# Quadratics

### Graphing Vertex Form By Mahimna

## Summary of Unit

- y=a(x-h)-k
- graphing vertex form
- finding the vertex form of a parabola
- translations (ex. horizontal translation, vertical translation, compression, and stretch)
- solving word problems
- step pattern

## Learning Goals

- I am able to find out the translation of a parabola from the equation

## Word Problem

A). What is the maximum height of the ball?

ans: 45

B). When did the maximum height occur

ans: 2 seconds

C). What is the height of the ball after one second

ans: set t=1 and then solve, the answer is 40.1 m

D). What is the initial height of the ball

ans: set t=0 then solve for h, and the answer is 25.4m

## Translations

## Graphing

## Factored Form

## Summary Of Unit

- Binomials
- Trinomials
- Complex trinomials
- Perfect square
- Difference of square

We also learned to graph an equation in factored form by finding the x-intercepts. Then using the x-intercepts and find your zeros to find the y value and x value.

## Learning Goals

- i am able to find the vertex of a parabola from a factored form equation

- i am able to solve a word problem that consists of factorable equation

## Complex Trinomials

## Perfect Square

## Difference of Square

## Finding the vertex and x-intercepts

## Word Problems

Solution:

H=105

105=-5t squared-10t+180

0=-5t squared-10t+180-105

0=-5t squared-10t+75 Factor out a -5

0=-5(t squared+2t-15)

-5(t+5) (t-3)

Therefore the rock reaches the ledge at 3 seconds

2. A ball is thrown vertically from the top of a cliff. The height of the ball is modeled by the function h= -5t squared-10t+75, where is h= height in meters and t= time in seconds. When will the ball reach its maximum height.

Solution:

h=-5t squared-10t+75 next factor using GCF

h=-5(t squared+2t-15) factor by grouping

h= -5(t-5) (t+3) add the two x-intercepts and divide by 2

5+3 divide 2 = 1 therefore t= 1 the x-intercepts have the opposite signs from the equation

h=-5(1) squared-10(1)+75 sub the x value into the original equation

h=-5-10+75

h=-15+75

h=65

The vertex is (1,65)

Answer: The ball will reach its maximum height in 1 second

## Standard Form

## Summary of Unit

- Use the quadratic formula and solve for the x intercepts
- Turn standard form to vertex form (Completing the square)
- Find the vertex by solving using the Quadratic formula
- Solve word problems that require max height, initial height, time of max height, and x-intercepts.
- we learned the that we can use the Discriminant to determine how many answers there are going to be.

## Learning goals

- I am able to graph a standard form equation using quadratic formula or completing the square

- I am able to complete the square and achieve vertex form

- I am able to find the x-intercepts by using the quadratic formula

## About Standard Form

## Quadratic Formula

Quadratic Formula:

## Using the Quadratic Formula

Example:

3x^2+2x+1 x=-b+- square root b^2-4(a)(c)/2(a)

x=-2+- square root 2^2-4(3)(1)/2(3)

x=-2+- square root 4-12/6 -2+-2.8/6

x= -2+2.8/6 equals 7.5

x= -2-2.8/6 equals -1.2

x= 7.5+(-1.2)/2 equals x=3.1

Now sub in x

3(3.1)^2+2(3.1)+1

86.4+6.2+1 y= 93.7

## Completing the Square

When completing the square you need to take the middle term divide it by 2 and square it. Then you add that term into the equation, but you just can't add random numbers in an equation, therefore you need to subtract it too. it will look like a(5xsquared+6x+2+9-9)+3.

If you have an a value, then you have to multiply the -9 with the a value when your taking the -9 out. Then you just turn the equation into a perfect square and you will have vertex form.

Example: x^2+6x+3

(x^2+6x)+3 6/2^2

(x^2+6x+9-9)+3

(x^2+6x+9)-9+3

(x^2+6x+9)-6

y=(x+3)^2-6

## Discriminant

## Graphing Standard Form

## Word problems

A). what is the height of the deck?

answer: Set t=0 therefore h=15

B). what is the maximum height reached?

Answer: x= -b+- square root b^2-4(a)(c)/2(a)

t=-100+- square root 100^2-4(-5)(15)/2(5)

t= -100+- square root 10,000+300/10

t=-100+-square root 10,300/10

t=-100+101.4/10 equals 7.1

t=-100-101.4/10 equals -0.04

t= 7.1+(-0.04)/2 therefore x=3.5

sub into original equation

h=-5(3.5)^2+100(3.5)+15

h= 306.2+350+15

h=671.2

The model rocket reach 671.2 meters.

C). How long did it take the model rocket to reach the max height.

Answer it took the model rocket 3.5 seconds to reach the maximum height.

2) There is a football that is thrown from the stadium onto the field. The ball's path can be modeled by the function h= t^2+6t+81. h is height of the ball in meters and t is time in seconds.

A). What is the height was the ball thrown from?

Answer: set t=0 therefore h=81

B). What is the maximum height the ball reached?

Answer: h= t^2+6t+81

h=(t^2+6t)+81 6/2^2

h=(t^2+6t+9-9)+81

h=(t^2+6t+9)-9+81

h=(t+3)^2+72

the ball reached the height of 72 meters.

C). what time did the ball reach the max height?

Answer: the ball reach the max height in 3 seconds.

## Reflection

## Quadratics

Mahimna Dave