By: Sharan Johel
Key features of Quadratic relations:
Maximum/minimum value: optimal value (k value). Labelled as: *y=k*
Optimal value (min/max value): The highest/lowest value the graph reaches. It is the highest/lowest point on the y-axis. This is the y –value of the vertex (labelled as y=)
Axis of symmetry: divides the parabola in half (h value). Labelled as: *x=h*
Y-intercept: When x=o and parabola crosses y-axis. Labelled as: *(0,y)*
X-intercepts: when y=o the parabola crosses the x-axis. Labelled as: *(x,0)*
Zeros: the zeroes are the same as x-intercepts or roots. Labelled as: *(x,0)*
Table of contents:
Introduction to quadratics:1) Analyzing quadratics (second differences)
2) All about parabolas (intro) ex: all the components to it
Types of equations:
- Standard form y= (ax^2+bx+c)
- Factored form y=a(x-r)(x-s)
- Vertex form y=a(x-h)^2+k
4) Vertex form (the properties)
- graphing from vertex form
- finding an equation when only given the vertex
5) Expanding binomials and multiplying binomials
6) Factored form:
- graphing from factored form
- common factoring
-factoring simple trinomials
-complex trinomial factoring
- binomial common factoring
- factoring by grouping
Factoring special quadratics:
- perfect squares
-difference of squares
7) Standard form:
- completing the square (moving from standard form to vertex form)
- solving quadratics using the quadratic formula
- graphing from standard form
And lastly learned lots of types of word problems such as expand and simplify problems, factoring etc.
There are many components to a parabola.
- The axis of symmetry (A.O.S) is the midpoint of the two zeros. (h value)
- Vertex (h,k)
- Optimal value is found by subbing the A.O.S value into the equation. (k value)
- X-intercepts (the zeros)
- Y-intercepts - when graph crosses the y-axis. (add picture of diagram)
Everything about parabolas:
- The zero of a parabola is where the graph crosses the x-axis
- The y-intercept of a parabola is where the graph crosses the y-axis.
- The Zeroes can also be called the x-intercepts or the roots.
- The vertex is the point where the axis of symmetry and the parabola meet. (its the point where the parabola is at its maximum or minimum value.
- The axis of symmetry (A.O.S) divides the parabola into 2 equal halves.
- The optimal value is the value of the y coordinate of the vertex
Forms of quadratic relations
What do we know from this equation?
- y-intercept = c value
- direction of opening ( a value)
Example: y=x^2+7x+12 ----> For this equation the y-intercept =12 and a value is 1 which means the parabola opens up.
Second form is factored form/intercept form; y=a(x-r)(x-s)
What do we know from this equation?
- the zeros/x-intercepts (r and s)
- direction of opening (a value)
Example: y=-2(x-1)(x+3) -------> for this equation we know the x-intercepts are 1 and -3 which are opposite of what is inside the brackets, and a=-2 (parabola opens down)
Third form is vertex form; y=a(x-h)^2+k
What do we know from this equation?
- The vertex (H,K)
- max/min optimal value=k
- The axis of symmetry is x=h
- direction of opening (a value)
Example: y=3(x-2)^2+5 --------->for this equation the vertex is (2,5) NOTE: the x value is always opposite of what is inside the bracket.
- a=3 (parabola opens up)
- A.O.S is x=2
Identifying the step pattern:
Graphing from vertex form and transformations
a= - direction of opening(up or down)
- stretch or compression
- step pattern
h= - horizontal translation ( left or right, if h is negative its right, and if positive its left)
k= - vertical translation (up or down, when k is negative it moves down and if positive it moves up)
Max/min (optimal value): y=k
Axis of Symmetry: x=h
Example: Describe the transformations that occurred to y=x(2) in order to become y=-2(x-3)(2)+6.
a= is -2 (vertical stretch by a factor of 2, opens down (reflection in the x-axis)
h= 3 - horizontal translation 3 units right.
k=6 - vertical translation 6 units up.
For this example we know that a=1, h=0, k=3
The step pattern we would use is 1,3,5, we use the normal step pattern because there is no a value in the equation.
Example 2: (graphing from vertex)
Step 1: First plot the vertex at (0,0)
Step 2: Move the vertex 3 units to the left (h=3)
Step 3: Move the vertex 6 units up (k=6) *now we have found our vertex (-3,6)
Step 4: Graph all points, remember we have an a value of 2, so the step pattern is going to change up a bit. Taking the a value, we now multiply it by 2.
Normal step pattern: 1,3,5
New step pattern: 1x2, 3x2, 5x2
= 2, 6, 10 .. now you can plot it and you have a parabola that should open up!
Finding the equation given the vertex (basics)
The equation would be : y=(x-3)2+5 *always remember to put the opposite h value, for example the vertex for this is (3,5 but when subbing the vertex into the equation, don't forget to put the positive 3 into a negative 3.*
Finding equations in vertex form
First identify the vertex which is (3,-1) and the point is (1,7)
Having that information given, you should sub it into the equation.
4a/4 = 8/4
Therefore the vertex is (3,-1) and a=2
Equation is: y=2(x-3)(2)-1
Introduction to factored form
Graphing from factored form
Note that r and s equal to the zeroes or x-intercepts.
- Axis of symmetry = r+s/2 and x=h.
a= direction of opening, stretch or compression, step pattern
when y=-3(x+7)(x+9), find the vertex,axis of symmetry and optimal value.
First you would need to find the zeroes:
x=-7 x=-3 *The zeroes are -7 and -3 because -7+7=0 and -3+3=0.*
Now that you found the zeroes, you need to find the axis of symmetry.
Formula for A.O.S (x=(r+s)/2)
Now that you found the zeroes/x-intercepts, now sub it into the equation and find the y value. (remember that you need to sub it in to the FIRST equation you were given).
*the zeroes are always opposite of what they actually were, when you sub it in the equation*
There we go! Now you know how to find the vertex, axis of symmetry, and optimal value. But... that's not it, we now need to make a graph using all the information we have.
(Our 2 zeroes are x=-7 and x= -3, our x value is -5 and y=12)
Common Factoring (adding brackets, opposite of expanding)
When you find the factors of two or more numbers, and then find some factors are the same “common”, then they are the "common factors".
Example: 12 and 30
• The factors of 12 are: 1, 2, 3, 4, 6 and 12
• The factors of 30 are: 1, 2, 3, 5, 6, 10, 15 and 30
So the common factors of 12 and 30 are: 1, 2, 3 and 6
Example 1: 8x+6
Step #2: *remember to write solution WITH brackets.*
8x/2 + 6/2 (dividing both the numbers with the GCF)
Therefore answer is: 2(4x+3) because 8x/2 = 4x and 6/2 = 3
Always put the GCF in front of the bracket.
Example 2: 14m+21n
Step #2: Divide both numbers with the GCF. ( 14m/7 + 21n/7 )
Therefore: 14m/7 is 2m and 21n/7 is 3n.
Below is a picture of factors for a number, and the GCF.
Factoring simple trinomials
x^2+7x+12 = (x+3)(x+4)
When factoring a polynomial of the form ax^2+bx+c (when a=1) we want to find; 2 numbers that have to ADD to give b and 2 numbers that MULTIPLY to give c.
Example: Factor x^2+6x+5
In order to factor this, we need to figure out ____x_____=5 and _____+_____=6
Now we know that 5x1= 6 and 5+1=6 so we have the numbers 5,1 so that would be our 2 digits we are going to use. Remember there are also many other numbers that can give us 6, like 3+3, but we know that isn't going to work because 3x3 does not equal to 5.
=(x+5)(x+1) *the order you put the numbers doesn't matter*
BELOW is a video explaining how to factor simple trinomials.
Example 2: n^2-13n+26
Knowing right off the bat, you should be able to identify that this example is NOT POSSIBLE.
More simple trinomial factoring
First you need to figure out the GCF of 15 and 18 which is 3.
Now we need to make the equation into a simple trinomial.
=3(x^2+5x+6) -----> now that we have made this into a simple trinomial we need to factor.
* (3)(x^2) = 3^2 *
* (3)(5x) = 15x *
* (3)(6) = 18 * Therefore we have common factored correctly.
Therefore the answer is: 3(x+2)(x+3)
- Some possible values can be 6 and 4 because, since k represents the sum we know that 6+4=10 so k=10.
Another option: k=12
The factors would be 12 and 2 , and since k represents the sum, we know that 12+2=12 so k can be 12.
Factoring complex trinomials
ax^2+bx+c (where a is NOT equal to 1)
Do you think you can factor the following?
3x^2+17x+10 ----> first you need to ask yourself if you can common factor, for this example you cannot.
Example 1: Binomial common factoring
---> *A binomial can be the common factor*
Answer: = (z-3)(3x+2)
Example 2: factor by grouping
- There is NO common factor in all terms
- We can group terms together that have a common factor
Remember that the terms in both the brackets should ALWAYS be the same.
Therefore the answer would be =(d+e)(f+g)
Factoring complex trinomials: Decomposition method
First you need to do (x)(x), then do (x)(2), after (1)(x) and finally (1)(2).
*collect like terms*
Example 1: (h+9)(h-5)
=h^2 - 5h + 9h - 45
=h^2 +4h - 45 ----> trinomial
=n^2 + 6n - 4n -24 *remember to always collect like terms*
=n^2 + 2n - 24
Special factoring (perfect squares, difference of squares)
Factoring Special Quadratics
Pattern is: 1^2, 2^2, 3^2 , 4^2, 5^2 .... (perfect squares)
y^2 - 25 -----> (y)^2 - (5)^2
Remember: formula is a^2 - b^2 =(a+b)(a-b)
CHECK YOUR ANSWER: y^2 - 5y + 5y - 25
* the -5y and 5y cancel out*
Therefore answer is correct.
*Remember to always ask if you can common factor, for this example it is a YES*
=3(4k^2 - 9c^2) - difference of squares
=3(2k^2 - 3c^2) - simplify
=3(2k+3c)(2k-3c) - ANSWER
Perfect square trinomials
*expand and simplify*
(x+2)^2 -----> (x+2)(x+2)
= x^2 + 2x + 2x + 4
= x^2 + 4x +4
example 2: (x+3)^2 ---> (x+3)(x+3)
* THESE ARE PERFECT SQUARE TRINOMIALS*
; a^2+2ab+b^2 = (a+b)^2
; a^2 - 2ab + b^2 = (a-b)^2
EXAMPLE: y^2 - 14y+49
16m^2 + 24m + 9 --> 16 and 9 are perfect squares!
First, you need to find the square root of 16m^2 which is 4m, then you need to find the square root if 9 which is 3. Now to see if those two numbers are correct you need to do 2(4)(3) which should be equaled to 24.
NOW you have found your a value and b value.
*a=4m and b=3
Therefore answer is =(4m+3)^2
Solving quadratic equations by factoring
y=ax^2+bx+c ---> standard form (set y=o)
Example 1: x^2 - 3x - 28=0 (simple trinomial)
_-7___+_4__=-3 * to solve for x you must let each bracket =0*
(These are called the ROOTS of the equation)
Example 2: m^2-8m=-15 ------> m^2-8m+15=0
Standard form: Graphing
To put this quadratic relation in factored form you need to FACTOR!
Reminder: Factored form is y=a(x-r)(x-s)
- the roots(x-intercepts) are the values of r and s.
Example 1: y=x^2+8x+15
STEP #1: Factor to get factored form!
So y=x^2+8x+15 will now be y=(x+5)(x+3) because... __5__x__3__=15 and __5__+__3___=8
STEP #2: Find the roots
x=-5 x=-3 ----> the roots (x-intercepts)
STEP #3: Find the A.O.S (Axis of symmetry)
x=-4 ----> this is the x-value of the vertex (h value)
STEP #4: Find the optimal value
Sub x=-4 into the ORIGINAL equation ( y=x^2+8x+15
y=-1 ---> this is the y-value of the vertex (k value)
Therefore: the vertex is (-4,-1), the x-intercepts are (-5,0) and (-3,0)
*Always see if you can common factor first, for this example you can*
y= -x^2 -2x+8
y= -1(x^2+2x-8) --> simple trinomial
y= -(x+4)(x-2) - factored form
NOW set y=0 to find x-intercepts
x=-4 x=2 ---> roots
NOW find A.O.S
x=-1 ---> x-value of the vertex
NOW find the optimal value
sub x=-1 into y=-x^2-2x+8
y=9 ---> y-value
Therefore: the vertex is (-1,9), the y-int is (0,8) and the x-intercepts are (-4,0) and (2,0)
Completing the square
What is a discriminant? A discriminant can help determine the number of solutions a quadratic equation has.
REMEMBER: If b^2 -4ac > 0 then the quadratic has 2 real roots.
If b^2 -4ac = 0 then the quadratic has 1 real root.
If b^2 -4ac < 0 then the quadratic has no real roots.
Below is a video which thoroughly explains all the key concepts of it.
Some connections made between a topic:
You can graph equations either in standard, factored or vertex form. From each of these equations you can easily find out the axis of symmetry(x value) , optimal value (y value) , and zeroes(x-intercepts). When finding all those things you will then begin to make your graph. When given an a value in the equation you can simply identify if the parabola is vertically compressed or stretched.
Reflection of the unit!
Quadratics for me was really easy at first but then started getting somewhat complicated. What I found easy was how to graph the parabola and figure out how to find the vertex. After we started learning lots of other new concepts, it just got to the point where I was really confused, and on top of that we started learning lots of new terminology, which was very new to me. I felt like I didn't do really well in the quadratics quiz #1 because I was really confused on the knowledge and thinking section. I felt like I needed to stop making small errors, sometimes I made those tiny mistakes here and there. I know I needed to get help in order to succeed during the test that was coming up. I think what brought my mark down was that it probably wasn't making sense to me or I just needed more practice. I was really sad with my mark, but I knew I shouldn't be sad for long because there was many more quizzes, tests, assignments etc, coming up.
Quiz #2 reflection
I was now done with my first test for quadratics, and now it was time to write my Quiz #2. For this section of quadratics it was all about factoring. Simple trinomial factoring, complex trinomial factoring, perfect squares, difference of squares, common factoring, how to find the roots of an equation etc. I felt like I did really well on this quiz because I find factoring really simple. I sometimes forget which method to use because there’s just so much to remember. I was really scared to write this quiz because I knew if I did bad, then I needed to improve as soon as possible, and practice really hard for the upcoming test. Once I saw my mark, I was the happiest person you could ever imagine. I knew even though I did well, doesn't mean I shouldn't study for the test. So far I think I am doing somewhat well in the Quadratics unit, I just find that there’s so much to learn and just so little time. I think I really need to improve on my application skills, because I struggle a lot with word problems. I know I can improve on some sections and I will continue to try my hardest. All in all, I think if I don’t try I won’t succeed, and that I should come ask for more help when needed. I know I will continue working hard for the upcoming lessons.