic function that models the given ordered pairs: (2,-1), (3,-2), (1,2). Here is how you turn 3 ordered pairs into a quadratic function.

First you must create a system of three equations. You create this system of equations by substituting the ordered pairs into the standard form of a quadratic equation f(x)=ax^2+bx+c. Using the reflexive and substitution properties of equality, I use the equation ax^2+bx+c=y. This will make solve the system of equations easier.

Substituting the ordered pairs gives the following equations:

(2,-1) gives a(2^2)+b(2)+c=-1

(3,-2) gives a(3^2)+b(3)+c=-2

(1,2) gives a(1^2)+b(1)+c=2

Simplifying these equations, maintaining the order from above and labeling them as equations A, B and C results with:

4a+2b+c=-1 (A)

9a+3b+c=-2 (B)

a+b+c=2 (C)

Now that you have a system of three equations, it is good to review the overall goal, to find the coefficients of a quadratic function that models the ordered pairs given. Okay, on with solving the system of 3 equations. You need to use the 3 equations and the elimination method to create a new system of equations with only 2 variables. You have to choose two pairs of equations and eliminate a variable. For this example, I will subtract C from A to create a new equation D and I will subtract C from B to create equation E.

(A) 4a+2b+c=-1

\underline{-(C) a+b+c=2}

(D) 3a+b=-3

(B) 9a+3b+c=-2

\underline{-(C) a+b+c=2}

(E) 8a+2b=-4

Equations D and E form a new system of equations.

3a+b=-3 (D)

8a+2b=-4 (E)

To solve this system, it would be easiest to subtract with E from equation D multiplied by 2.

2(3a+b=-3) (D)

\underline{-(8a+2b=-4)} (E)

6a+2b=-6 (D)

\underline{-(8a+2b=-4)} (E)

-2a=-2  Divide both sides by -2 and

a=1

You now have the coefficient of the quadratic term. Using a = 1, you substitute into either equation D or E. For this problem, I used equation D.

a=1

3a+b=-3

3(1)+b=-3  Substitute 1 for a.

3+b=-3  Simplify.

b=-6  Subtract 3 from each side.

This is the coefficient of the linear term. Now use both a = 1 and b = -6 to find the value of the constant. You must use one of the original equations A, B or C. I used equation C because it is very simple.

a=1 and b=-6

a+b+c=2 (C)

1+(-6)+c=2  Substitute 1 for a and -6 for b.

-5+c=2  Simplify.

c=7  Add 2 to both sides.

Finally, we have all of the coefficients a = 1, b = -6 and c = 7 and can write the quadratic function that models the given points.

a=1, b=-6, c=7

f(x)=ax^2+bx+c

f(x)=x^2-6x+7 This is the solution.

But how do you know it is correct? You should check your work by substituting the values a = 1, b = -6 and c = 7 into one of the equations A, B or C. Since I already used C I will use equation A.

a=1, b=-6, c=7

4a+2b+c=-1 (A)

4(1)+2(-6)+7=-1  Substitute.

4+(-12)+7=-1  Simplify.

-8+7=-1  Simplify.

-1=-1  Simplify and it checks.

This is just one example of finding the equation of a quadratic function from three points. Also, because the three original equations started with all three variables, it was a more complicated type.

Example Find the Quadratic Function That Models 3 Ordered Pairs For You To Do

Find the quadratic function that models the given ordered pairs:

(-2,-17), (-1,10), (5,-10).

Click here for the solution.

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