# Quadratics

### Graphing Vertex Form, Factored Form and Standard Form

## Graphing Vertex Form

## Learning goals?

**By the end of this unit, I should be able to...**

**Differentiate between what a,h and k values represent****Solve a word problem by either using an equation to figure out information or by creating an equation to figure out information****Graph using table of values, mapping notation or step pattern**

## What is graphing with vertex form?

__A quick summary...__

- y=x^2 is the base function

- The vertex is always the (h,k) values- The axis of symmetry is (x=h)

- 3 methods of graphing: Table of values, Mapping notation, and step pattern

- To find y-intercept, sub x=0.

- "a" value = stretch or compress; open up or down

- "h" value = horizontal translation; always opposite sign then what's in the bracket

- "k" value = vertical translation

## "a" value The "a" value in vertex form represents whether a parabola opens up or down and whether it is vertically stretched or compressed. If "a" is positive it opens up and if it is negative is opens down. If "a" is greater than 1 then the parabola is vertically stretched. Lastly, if "a" is less than 1 then the parabola is vertically compressed. | ## "h" value The "h" value in vertex form represents how much a parabola is horizontally translated. It also represents the axis of symmetry. If "h" is negative in the equation then it is translated that many units to the right. If "h" is positive in the equation then it is translated that many units to the left. | ## "k" value The "k" value in vertex form represents how much a parabola is vertically translated. It also represents the optimal value. If "k" is positive then it is translated that many units upwards. If "k" is negative it is translated that many units downwards. |

## "a" value

## "h" value

## Practice Word Problems

## example 1

The path of a soccer ball is modelled by the relation ℎ= −2(t−4)^2+33, where h is the height, in metres, of the soccer ball above the ground and t is the time, in seconds, since the ball was kicked.

a) What is the maximum height reached by the ball after it was kicked?

b) How many seconds after it was kicked did the ball reach this maximum height?

c) How high was the ball above the ground when it was kicked?

## solve

a) The maximum height of the ball is 33 meters. This is because the maximum point is also the vertex of the parabola and the vertex is always "h" and "k" values. "k" is always the y value and in this problem the height is the y value, therefore the maximum height is 33 meters.

b) The ball reached the maximum height at 4 seconds. This is because of the vertex. "h" is always the x value and in this problem the time is the x value, therefore the time when the maximum height occurred is at 4 seconds.

c) In order to solve this you need to substitute 0 into "t" and solve for "h".

ℎ = −2(t−4)^2+33

ℎ = −2(0−4)^2+33

ℎ = −2(16) +33

ℎ = -32 +33

ℎ = 1

Therefore the ball was 1 meter above the ground when it was kicked.

## Example 2

a) Determine an equation to model the path of the ball

b) What is the height of the ball when it reaches a horizontal distance of 5 meters?

## solve

y=a(x-h)^2+k

2=a(0-4)^2+34

2-34=16a

-32/16=16a/16

a= -2

Now that we know the "a" value we can complete the equation.

The equation is__ y=-2(x-4)^2+34__

b) Since it is asking for the horizontal distance then we have to sub x=5. The we solve for y.

y=-2(5-4)^2+34

y= -2+34

y=32

Therefore the height of the ball when it reaches a horizontal distance of 5 meters is 32 meters.

## Factored Form

## Learning Goals

- I should be able to factor simple trinomials, complex trinomials, difference of squares and perfect squares
- I should be able to find the x-intercepts of a factor
- I should be able to find the vertex and graph using the 3-point method

## Summary of the Unit

- Types of Factoring:
- Greatest Common Factor
- Simple factoring (a=1)
- Complex factoring
- Special case - Difference of squares
- Special case – Perfect square

- x- intercepts can be found by solving for x in the factor and once there are 2 x-intercepts we can find the average of both and figure out the axis of symmetry.

- After the axis of symmetry is found, that value of x will be substituted into the original equation to solve for y.

- That is how we figure out the vertex, it is the value of y and the AOS.

- Once the 2 x intercepts are plotted and so is the vertex then that is the completed 3 point method

## Word Problem

a)What is the initial height of the toy rocket?

b)When will the rocket hit the ground

c) What was the maximum height?

d) How long does it take the rocket to reach its maximum height?

## Answers

b) The rocket hits the ground at 7 seconds.

-t^2+4t+21

-(t^2-4t-21)

-7,3

-(t^2 + 3t)-(7t-21)

-[t(t+3)-7(t+3)]

(t-7)(t+3)

t-7=0

t=7

t+3=0

t=-3

c) The maximum height is 25 meters.

AOS= 7-3/2 = x=2

sub x into original equation.

-t+4t+21

-2^2+4(2)+21

-4+8+21

y = 25 meters

d) It takes the rocket 2 seconds.

AOS=2

## Standard Form

## Learning Goals

- I am able to rewrite each relation in the form y=a(x-h)^2+k by completing the square
- I am able to find the x-intercepts by substituting the values from the standard form into the quadratic formula
- I am able to determine the number of solutions a quadratic relation has by the discriminant

## Summary of the Unit

y = ax^2 + bx + c

- The c value is always the y-intercept
- To convert from standard form to vertex form you must complete the square
- In order to find the x-intercepts, use the quadratic formula
- After completing the squares the vertex is given and after using the quadratic formula the x-intercepts are given. From this given information the 3 - point method can be used to graph
- AOS= adding both x values and then dividing by 2
- If Discriminant is greater than 0, then there are 2 x-intercepts
- If Discriminant is less than 0, then there are 0 x-intercepts
- If Discriminant is equal to 0, then there is only 1 x-intercept

## Completing the Square

## Quadratic Formula

## Word Problem

a) What is the initial height of the basketball?

b) What is the maximum height reached by the basketball and at what horizontal distance does this occur at?

## Solutions

b) h = -0.25d^2 + 2d + 1.5

a = -0.25 b=2 c=1.5

h = -0.25 (d^2 - 8) + 1.5

h = -0.25 (d^2 - 8 +16 - 16) + 1.5

h = -0.25 (d^2 - 8 + 16) + 1.5 + 4

h = -0.25 (d-4)^2 + 5.5

Vertex (4, 5.5)

Therefore, the maximum height reached by the basketball is 5.5 meters and it occurs at a horizontal distance of 4 meters.