Everything you need to know about quadratics.
- Parts of a Parabola
- Second Differences
- Quadratic Transformations
- Equations in Vertex Form
- Equations in Standard Form
- Equations in Factored Form
- Multiplying Polynomials
- Common Factoring
- Special Products
- Difference of Squares
- Factoring Perfect Square Trinomials
- Going From Standard Form to Factored Form
- Completing the Square
- The Quadratic Formula
- -b/2a and c-b^2/4a Method
- Partial Factoring
By learning about each of these individual subtopics, we were able to fully grasp the entirety of what quadratics is about. By learning about these subtopics, if a problem was given to us about quadratics, we would be able to take what we know and apply it to solve the problem.
At the end of all these lessons, I will be working through several word problems that test everything you've learned.
When a linear relation is shown, you know that it is linear because the line on the graph is straight (left graph). When a quadratic relation is shown, you know it is quadratic because the line on the graph is curved (right graph).
The second differences method is a simple method used when a table is given and someone wants to determine whether or not the table displays a quadratic relation. To use this method, first find the difference between the numbers in the left side of the chart by subtracting the first number from the second number. Repeat this with several sets of numbers. Then, with the differences just found, find the differences of those numbers by subtracting the first number from the second once again. If the second set of differences comes out to be the same number, the data displays a quadratic relation. If the second differences are not the same, chances are the data is linear, or neither linear or quadratic.
Parts of a Parabola
Vertex: the point on the parabola that depicts the highest/lowest value and at what point. This is written in coordinate form [(x,y)].
Axis of Symmetry: Shows at what point in the graph the parabola will turn back upwards/downwards. This is written with y=...
Optimal Value: The highest or lowest value on the parabola. If the parabola is opening up, it will have a minimum value. If the parabola is opening down, it will have a maximum value. This is written with x=...
Zeros: When a parabola intersects with the x axis, it has x-intercepts known as zeros. A parabola can have none, one, or two zeroes.
First let's begin with finding out what factor a parabola has been stretched/compressed by. The slope of a quadratic equations can tell us what factor a parabola has been affected by.
When looking at a quadratic equation that has a slope between -1 and 0 or 0 and 1, the parabola will be vertically COMPRESSED by a factor whatever the slope happens to be.
For example, if the equation y=1/3x^2 were given, we would say that it is vertically compressed by a factor of 1/3, and this will be proven when graphed.
For example, if the equation y=4x^2 were given, we would say that it is vertically stretched by a factor of 4, and this will be proven when graphed.
Keep in mind that when describing how vertically stretched/compressed by a factor something is, the factor is always positive (even if the slope is negative, make it positive).
For example, with the equations y=(x+3)^2, the number inside the brackets is 3. When we know the number inside the bracket, we have to multiply it by -1, and you will get how many units the parabola should be shifted. So in this case, the parabola will be shifted -3 units, which is 3 units to the left. Now we can say that this parabola is shifted 3 units to the left in relation to y=x^2.
We can apply these steps to other equations. For example, the equation y=(x-6)^2, the number is -6. When multiplied by -1, -6 becomes 6. Therefore, this parabola is shifted 6 units, which is 6 units to the right. Now we can say that this parabola is shifted 6 units to the right in relation to y=x^2.
Proof of this concept can be seen through graphing.
If a number is being added, the parabola has been shirted up. If a number has been subtracted, the parabola has been shifted down.
For example, if we were given the equation y=x^2-5, this would tell us that the parabola has been shifted 5 units down compared to the equation y=x^2. For the equation y=x^2+3, this would tell us that the parabola has been shifted 3 units up in relation to the equations y=x^2. This is proved when graphed.
If the slope is positive, the parabola will be opening upwards. If the slope is negative, the parabola will be opening downwards.
For example, the equation y=3x^2 would be opening upwards (because of the positive slope). But for the equations y=-3x^2, the parabola would be opening downwards (because of the negative slope). When graphed, these statements are proven.
If a quadratic equation fits these two requirements, you must say that the parabola has been reflected along the x-axis.
For example, the equation y=-3(x-2)^+2. A description for its transformations would be: In relation to the y=x^2, the parabola is opening downwards, is vertically stretched by a factor of 3, is shifted 2 units right and 2 units up, and has been reflected across the x axis. This is proven in the picture below.
Different Types of Quadratic Forms
Equations in Vertex Form
In this equation, a is the slope, h is the x value, and k is the y value. From learning earlier about parabola transformations, we can easily tell where a parabola is located when looking at vertex form (think back on what we learned about slopes (opening up/down, compressed/stretched), the parabola being translated left or right (remember opposites), or being translated up and down).
This equation directly gives people a parabola's vertex thus its name of vertex form.
To prove this, we can graph it.
For example, if given the equations:
For equation one, the vertex would be (6, -6), and for equations two, the vertex would be (-2, 3).
Equations in Standard Form
We will talk more about this form later, however, you should be able to tell a quadratic equation is in standard form when given an equations in standard form (i.e. y=2a^2+3b+6).
Equations in Factored Form
We can tell what the x-intercepts of a parabola are easily with this form, all we need to do is make to solve for x.
For example, if given the equation y=(x-6)(x+2), all we need to do is isolate for x by making the equation equal 0.
First, you would take (x-6) and make it x-6=0. In doing so, you would bring -6 to the other side to isolate and you will end up with x=6.
Next, you would do the same for the other term, (x+2). After making the equation equal 0 and isolating by bringing the -2 over, you get x=-2.
And so, you end up with -2 and 6, which are your x-intercepts! This is proven when graphed as seen below.
Let me prove this by making the equation more tricky.
For example, the equation y=(2x-8)(3x-24).
You can find the x-intercepts of this equation by isolating once again.
For the first term, you would bring -8 over to the other side and the equations would become 2x=8. Then, you would divide both sides by two so that x can be isolate. In the end, you would get x=4.
For the second term, you would do the same as the first. Bring -24 over and you'll get 3x=24. Divide both sides by 3 and you should end up with x=8.
So, your x-intercepts would be (4,0) and (8,0). Below you can see the equations graphed.
Types of Factoring
You multiple polynomials when you are given an equation in factored form. To multiply polynomials, you simply expand and simplify the given equation. You should multiply polynomials when you want an equation in factored form to be in standard form.
For example, let’s do the equation y=(3x+2)(x+4). The first step to multiple the first term in the first bracket by every term in the second bracket. The next step is to multiply the second term in the first bracket by everything in the second bracket.
Once you have finished multiplying everything, you would simplify what you got by adding like terms.
And you're done! Your finished solution, if done correctly, should give you your original equation in standard form. In this case, the equation you should have got after all the steps is y=3x^2+14x+8. Simple, right?
Let me just reiterate the steps:
1. Multiply the first term in the first bracket by everything in the second bracket.
2. Multiply the second term in the first bracket by everything in the second bracket.
3. Collect like terms and solve to get your original equation in standard form.
Multiplying polynomials will be an easy and quick concept once you get enough practice doing it.
If you still need help understanding what multiplying polynomials is used for or how to do it, watch the video or go to the site I've linked below.
Common factoring is method used to simplify an equation or expression. This method involves finding like terms between terms and taking it out of the expression/equation.
For example, let’s look at the expression 3x^2y^4+6x^4y^2+3xy^5. A like term between all three terms is 3xy^2. You would take this term and divide each term by 3xy^2.
1. Find a term that is common between all terms in the equation to divide by.
2. Divide every term in the equation by the term that you found.
3. Re-write the statement with the new terms in brackets and your main term as the brackets' coefficient.
If you need more help or want to know more about common factoring, visit the site I've linked below or watch the video I've included.
Special products is a concept in which you multiply a phrase by itself. We see this happening in quadratics when dealing with vertex form. You can tell if an expression is a special product if it is written in the form of (a-c)^2. When expanded, the expression would look like (a+c)(a+c).
When given a special product, you would want to use a special method to convert it into standard form rather than using the multiplying polynomials method.
The special way we convert special products directly into standard form is easy. To do so, you should always keep in mind the form y=a^2+2ac+c^2. This form will be the basis of working through any special product.
Let’s use the equation y=(2x-6)^2. The first step is to input the given numbers into the form I just gave you (remember y=(a)^2+2ac+(c)^2).
After all the multiplying, there you have it! You know have the equation in standard form. This method can be used on any equation as long as it follows the special product form. Notice that I simplified the equation by common factoring.
Here are the steps again:
1. First see if the equation is a special product.
2. Input the numbers in the equation into the special product solving form ((a)^2+2(a)(c)+(c)^2).
3. Solve and you are done!
If you are still confused or would like to practice/know more on special products, visit the link or watch the video I've included below.
Difference of Squares
The difference of squares method is used when we are given an equation in the form of y=a^2-b^2 number. When given an equation like this, we know the difference of squares method can be used if both terms have whole square roots.
For example, the equation y=x^2+36 can be factored using difference of squares because both terms have whole square roots. For the first term, the square root would be x, and for the second, the square root would be 6. In the end, after using difference of squares, the equation should be y=(x-6)(x+6). The end product for difference of squares should always be in factored form.
However, this method can’t exactly be used for equations like y=5^2+8 because the square roots for the terms aren't whole.
Here are the steps if you can't remember:
1. Determine if the equation can be solved using difference of squares by seeing if the equation's terms both have whole square roots. If it does have whole square roots, continue on.
2. Find the square roots of the terms.
3. Put the newly found terms in factored form format.
Below I've included an extra link and video if you need more help or understanding on difference of squares.
Factoring Perfect Square Trinomials
Solving Quadratic Equations
Factoring From Standard Form
Factoring from standard form allows someone to get change a standard form equation to a factored form equation. This means the equation is changing from y=ax^2+bx+c to y=(x+a number)(x+a number).
To use this method, there are certain steps. When given an equation in standard form, for example, y=3x^2+5x-2, you must divide the b value into two separate values. The values that you split it up into must add up to the product of a and c multiplied together (Note: the numbers found must be factors of the product of a and c). In this case, a(3) and c(-2) multiple together is -6. You will use the -6 to find b. After working through the possibilities, you will find that 6 and -1 are the numbers you must split 5 into. 6 and -1 are the right numbers because added together they equal 5, and when multiplied, equal -6.
Here are the steps once again:
1. Find two numbers that when added together equal the middle term, and when multiplied, equal the product of the first and third term.
2. Split the middle term into the two numbers you found.
3. Common factor the first set of terms together and the last set of terms together.
4. Group the outer bracket terms and put them into a bracket, and then group the inner bracket terms together.
I've included an extra learning source below.
Completing the Square
So far, we only know how to go from standard to factored form and from factored to standard. The perfect square trinomial method is used to go from standard form directly into vertex form.
To showcase this method, let’s use the example y=x^2+2x-1. The first step is to separate the first and second terms from the third with brackets. The second step is to take the b value, in this case 2, and divide it by 2 and then square it. After doing so, you’ll get 1. You will then take this 1 and add it into the bracket as well as subtract it.
Once you have done all these steps, you can now put this equation into vertex form. Remember that vertex form follows the format of y=(x-h)^2+k. The x value will be whatever the ax value is, the h value will be the opposite of half of the b value, and the k value will be the number outside of the bracket. The finished answer is shown above.
The equation we just did together has an a value of 1. When we have an equation in standard form that has an a value other than 1, there are a few additional steps to be done. For example, the equation, y=2x^2+2x+3. The first additional step would be to common factor the first two terms.
1. Common factor the first two terms if needed.
2. Find the square root of the second term in the bracket and multiply by 2. Add and subtract the product inside the bracket.
3. Take the negative c value outside of the bracket.
4. If necessary, multiply the negative value just taken out of the bracket and the bracket accordingly.
5. Add the outer terms together.
6. Input the numbers into vertex form. The a value will be the coefficient of the bracket, the h value will be half of the b value in the bracket, and the k value will be the number on the outside of the bracket.
Below I've included extra help and learning links.
The Quadratic Formula
The quadratic formula is used when an equation is given in standard form and the terms in the equation are either: not whole numbers, not factorable, or are really big. The quadratic formula allows people to find the x-intercepts of an equation without standard form.
What is the quadratic formula? It is x=-b+-Öb^2-4ac/2a.
In order to use this formula, you would plug the a,b, and c values into the formula, much like the special product technique, and then solve up until a certain point.
Let’s do a question together. Let’s do the equation 0=5x^2-7x+2. First off, the is the perfect equation to use the quadratic formula on because it is not factorable. Okay, so the first step is to plug all the values into the equation. Do whatever the equation tells you to do (remember BEDMAS). When you get to the point where it is a number plus minus another numbers divided by another number, that’s where you must solve for two equations. The first equation involves you solving by adding the second number to the first number and then dividing, and the second equations involves you subtracting the second number to the first number and then dividing. The numbers you get from each equations are the x-intercepts for the equation. In this case, the x-intercepts will be (1,0) and (0.4,).
Here are the steps again:
1. Input the numbers into the quadratic formula.
2. Solve up until you only have 3 terms left.
3. Make two equations in which one adds the terms, and the other subtracts the terms.
I've included some helpful links related to the quadratic formula below.
-b/2a and c-b^2/4a Method
The -b/2a method allows you to find the h value in vertex form, and the c-b^2/4a allows you to find the h value in vertex form. You can use this method when you are given an equation in standard form. You use this method by inputting the numbers into the two formula.
For example, the equation y=x^2+7x+10. When you input the numbers accordingly in the first equation and solve correctly, you would get your h value. Then, when you input the numbers accordingly in the second equation and solve correctly, you should get your k value. Now all you have to do is put the numbers inside the vertex form structure, and you’re done! This method allows you to easily go from standard to vertex form (much like the completing the square method).
Unlike previous methods that focus on finding the vertex or x-intercepts of an equation, the partial factoring method allows users to find points on a graph where y values are the same. From there, you can go and find the vertex of an equation easily.
Let’s use the equation 0=x^2+6x-10. The first step would be to remove the c value.
Lastly, you would solve for x for the term outside the brackets and the terms inside the brackets. The answers you got are the x values in which -10 will be the y coordinate (in this case (0,-10) and (-6,-10).
With these x values, you can find the midway between them and then find the vertex from there.
This goes for any set of x values that you get from partial factoring.
Quadratic Word Problems
Word Problem One
Word Problem Two
To find how long it takes for the rocket to fall to the ground, you have to find the x-intercepts of the rocket's path. To do so, you would use the quadratic equations since decimals are involved. Once you get the intercepts, the answer will be the positive number because time can't be negative.
To find when the rocket is above 95.7m, you would make one side of the equation equal 95.7. Because you are trying to solve for y, you have to bring 95.7 over to the other side to solve for y. After simplifying your new equation, you again, have to use the quadratic equation to find the new x-intercepts which will tell you when the rocket will be above 97.5m.
To find the maximum height of the rocket, you have to find the vertex of the rocket's path. To find the vertex, I used the -b/2a and c-b^2/4a method. This directly gave me the vertex.