Quadratic Relations
All you need to know about Quadratics!
Intro to Quadratics
Identifying Quadratic Relations in Different Forms/Formulas
- Vertex Form: a(x-h)²+k
What does each variable represent?
- a is the vertical stretch (Either opens up or down).
- h moves the parabola left or right (The x coordinate & is always the opposite of the sign given for example -4 would be +4 & +3 is -3).
- k moves the parabola up or down (The y coordinate).
- The "²" sign is what makes the parabola a U-Shaped line.
e.g. y=(x-1)²+3
Three types of transformations:
- Vertical Stretch
- Vertical Compression
- Vertical Reflection
Optimal Value: Highest point or Lowest point of the parabola, can be found on y-axis (Represents y coordinate).
- Factored Form: y=a(x-r)(x-s)
What does each variable represent?
Axis of Symmetry: The mid-point of the equation
Optimal Value: The max or min of the equation
e.g. y=1(x+3)(x-9)
- Standard Form: y=ax²+bx+c
Also solved as ax²+bx+c = 0
The vertex has the x-coordinate x = -b/2a
The y-coordinate of the vertex is the maximum or minimum value of the function.
The y-intercept of the equation is c.
e.g. y= 2x²+4x+10
Finding the x-intercepts in Vertex Form
a) y= -4(x+2)²+16
To find the y-intercept, set x=0
y= -4[(0)+2]²+16
y= -4(2)²+16
y= 0
Therefore y-intercept is (0,0)
To find the x-intercepts, set y=0
0= -4(x+2)²+16
-isolate for x, BEDMAS backwards
(Bring the 16 to the left side)
-16= -4(x+)² (now subtract 16)
-16/-4=-4(x+2)²/-4 (now divide -4)
√4=(x+2)² (now square root)
x+2= +/- √4
x= +/- √4-2 (now subtract 2)
(Now you are going to find the x-intercepts)
x= + √4-2 or x= - √4-2
x=2-2 x=-2-2
x=0 & x=-4
Graphing using the Step Pattern
Point one: Over one and up one
Point two: Over two and up four
*Always multiply the stretch given with the second step.
Examples
1. y=x²
Vertex: (0,0)
Step Pattern:
Point one: over 1 and up 1
Point two: over 2 and up 4
Vertex: (3,4)
Step pattern:
2x1=2 & 2x4=8
Point one: over 1 and up 2
Point two: over 2 and up 8
Determine the Equation for Vertex form from a Graph
The parabola opens upward, so we know "a" will be positive.
y=a(x-h)²+k
y=a(x-2)²-1
The parabola passes through point (0,3). Substitute x=0 and y=3 and solve from there.
3=a(0-2)²-1
3=a(-2)²-1
3=a(4)-1 (Bring -1 to the other side and change the sign)
3+1=4a
4=4a
4/4=4/4a (Divide to cancel the "4" from the "a" value)
1=a
The equation is y=1(x-2)²-1
Graphing Factored Form
How to Find each Variable:
- X Value:
There are two x-intercepts in factored form, here they are -3 & +9.
Finding the x value a.k.a the axis of symmetry: Add the x values and divide by 2
Ex. x= -3 and x= 9
-3+9/2
x= 3
- Y Value:
To find the Y value a.k.a the optimal value: Sub the x value in the original equation
Ex. y=0.5(x+3)(x-9)
y=0.5(3+3)(3-9)
y=0.5(6)(-6)
y= -18
Therefore, the vertex for this graph is (3,-18).
Common Factoring
Examples:
1) Factor the number 6
6= 3x2
2) Factor (2x+6)
- You have to see what is common in this equation.
So you should get that 2 is common because it can be multiplied to 6 and 2x
Finally answer would be 2(x+3)
3) 8x+6
- Find whats common or find GCF (greatest common factor)
GCF= 2
- Just write down the solution with brackets
Finally answer: 2(4x+3)
4) 12x3y-6x2y+18x2y
GCF= 6x2y
Solution:
= 6x2y(2x-1+3)
= 6x2y(2x+2)
Trinomials
What is a "trinomial"?
Trinomials are factors that have 3 values in them and can be followed by this formula (x²+bx+c).EX. 1) x²+3x+4x
Ex. 2) 2x²+8x+9
Ex. 3) 3x²-5x+4
Simple Trinomial
- Simple Trinomials are trinomials that have an "a" value of 1
- Many polynomials, such as 12x²+7x+2, can be written as the product of two binomials of the form (x + r) and (x + s)
- When factoring a polynomial of the form ax²+bx+c (when a = 1), we want to find:
(1) 2 numbers that ADD to give b
(2) 2 numbers that MULTIPLY to give c
To factor these trinomials, use a table to help you until you become familiar with the procedure
Example. Factor: x²-8x+12
=(x-6)(x-2)
Examples:
Factor the following trinomials using the procedure above.
a) m²-13m+26
13x2=26
13+2= -13
- not possible
b) h²-7h-18
-9x2= -18
-9+2= -7
=(h-9)(h+2)
Complex Trinomials
ax²+bx+c
(where a=1)
Can you factor the following?
3x²+17+10
- No common factor
- a=3
Example 1: Binomial Common Factoring
Factor: 3x(z-2)+2(z-2)
(Hint: a binomial can be the common factor)
= (3x+2)(z-2)
Example 2: Factor by graphing
Factor: (df+ef)+(dg+eg)
(There is no common factor in all terms)
(We can graph terms together that have a common factor)
= f(d+e)(f+g)
How to Solve Complex Trinomials
A) factor
3x²+8x+5
Product of a multiplied c
= 3 x 5
=15
3x5=15
3+5=8
* Rewrite the middle terms with the 2 factors
3x²+3x+5x+5
Not 3x²+5x+3x+5
* Factor by grouping
(3x²+3x)+(5x+5)
= 3x(x1)+5(x+1)
* Think of the binomial (x+1) as
= (x+1)(3x+5)
EXAMPLES OF COMPLEX TRINOMIAL
Example 1) 5x²-14x+8
P: 5x8= 40
-10 x (-4)=40
-10 + (-4)= -14
= (5x2-10x)(-4x+8)
= 5x(x-2)-4(x-2)
= (x-2)(5x-4)
Example 2) 8a²+10x-3
p; 8 x (-3)= -24
12x(-2)= -24
12+(-2)= 10
=(8a2-2a)+(12a-3)
=2a(4a-1)+3(4a-1)
=(4a-1)(2a+3)
HOW TO GRAPH COMPLEX TRINOMIAL
When solving an equation that requires factoring, ONE SIDE MUST EQUAL ZERO
Example 1: 5x²-14x+8
Step 1: Factor
P: 5x8= 40
-10 x (-4)=40
-10 + (-4)= -14
= (5x2-10x)(-4x+8)
= 5x(x-2)-4(x-2)
= (x-2)(5x-4)
Step 2: Find x-intercepts
(Now to solve for for x, you must let each bracket t zero)
First x-intercept
x-2=0
x=2
Second x-intercept
5x-4=0
5x=4 (Now divide 5 to find x)
5x/5=4/5
x=4/5
Example 2: 8a²+10x-3
Step 1: Factor
p; 8 x (-3)= -24
12x(-2)= -24
12+(-2)= 10
=(8a²-2a)+(12a-3)
=2a(4a-1)+3(4a-1)
=(4a-1)(2a+3)
Step 2: Find x-intercepts
First x-intercept
4a-1=0
4a=1
4a/4=1/4
a=1/4
Second x-intercept
2a+3=0
2a= -3
2a/2= -3/2
a=-3/2
How to go from Factored form to Standard form
Expand and Simplify:
y=(x+6)(x+4)
y=x²+4x+6x+24
y=x²+10x+24
Therefore the standard form of y=(x+6)(x+4) is y=x²+10x+24
Example 1) y=(x+9)(x+4)
Expand and Simplify:
y=(x+9)(x+4)
y=x²+4x+9x+36
y=x²+10x+36
Therefore the standard form of y=(x+9)(x+4) is y=x²+10x+36
Graphing Standard Form
Check to see if the coefficient in front of the x² value can be divided equally and if you can divide the equation by that number
Now you should get a number like y=2(x2-2x-8) for example
You should now break the numbers in the bracket to make two numbers that multiply each other ex. y=2(x-4)(x+2) now find the x-intercepts, find the axis of symmetry & find the optimal value
And there you should get a parabola
EXAMPLES OF GRAPHING STANDARD FORM
Example 1) y=2x²-4x-16
Part 1
Make the equation into factored form
y=2x²-4x-16
y=2(x2-2x-8)
y=2(x-4)(x+2)
Part 2
Now graph it in the factored form
a) State the zeros
y=2(x-4)(x+2)
0=2(x-4)(x+2)
x-4=0 and x+2=0
x=4 and x= -2
b) State the axis of symmetry
x=4+(-2)/2
x= 2/2
x= 1
c) State the optimal value
y=2(x-4)(x+2)
y=2(1-4)(1+2)
y=2(-3)(3)
y=2(-9)
y= -18
d) State the direction of opening
The direction of opening is up
Example 2) y=2x²+8x-24
Part 1
Make the equation into factored form
y=2x²+8x-24
y=2(x2+4x-12)
y=2(x+6)(x-2)
Part 2
Now graph it in the factored form
a) State the zeros
y=2(x+6)(x-2)
=2(x+6)(x-2)
x+6=0 and x-2=0
x= -6 and x= 2
b) State the axis of symmetry
x= -6+2/2
x= -4/2
x= -2
c) State the optimal value
y=2(x+6)(x-2)
y=2(-2+6)(-2-2)
y=2(4)(-4)
y=2(-16)
y= -32
d) State the direction of opening
The direction of opening is up
Standard Form Word Problem Example
Perfect Squares
For example, (x + 1) × (x + 1) = x2 + x + x + 1 = x2 + 2x + 1 and x2 + 2x + 1 is a perfect square trinomial
EXAMPLES OF PERFECT SQUARES
Example 1) 9x²+24xy+16y²
=(3x)²+2(3x)(4y)+(4y)²
= (3x+4y)(3x+4y)
=(3x+4y)²
Example 2) 4m²-20mn+25n²
= (2m)²-2(2m)(-5n)+(-5n)²
= (2m-5n)(2m-5n)
= (2m-5n)²
Example 3) 49a²+42ab+9b²
= (7a)²+2(7a)(3b)+(3b)²
= (7a+3b)(7a+3b)
= (7a+3b)²
Differences of Squares
For example, (x+8)(x+8) = x²- 64 and x²+64 is a difference of square trinomial
EXAMPLES OF DIFFERENCE OF SQUARE
Example 1) 9x²-49
= (3x)²-(7)²
= (3x+7)(3x-7)
Example 2) 81a²-4
= (9a)²-(2)²
= (9a+2)(9a-2)
Example 3) 4h²-49
= (2h)²-(7)²
= (2h-7)(2h+7)
Example 4) 144y²-169
= (12y)²-(13)
= (12y-13)(12y+13)
Completing the Square
Examples of Completing the Square
Quadratic Formula
How to solve for x-intercepts using quadratic formula
Word Problems for the Quadratic Formula
Axis of Symmetry H= -B/2A
WHEN CAN WE USE AXIS OF SYMMETRY:
You can use Axis of Symmetry when you are asked to find the vertex in a standard form. You have to use the Axis of Symmetry when you are using the quadratic formula and the discriminants values are negative.
Steps to find the Vertex in the Axis of Symmetry in Quadratic Equation
Connections between topics
Standard Form
- How to go from Standard Form to Factored Form
Vertex Form
- How to go form Vertex Form to Standard Form
- How to go from Vertex Form to factored Form
Factored Form
- How to go from Factored Form to Standard Form
- How to go from Factored From to Vertex Form