By: Dilveer Dhaliwal

How quadratics help us in reality

Quadratics consist of parabolas which are basically curved lines, so we are going to be leaving linear relations behind. We might not realise this at first but quadratic relations are all around us ranging from mountains to bridges even the McDonalds sign we see everyday.

How to determine Quadratic Realtions

  • Recall that linear relations had first differences


  • Quadratic relations have SECOND differences
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  • The vertex form is one way that a quadratic relationship can be written as it is in the form y=a(x-h)² +k and the basic form of it is y=x².
  • An example for this is y= 2(x-3)² +5
  • The "x" and "y" do not have a value because they are variables for the x and y intercept(s) which will have different values according to the relation

What each variable represents in the Vertex Form

  • In the vertex form y=a(x-h)² +k each variable has a part to do
  • The a tells us if there is a stretch/expansion or compression to the parabola.
  • The h tells us the x value of the vertex and the axis of symmetry.
  • The k tells us the y value of the vertex.


  • The transformations that can occur to a parabola are vertical or horizontal translations, vertical stretches and reflections.
  • As we know the a represents vertical stretch and compression if the value is more than 1 than it is a compression and if it is less than 1 it will be a stretch/expansion
  • If a is a negative value then there will be a reflection on the x-axis (the parabola will be flipped; downwards)
  • The h tells if the parabola moves left or right (horizontal stretch). If the h is negative then it goes towards the right and if it is positive it goes towards the left. Basically it will move the opposite of its sign.

  • The k informs us if the parabola moves up or down (vertical stretch) so if it is negative the graph will move downwards and if it is positive it will move upwards.

Graphing from Vertex Form using the Step Pattern

  • y=x² this is the basic parabola.
  • The original step pattern is "over 1 up 1 , over 2 up 4" this pattern can be used for any parabola as long as it is the basic (original) one
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Graphing from Vertex Form when not in Basic Form

  • If the vertex form has the "a" in front of it, it is not a basic parabola anymore, it will either have a stretch or compression and will open either up or down
  • The value of a will decide if the step pattern will change

An example...

  1. First, you will graph the vertex
  2. Then multiply the original step pattern by the value that a had been given
  3. Then plot the points (vertex, 4 step pattern points).

  • It does not matter if there is a decimal, fraction, or a negative number for a, you still have to multiply that number with the original step pattern.
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How to find the Axis of Symmetry (AOS)

  • The way it is written is x=h, so it would be the h in the vertex form
  • The axis of symmetry is the middle point of the zeros; the x intercept of the vertex
  • So the AOS of the parabola above will be -2 because it is the midpoint of the parabola and the it is the x-intercept of the vertex

How to find the Optimal Value

  • It is written as y=k, so it would be the k in the vertex form
  • The optimal value is either the highest point or the lowest point on the parabola; the y-intercept of the vertex
  • So the optimal value of the parabola above would be -6 because in this case it is the lowest point on the parabola and y-intercept of the vertex

X-intercepts and Zeros

  • To find the x-intercept/zeros in vertex form you would have to set y=0
  • For instance let's say you have to find the zeros in the equation y=-3(x+2)²+27 the y would be replaced with 0
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  • We do negative and positive square root because the number being squared could be either one
  • As a result the x-intercepts would be 1 and/or -5 depending on the relation


  • The factored form is another way that a quadratic relation can be written as and the form is y=(x-r)(x-s)
  • An example of this form is y=0.5(x+3)(x-9)

Zero and X-intercepts

  • Zeros can be found by setting each factor equal to 0 meaning x-r=0 and s-r=0
  • For example in the relationship y=0.5(x+3)(x-9) the zeros would be found by taking each factor (x+3) and (x-9) and make them equal to zero
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  • Therefore, the zeros/x-intercepts are -3 and 9 for this relation so when you graph the zeros these are the values you would use

Axis of Symmetry (AOS)

  • To find the axis of symmetry you have to add the two x-intercepts and divide it by 2
  • Therefore the formula or method you would use is x= (r+s)/2
  • For the example we have been using previously we would find the x-intercepts (-3 and 9) add them then divide the answer by 2
  • So it would be





  • For that reason, the axis of symmetry would be 3

Optimal Value

  • In order to find the optimal value you would have to substitute the axis of symmetry into the equation as x since it is a x value on the parabola
  • If we use the example we have been using so far we would substitute x=3 into the equation y=0.5(x+3)(x-9)
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  • Therefore the optimal value for the example is -18


  • Know that you know how to find the x-intercepts, axis of symmetry and optimal value you can plot 3 points
  1. The first x-intercept which is x=-3
  2. The second x-intercept which is x=9
  3. The vertex which is (axis of symmetry;x, optimal value;y) (3,-18)

Word Problem

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  • Last but not least, the standard form is also a way to write a quadratic relation and the form is y=ax²+bx+c
  • An example of this form is y=3x²+15x+18

Quadratic Formula

  • The quadratic formula is
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How to find the zeros using the Quadratic Formula

  • In order to use the formula the standard form should equal to 0 since you are solving for the zeros
  • Since we know the formula and have the question all you have to do is plug in the appropriate values just like in the example below
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  • The square root is both positive and negative since there would be a negative and positive square root to the number which in the end would give us two zeros
  • If you want to keep the exact answer leave it the way it is in step 4 but if you want to take it down to a fraction/decimal then you can continue and add the first time and subtract the next when solving


  • The discriminant is the number that is being square rooted (b²-4ac) this number tells us if there are any zeros/solutions and if there are, how many there are.
  • If the discriminant is a negative there will be no zeros because a negative number cannot be square rooted
  • If the discriminant is 0 then there is only one zero because it does not matter if you add or subtract a 0 the answer will be the same
  • If the discriminant is greater than 0 then it will have two zeros, just as in the example above, since the discriminant is 9 there were two zeros

Axis of Symmetry

  • The formula for this is
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  • To find the axis of symmetry all you have to do is fill in the following formula just as shown in the example below
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  • It is as simple as that and it hardly takes anytime so the axis of symmetry would be x=0.7 for this parabola

Optimal Value

  • To find the optimal value (y-value;y-intercept in vertex) all you have to do is substitute the axis of symmetry into the original equation to get y
  • An example for this is...using the axis of symmetry from above
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  • So they optimal value for the example we have been using would be -0.45

Word Problem

This is how to do the following word problem...

Completing the Square

  • Completing the square helps us turn from Standard form to Vertex form (y=ax²+bx+c to y=a(x-h)²+k
  • There are a few steps and they are as follows:
  1. Take the c out and make it k since they both carry the same value and represent the same thing on the graph
  2. Then put brackets around the remaining two numbers
  3. Then factor out the GCF from the number in the brackets; after divide b by two and square to get the number you need to add and subtract
  4. Then add and subtract that number from the brackets
  5. When taking the subtraction number out of the brackets multiply it by the a value;then you will be able to make a perfect square
  6. Make the perfect square (write x and divide b by 2 then put squared outside of the bracket and write the vertex form

Just as in the example below

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  • Therefore the vertex form would be y=2(x+2)²

Factoring to turn to Factored Form

Factoring-4 terms by Grouping

  • If you ever end up with four terms to factor it is very simple
  1. We group the first two and second terms
  2. Then we factor out the GCF
  3. Then we can common factor just as shown above
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  • So basically we grouped the terms into binomials first two and second two, then we found the GCF and factored it out and finally we common factored
  • This gave us (x²+3)(x²+7)

Factoring Simple Trinomials

  • The way that simple trinomials are written is x²+bx+c
  • An example for this is w²+5w+6
  • There are six steps to get into the factored form and they are
  1. Make sure the format is x²+bx+c
  2. Find two numbers that give the product of 'c'
  3. Add the two number to see if they give the sum of 'b'
  4. Factor them in for 'bx'
  5. Do four terms by grouping method
  6. Then we do common binomials

An example for this is

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  • Therefore w­­²+5w+6 in factored form is (w+2)(w+3)

Factoring Complex Trinomials

  • Complex trinomials will include a value with the a making it complex
  • An example of this is 8x²+6x-5
  • In order to factor this we do the following:
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Perfect Squares

  • Perfect Squares make expanding easier
  • There are 3 steps to do them:
  1. Square the first term
  2. Double Product
  3. Square second term
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  • But you can only do perfect squares if what you are expanding is either for example (a+b)² or (a+b)(a+b) since you are multiplying the same things that would end up being squared, or (a-b)(a-b) and (a-b)² but of course the variables and values would be different and they can be a part of a larger relation.

Factoring Perfect Square

Factoring Perfect Squares
  • You may also use the acronym


Doesn`t Pull


which stands for

Square, Double Product and Square

Difference of Squares

  • Let's say you don't have a perfect square but something like it for example (y-7(y+7) the only different thing are the signs and this method is called difference of squares
  • When we do difference of squares there are only two steps
  1. Square the first term
  2. Square the second term

  • For example
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  • The subtraction sign will always stay as subtraction, hence the name difference of squares
  • This method can be applied anywhere as long as the relation you are expanding is in the format (a+b)(a-b) or (a-b)(a+b)

Factoring Differnece of Squares

Factoring Difference of Squares

Difference of Squares factors into a product of sum and difference

Common Factoring

  • Common Factors are when we factor out the GCF in an expression
  • There are only three steps to this
  1. Find GCF
  2. Divide expression by GCF
  3. The GCF number will go outside the brackets as well




Since the GCF was 2 we divided the whole expression by 2 and then put the GCF outside of the brackets as well which factored into 2(x+2y)

OR you can make a list on all the factors of each number and find the common one(s) and find the GCF that way; as shown in the example below

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Binomial Factors

  • This method is known as binomial factors because we are putting the expression into only two parts; an easier way to expand
  • For example
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  • Basically we did three things to get the answer we got
  1. Since (2s+5t) are the same we can write them just once
  2. Then we have 3s and -7t left which can be put in another set of brackets
  3. Both are multiplied so when you need to expand to get back to what you started you may do that

Hence we get (2s+5t)(3s-7t) as our factored form


In the quadratics unit I have learned many new things and how they can be used in real life, I also found a few things to be difficult while others were easier. To start off, I found a few things difficult like understanding complex trinomials. At first I found it difficult to try and find all the possibilities that a could have. Later, once I got the hang of it it was not as hard as I thought it was. Another difficulty I had in this whole unit was when we had to find the amount of x-intercepts a parabola had by looking at the equation. I had difficulties because I never really knew how find the x-intercepts by just looking at the vertex form. Then later on we learned that all we had to do was substitute y as 0 and solve for x and that x would give us the two x-intercepts. I also used to get confused between perfect squares and difference of squares because I would get confused perfect squares for difference of squares and vice versa. So the method I used to remember which is which is that difference of squares is a product of addition and subtraction which would eliminate that, that is not a perfect square making it difference of squares. The things I could do with ease were simple trinomials and expanding especially on the assessment below.
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On this assessment I did not do the best but I did get how to expand and do simple trinomials as well as the fact that I improved on complex trinomials. These are the things I found difficult and easy. I learned that quadratics are all around us, there are mountains,bridges and many more. I now know that we do not have to measure the maximum height of something we can just use quadratics for it. All I would have to do is get a few measurements and use the appropriate quadratics formulas and I would get the maximum height, at what time it was there and so forth. Therefore in the future I do not need to get a measuring tape and take 10 minutes trying to get one length accurate when I can just get the necessary lengths and get other measurements using quadratics in those 10 minutes. In conclusion, I found quadratics to be okay not to hard and not to easy and I also learned how quadratics help in real life.

Word Problem

Now you try solving this word problem.

When you're done you can check your answers which are right next to the question