Dilations: Scale Factor & Points Other than Origin

Hi there and welcome to MathSux! Today we are going to break down dilations; what they are, how to find the scale factor, and how to dilate about a point other than the origin. Dilations are a type of transformation that are a bit different when compared to other types of transformations out there (translations, rotations, reflections). Once a shape is dilated, the length, area, and perimeter of the shape change, keep on reading to see how! And if you’re looking for more transformations, check out these posts on reflections and rotations. Thanks so much for stopping by and happy calculating! 🙂

What are Dilations?

Dilations are a type of transformation in geometry where we take a point, line, or shape and make it bigger or smaller, depending on the Scale Factor.

We always multiply the value of the scale factor by the original shape’s length or coordinate point(s) to get the dilated image of the shape. A scale factor greater than one makes a shape bigger, and a scale factor less than one makes a shape smaller. Let’s take a look at how different values of scale factors affect the dilation below:

Scale Factor >1 Bigger

Scale Factor <1 Smaller

Scale Factor=2

 In the below diagram the original triangle ABC gets dilated by a scale factor of 2.  Notice that the triangle gets bigger, and that each length of the original triangle is multiplied by 2.

Dilations

Scale Factor=1/2

Here, the original triangle ABC gets dilated by a scale factor of 1/2.  Notice that the triangle gets smaller, and that each length of the original triangle is multiplied by 1/2 (or divided by 2).

Dilations

Properties of Dilations:

There are few things that happen when a shape and/or line undergoes a dilation.  Let’s take a look at each property of a dilation below:

1. Angle values remain the same.

2. Parallel and perpendicular lines remain the same.

3. Length, area, and perimeter do not remain the same.

Now that we a bit more familiar with how dilations work, let’s look at some examples on the coordinate plane:

Example #1: Finding the Scale Factor

Step 1: First, let’s look at two corresponding sides of our triangle and measure their length.

Dilations

Step 2: Now, let’s look at the difference between the two lengths and ask ourselves, how did we go from 3 units to 1 unit?

Remember, we are always multiplying the scale factor by the original length values in order to dilate an image. Therefore, we know we must have multiplied the original length by 1/3 to get the new length of 1.

Dilations

When it comes to dilations, there are different types of questions we may be faced with.  In the last question, the triangle dilated was done so about the origin, but this won’t always be the case.  Let’s see how to dilate a point about a point other than the origin with this next example.

Example #2: Dilating about a Point other than the Origin

Dilations

Step 1: First, let’s look at our point of dilation, notice it is not at the origin! In this question, we are dilating about point m!  To understand where our triangle is in relation to point m, let’s draw a new x and y axes originating from this point in blue below.

Dilations

Step 2: Now, let’s look at coordinate point K, in relation to our new axes.

Step 3: Let’s use the scale factor of 2 and the transformation rule for dilation, to find the value of its new coordinate point. Remember, in order to perform a dilation, we multiply each coordinate point by the scale factor.

Step 4: Finally, let’s graph the dilated image of coordinate point K. Remember we are graphing the point (6,4) in relation to the x and y-axis that stems from point m.

Dilations

Check out these dilation questions below!

Practice Questions:

1) Plot the image of Point Z under a dilation about the origin by a scale factor of 2.

2) Triangle DEF is the image of triangle ABC after a dilation about the origin. What is the scale factor of the dilation?

Dilations

3) Point L is dilated by a scale factor of 2 about point r. Draw the dilated image of point L.

Dilations

4) Line DE is the dilated image of line AB. What is the scale factor of the dilation?

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions below. Thanks for stopping by and happy calculating! 🙂

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What is the Discriminant?

Hi everyone and welcome to MathSux! In this post, we are going to answer the question, what is the discriminant? Before going any further, if you need a review on what the quadratic equation or imaginary numbers are, check out each related link! Also, don’t forget to check out the video and practice questions below. Happy calculating! 🙂

What is the Discriminant?

The discriminant is a formula we can use that tells us more about a quadratic equation including:

  1. The number of solutions a quadratic equation has.
  2. The “nature” of the roots of the solution (rational/irrational or real/imaginary).

Discriminant Formula:

The discriminant formula may look familiar! It is part of the quadratic formula and we have seen it before, using the very same coefficients a, b, and c from the quadratic equation.

What is the Discriminant?

How does it Work?

When we find the value of the discriminant of any quadratic equation, it will give us a value that tells us how many solutions (or roots) a quadratic equation has.  Remember when we say “roots” what we really mean are the x-value(s) of the quadratic equation that hit the x-axis. This value will also tell us if the solutions to the quadratic equation are rational/irrational or real/imaginary. Take a look at how it all breaks down below:

Now that we are familiar with the rules, let’s take a look at an Example:

Step 1: First let’s write out our quadratic equation and identify the coefficients a, b, and c so they are ready to be plugged into our discriminant formula.

Step 2: Now let’s write out and fill in our formula using the coefficients and solve.

What is the Discriminant?

Step 3: Now let’s analyze our answer! Since, we got a discriminant value of 36, notice that it is a positive perfect square! If we look back at our discriminant table, this tells us that our quadratic equation is going to have 2 real and rational solutions.

Practice Questions:

Find the discriminant, number of solutions and nature of the roots of the following quadratic equations:  

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions below or check out the video above. Thanks for stopping by and happy calculating! 🙂

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Factor By Grouping Examples: Algebra

Hey there math peeps and welcome to MathSux! In today’s post we are going to cover factor by grouping examples, a surprisingly cool and easy factoring method used to factor quadratic equations when “a” is greater than one. It can also be used to factor four term polynomials. We are going to look at an example of each below. If you have any questions, please don’t hesitate to check out the video and try the practice problems at the end of this post. Thanks for stopping by and happy calculating! 🙂

What is Factor by Grouping?

Factor by Grouping is a factoring method that groups common factors of an algebraic expression together.  Many times, we use factoring to find the x-values of a quadratic equation when the coefficient “a” is greater than 1.

When should we use Factor by Grouping?

1) If the first coefficient in a quadratic equation, a, is greater than 1:

Factor By Grouping examples

2) When there is a polynomial with 4 terms:

Factor by Grouping Examples:

Ready for an Example?  Let us look at how to factor a quadratic equation when a is greater than one.

Factor By Grouping examples
Factor By Grouping examples

Now, let’s take a look at another type of example, that can be solved with the help of factor by grouping!

Notice that this question is actually easier to solve than the last! The polynomial above, is already split into 4 terms, therefore, we can jump ahead, skipping the product/sum steps we did in the previous example!

Factor By Grouping examples

Ready to try practice questions on your own? Check them out below to master Factor by Grouping!

Practice Questions:

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions below. Thanks for stopping by and happy calculating! 🙂

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Looking to learn about different Factoring methods? Check out the posts below!

GCF, Product/Sum, Difference of Two Squares, Quadratic Formula

Quadratic Formula, Product/Sum, Completing the Square, Graph

Completing the Square

Free High School Math Resources

Free High School Math Resources

Greetings math friends! Today’s post is for the New York state teachers out there in need of a lesson boost. In this post, we’ll go over what MathSux has to offer for free high school math resources including videos, lessons, practice questions, etc. Remember everything you see here is 100% free and designed to make your life (and your students’ life) easier.

Signing up with MathSux will get you access to FREE:

1) Math Videos

2) Math Lessons

3) Practice Questions

4) NYS Regents Review

Everything designed here aligns with the NYS Common Core Standards for Algebra, Geometry, Algebra 2/Trig. and Statistics.

I am a NYS math teacher that creates free math videos, lessons, and practice questions every week, right here, for you! On the YouTube channel, you’ll also find NYS Common Core Regents questions reviewed one question at a time.

Featured on Google Classrooms around the world, MathSux.org is a great resource especially now, in the time of COVID and zooming and schooling from home. I hope you stick around and find these resources helpful.

And if you’re looking to get the latest MathSux.org videos and emails straight to your inbox, don’t forget to sign up on the right hand-side of the website. Thanks so much for stopping by and happy calculating! 🙂

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Constructing a Perpendicular Bisector

Hi everyone and welcome to MathSux! In this post we are going to be constructing a perpendicular bisector, a line that cuts a line segment in half and creates four 90º angles. It’s a super fast and super simple construction! If you’re looking for more constructions, don’t forget to check more out here. Thanks so much for stopping by and happy calculating! 🙂

What is the Perpendicular Bisector of a line ?

  • Cut’s our line AB in half at its midpoint, creating two equal halves.
  • This will also create four 90º angles about the line.
Constructing a Perpendicular Bisector

What is happening in this GIF?

Step 1: First, we are going to measure out a little more than halfway across the line AB by using a compass.

Step 2: Next we are going to place the compass on point A and swing above and below line AB to make a half circle.

Step 3: Keeping the same distance on our compass, we are then going to place the point of the compass onto Point B and repeat the same step we did on point A, drawing a semi circle.

Step 4: Notice the intersections above and below line AB!? Now, we want to connect these two points by drawing a line with a ruler or straight edge.

Step 5: Yay! We now have a perpendicular bisector! This cuts line AB right at its midpoint, dividing line AB into two equal halves.  It also creates four 90º angles.

Still got questions? No problem! Don’t hesitate to comment with any questions below or check out the video above. Thanks for stopping by and happy calculating! 🙂

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Want to see how to construct a square inscribed in a circle? Or maybe you want to construct an equilateral triangle? Click on each link to view each construction!

Square Inscribed in a Circle Construction

Greetings math friends and welcome to MathSux! In this week’s post, we are going to take a step by step look on how a square inscribed in a circle construction works! We got videos, we got GIF’s, and we got a step by step written explanation below, the choice of learning this construction is up to you! Happy Calculating! 🙂

Square Inscribed in a Circle Construction

How to Construct a Square Inscribed in a Circle:

Step 1: Draw a circle using a compass.

Step 2: Using a ruler, draw a diameter across the length of the circle, going through its midpoint.

Step 3: Open up the compass across the circle. Then take the point of the compass to one end of the diameter and swing the compass above the circle, making a mark.

Step 4: Keeping that same length of the compass, go to the other side of the diameter and swing above the circle again making another mark until the two arcs intersect.

Step 5: Repeat steps 3 and 4, this time creating marks below the circle.

Step 6: Connect the point of intersection above and below the circle using a ruler. This creates a perpendicular bisector, cutting the diameter in half and forming 90º angles.

Step 7: Lastly, use a ruler to connect each corner point to one another creating a square.

Still got questions? No problem! Don’t hesitate to comment with any questions. Happy calculating! 🙂

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Looking for something similar to square inscribed in a circle construction? Check out this post here on how to construct and equilateral triangle here!

Simultaneous Equations: Algebra

Happy new year and welcome to Math Sux! In this post we are going to dive right into simultaneous equations and how to solve them three different ways! We will go over how to solve simultaneous equations using the (1) Substitution Method (2) Elimination Method and (3) Graphing Method. Each and every method leading us to the same exact answer! At the end of this post don’t forget to try the practice questions choosing the method that best works for you! Happy calculating! 🙂

What are Simultaneous Equations?

Simultaneous Equations are when two equations are graphed on a coordinate plane and they intersect at, at least one point.  The coordinate point of intersection for both equation is the answer we are trying to find when solving for simultaneous equations. There are three different methods for finding this answer:

We’re going to go over each method for solving simultaneous equations step by step with the example below:

Method #1: Substitution

The idea behind Substitution, is to solve for 1 variable first algebraically, and the plug this value back into the other equation solving for one variable.  Then solving for the remaining variable.  If this sounds confusing, don’t worry! We’re going to do this step by step:

Step 1: Let’s choose the first equation and move our terms around to solve for y.

Simultaneous Equations

Step 3: The equation is set up and ready to solve for x!

Simultaneous Equations

Step 4: All we need to do now, is plug x=3 into one of our original equations to solve for y.

Simultaneous Equations

Step 5: Now that we have solved for both x and y, we have officially found where these two simultaneous equations meet!

Simultaneous Equations

Method #2: Elimination

The main idea of Elimination is to add our two equations together to cancel out one of the variables, allowing us to solve for the remaining variable.  We do this by lining up both equations one on top of the other and adding them together.  If variables at first do not easily cancel out, we then multiply one of the equations by a number so it can. Check out how it’s done step by step below!

Step 1: First, let’s stack both equations one on top of the other to see if we can cancel anything out:

Simultaneous Equations

Step 2: Our goal is to get a 2 in front of y in the first equation, so we are going to multiply the entire first equation by 2.

Simultaneous Equations

Step 3: Now that we multiplied the entire first equation by 2, we can line up our two equations again, adding them together, this time canceling out the variable y to solve for x.

Simultaneous Equations

Step 4: Now, that we’ve found the value of variable x=3, we can plug this into one of our equations and solve for missing unknown variable y.

Simultaneous Equations

Step 5: Now that we have solved for both x and y, we have officially found where these two simultaneous equations meet!

This image has an empty alt attribute; its file name is Screen-Shot-2020-12-30-at-10.21.46-AM.png

Method #3: Graphing

The main idea of Graphing is to graph each a equation on a coordinate plane and then see at what point they intersect.  This is the best method to visualize and check our answer!

Step 1: Before we start graphing let’s convert each equation into y=mx+b (equation of a line) form.

Equation 1:

Equation 2:

Step 2: Now, let’s graph each line, y=3x-4 and y=-x+8, to see at what coordinate point they intersect.

Simultaneous Equations

Need to review how to draw an equation of a line? Check out this post here! Notice we got the same exact answer using all three methods (1) Substitution (2) Elimination and (3) Graphing.

Ready to try the practice problems on your own?! Check them out below!

Practice Questions:

Solve the following simultaneous equations for x and y.

Solutions:

  1. (1, 3)
  2. (4,5)
  3. (-1, -6)
  4. (3, -3)

Want more MathSux?  Don’t forget to check out our Youtube channel and more below! And if you have any questions, please don’t hesitate to comment below. Happy Calculating!

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30 60 90 Special Triangles: Geometry

Hi everyone and welcome to MathSux! In this post we are going to break down 30 60 90 degree special triangles. What is it? Where did it come from? What are the ratios of it’s side lengths and how to do we use them? You will find all of the answers to these questions below. Also, don’t forget to check out the video below and practice questions at the end of this post. Happy calculating! 🙂

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What is a 30 60 90 Triangle and why is it “Special”?

The 30 60 90 triangle is special because it forms an equilateral triangle when a mirror image of itself is drawn, meaning all sides are equal!  This allows us to find the ratio between each side of the triangle by using the Pythagorean theorem. Check it out below!

30 60 90 Special Triangles

Now let’s draw a mirror image of our triangle.  Next, we can label the length of the new side opposite 30º “a,” and add this new mirror image length with the original we had to get, a+a=2a.

30 60 90 Special Triangles
30 60 90 Special Triangles

If we look at our original 30 60 90 triangle, we now have the following values for each side based on our equilateral triangle:

30 60 90 Special Triangles
30 60 90 Special Triangles

Now we can re-label our triangle, knowing the length of the hypotenuse in relation to the two legs. This creates a ratio that applies to all 30 60 90 triangles!

30 60 90 triangle side lengths

How do I use this ratio?

30 60 90 triangle side lengths

Knowing the above ratio, allows us to find any length of any and every 30 60 90 triangle, when given the value of one of its sides.

Let’s try an Example:

30 60 90 triangle side lengths

-> First let’s look at our ratio and compare it to our given triangle.

30 60 90 triangle side lengths

->Notice we are given the value of a, which equals 4, knowing this we can now fill in each length of our triangle based on the ratio of a 30 60 90 triangle.

30 60 90 triangle side lengths
30 60 90 triangle side lengths

Now let’s look at an Example where we are given the length of the hypotenuse and need to find the values of the other two missing sides.

30 60 90 triangle side lengths

->First let’s look at our ratio and compare it to our given triangle.

30 60 90 triangle side lengths

-> Notice we are given the value of the hypotenuse, 2a=20. Knowing this we can find the value of a by dividing 20 by 2 to get a=10. Once we have the value of a=10, we can easily find the length of the last leg based on the 30 60 90 ratio:

30 60 90 triangle side lengths
30 60 90 triangle side lengths

Now for our last Example, when we are given the side length across from 60º and need to find the other two missing sides.

30 60 90 triangle side lengths

->First let’s look at our ratio and compare it to our given triangle.

30 60 90 triangle side lengths

-> In this case, we need to use little algebra to find the value of a, using the ratio for 30 60 90 triangles.

30 60 90 triangle side lengths

Now that we have one piece of the puzzle, the value of a, let’s fill it in our triangle below:

Finally, let’s find the value of the length of the hypotenuse, which is equal to 2a.

Practice Questions:

Find the value of the missing sides of each 30 60 90 degree triangle.

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above. Happy calculating! 🙂

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Looking to review 45 45 90 degree special triangles? Check out this post here!

Trig Functions (Amp, Freq, Phase Shifts): Algebra 2/Trig.

Hi everyone, and welcome to MathSux! In this post we are going to break down how to graph trig functions by identifying its amplitude, frequency, period, and horizontal and vertical phase shifts. Fear not! Because we will breakdown what each of these mean and how to find them, then apply each of these changes step by step on our graph. And if you’re ready for more, check out the video and the practice problems below, happy calculating! 🙂

*For a review on how to derive basic Trig functions (y=sinx, y=cosx, and y=tanx), click here.

What are the Different Parts of a Trig Function?

Trig functions

Amplitude: The distance (or absolute value) between the x-axis and the highest point on the graph.

Frequency: This is the number of cycles that happen between 0 and 2π. (Α “cycle” in this case is the number of “s” cycles for the sine function).

Period: The x-value/length of one cycle. (Α “cycle” in this case is the number of “s” cycles for the sine function). This is found by looking at the graph and seeing where the first cycle ends, or, by using the formula:       

Horizontal Shift: When a trigonometric function is moved either left or right along the x-axis.

Vertical Shift: When a trigonometric function is moved either up or down along the y-axis.

Let’s try an Example, graphing a Trig Function step by step.

Step 1: First let’s label and identify all the different parts of our trig function.

Trig functions

Step 2: Now let’s transform our graph one step at a time.  First let’s start graphing y=cos(x) without any transformations.

Trig functions

Step 3: Let’s add our amplitude of 2, the distance to the x-axis.  To do this our highest and lowest points on the y-axis will now be moved to 2 and -2 respectively. 

Trig functions

Step 4: Next, we do a horizontal phase shift to the left by (π/2).  To do this, we look at where negative (π/2) is on our graph at (-π/2) and move our entire graph over to start at this new point, “shifting” it to (π/2).

Trig functions

Step 5: For our last transformation, we have a vertical phase shift up 1 unit.  All this means is that we are going to shift our entire graph up by 1 unit along the y-axis.

Trig functions

Practice Questions:

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above for an in depth explanation. Happy calculating! 🙂

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Graphing Linear Inequalities: Algebra

graphing linear inequalities

Hi and welcome to MathSux! In this post, we are going to go over the rules for graphing linear inequalities on a coordinate plane when it comes to drawing lines, circles , and shading, then we are going to solve an example step by step. If you have any questions, check out the video below and try the practice questions at the end of this post! If you still have questions, don’t hesitate to comment below and happy calculating! 🙂

Graphing Linear Inequalities:

When graphing linear inequalities, we always want to treat the inequality as an equation of a line in  form y=mx+b….with a few exceptions:

Graphing Linear Inequalities
inequality shading above or below y-axis

Now that we know the rules, of graphing inequalities, let’s take a look at an Example!

graphing inequality example

Step 1: First, let’s identify what type of inequality we have here.  Since we are working with a > sign, we will need to use a dotted line and open circles when creating our graph.

graphing inequality example

Step 2: Now we are going to start graphing our linear inequality as a normal equation of a line, by identifying the slope and the y-intercept only this time keeping open circles in mind.  (For a review on how to graph regular equation of a line in y=mx+b form, click here)

graphing inequality example
graphing linear inequalities

Step 3: Now let’s connect our dots, by using a dotted line to represent our greater than sign.

graphing linear inequalities

Step 4: Now it is time for us to shade our graph, since this is an inequality, we need to show all of our potential solutions with shading.  Since we have a greater than sign, , we will be shading above the y-axis.  Notice all the positive y-values above are included to the left of our line.  This is where we will shade.

graphing linear inequalities

Step 5: Check!  Now we need to check our work.  To do that, we can choose any point within our shaded region, if the coordinate point we chose hold true when plugged into our inequality then we are correct!

Let’s take the point (-3,2) plugging it into our inequality where x=-3 and y=2.

graphing linear inequalities

Practice Questions:

graphing linear inequalities

Solutions:

graphing linear inequalities
graphing linear inequalities

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above for an in depth explanation. Happy calculating! 🙂

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