# The Quadratics Mission

### Everything you need to know about Quadratics

## Learning Goals

can solve problems involving Quadratics Equations

can solve for axis of symmetry and vertex

## Summary Of Quadratics

**Quadratics deals with equations involving quadratic polynomials, solving for the Axis of Symmetry and Axis, solving for X or Y. There are three ways to solve Quadratic Equations, one is by factoring it, and solving by graphing, and also solving by completing the square.**

## Vertex Form y = a (x-h)² + k

y= -x ² + 25

set y to 0 to get your two vertex intercepts:

0= -x ² + 25

x ² = 25

x= +- 25 (plus minus)

x= 5 x=-5

x intercepts = 5 & -5

vertex x-=5-5/2= 0/2 = 0

y =x ² +25

= -0 + 25

y= 25

vertex (0,25)

## Vertical and Horizontal Translations

## Vertical Translations A vertical translation moves the graph up or down. | ## Horizontal Translations A horizontal translation moves the graph left or right. |

## Finding X & Y Intercepts

example: y= 2x-4

0 = 2x - 4

2x =4

2x/2=4/2

x=2

x intercepts (2,0)

set y=0 to solve for x

example y=2x-4

y= 2(0) -4

y=0-4

y=-4

y intercepts (0,-4)

## Word Problem using Vertex Form

1. What is the maximum height of the ball, in seconds?

x=k

k=4

Therefore the ball reached its maximum height in 4 seconds.

2. What was the maximum height in meters?

y=h

h=52.5

The maximum height of the ball in meters is 52.5 m.

3. How long did it take the ball to reach its maximum height? What was the intial height of the ball?

Set x to 0 and solve for your y, this question is basically asking what is the height of the ball when the time is set to 0.

h = 12 (t - 4)² + 52.5

h = 12 (0 - 4)² + 52.5

h = 12 (-4)² + 52.5

h = 12 (16) + 52.5

h = 192 + 52.5

h = 244.5

Therefore, the initial height is 244.5 meters.

4. When did the ball hit the ground?

Set your y, which in our case is h, to 0 to solve for t.

h = 12 (t - 4)² + 52.5

0 = 12 (t - 4)² + 52.5

-52.5/12= 12 (t-4)² /3.5

-√43.75=√t-4

Positive

66.14=t-4

66.14+4=t

70.14=t

Negative

-66.14+4= t

-62.14=t

What is the height of the ball at 2.5 seconds?

h = 12 (t - 4)² + 52.5

Sub 4.5 to where t is, because T is Time in seconds.

h = 12 (2.5- 4)² + 52.5

h = 12 (-1.5)² + 52.5

With BEDMAS, solve your exponents first.

h = 12 (2.25) + 52.5

h =27 + 52.5

h=79.5

Therefore, the height of the ball at 2.5 seconds is 79.5 meters.

## Types of Factoring: GCF

Take the most greatest factor which in this case is 2 and put the common coefficient in the front, then take the most common variables and the exponent that's common with all the factor along with it.

2x^2y^3(1+5x^3y^5+3xy)

## Factoring by Grouping

3x² + 9x

See what is common in this equation.

3x² + 9x

3 is common as a GCF and x is common as a variable.

now put it in groups with what is common!

3x(1x+3x)

Now, if we put distributive property what we have put in groups, the equation will turn into the original equation.

## Simple Trinomial Factoring

take the common variables first and break them down into two brackets:

(x )(x)

then look for factors that subtract with eachother to equal to 2 but multiply to -15.

which are 5 & -3.

so ( x+5)(x-3)

## Complex Trinomial Factoring

a=6

b=5

c=4

multiply a & c to get b:

ac= -24

find two numbers that equal to 5, which is 3 & 8 (3-8)

6x^2+(3-8)x -4

6x^2 + 3x - 8x -4

3x(2x-1) - 4(2x+1)

(3x-4)(2x-1)

## Difference of Squares

**Difference of Squares the difference of two squares is a squared (multiplied by itself) number subtracted from another squared number.**

So for example:

x^2 - 16

=*x^2* – 16

find a number that squares 16, which is 4x4=16

= x*^2*–4^2

= ( x-4) (x+4)

The final answer is what you would get again if you solved again to get x^2-16

## Perfect Square A^2 + 2(A)(B) + B^2

**Perfect Square is a number that can be solved as a number in the middle by two equal integers.**

EX:

9x^2 + 24xy + 16y^2

Find two numbers that square A & C, leave B alone!

(3x)^2+24xy+(4y)^2

A & C should equal to B, but leave out the squares.

3x^2+2(3x)(4y)+16y^2

Squaring the A & B terms then putting it in the middle should always equal to your middle term.

Now, put them in two brackets.

(3x+4y ) (3x+4y)

Putting them into brackets, then solving it would equal to your original equation, and that is what solving a Perfect Square is.

## Standard Form y= ax^2 + bx + c

In connection, solving in Factored Form can be similar to solving in Standard Form. which is that the value of A gives you the direction of opening, but in this case the y-intercept would be our variable C and keep in mind that Maximum= the parabola goes downwards and is negative and Minimum= the parabola goes upwards and it is positive.

Here is an example:

Putting in the given equation "0 = 5x^2 - 7x +2" into Standard form. I put y=0 because I am trying to find two x intercepts.

So, now multiply 5 & 2 and find two number that gets you the B variable which in this equation is 7.

5x2=10

10= -5 - 2 = -7

Expand the original equation:

5x^2 - 5x - 2x +2 = 0

Put them in brackets to get your two x intercepts.

(5x^2-5x)- (2x+2) = 0

Find two GCFs that multiply to the original equation:

5x(x-1) - 2 (x-1)=0

which leaves us with two brackets again:

(x-1)(5x-2)=0

x-1=0 5x-2=0

x= 1 x= 2/5

Therefore my two x intercepts are 1 and 2/5.

## Quadratic Formula

## The Discriminant Formula

## Quadratics Reflection

Quadratics was a fun and challenging unit that I've learned. I was able to understand how to solve different types of factoring, squaring, graphing, finding x and y intercepts and finding the parabolas.

Going more into the unit, learning all the terms and how to solve word problems and finding the parabolas were very challenging in my opinion, rather than the factoring (which was my most favourite thing to do in this unit). You definitely have to put time in to learn all the concepts in this unit to understand everything that you have to do in Quadratics.