## 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

## 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 ball is thrown up into the sky from a height of 2 meters off the ground. It reaches a maximum height of 34 meters at a horizontal distance of 4 meters.

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

a) We know that the vertex is (4,34). This is because the maximum or minimum point is always the "k" value and we know the the horizontal distance is the same as the horizontal translation which is the "h" value. The ball is thrown from a height of 2 meters. This means that 2 is the y intercept and that the x - value is 0. Therefore, this becomes another coordinate which is (0,2). Now we can sub these values into y=a(x-h)^2+k and solve for "a".

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.

## 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

Factoring Using the Great Common Factor, GCF - Example 1
Factoring Complex Trinomials
Factored Form - Difference of Squares and Perfect Squares

## Word Problem

The height, h, in meters, of a toy rocket at any time, t, in seconds, during its flight can be estimated using the formula h= -t^2+4t+21.

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?

a) The initial height of the toy rocket is 21 meters.

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

## 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

❖ Completing the Square - Solving Quadratic Equations ❖

## Word Problem

The path of a basketball after it is thrown in the air is given by the following equation: h = -0.25d^2 + 2d + 1.5, where h is the height and d is the horizontal distance in meters.

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

a) The initial height of the basketball is 1.5 meters because the c value is always the y intercept

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.

## My reflection/assessment

Throughout this quadratics unit I have learned a lot about the different ways of graphing through vertex, factored and standard form. Each of these quadratic relations provide information needed to graph. Vertex form gives the vertex with the h and k values. Factored gives the x-intercepts with the r and s values. Standard form provides the y - intercept and once you complete the square you get the vertex and once you use the quadratic formula you get the x-intercepts. I have learned that factoring connects to graphing because it gives the x - intercepts and then we can find the axis of symmetry and substitute into the original equation and solve for y. This gives us 3 points that we can use to graph. I have also learned that vertex form gives the vertex and we can use methods such as mapping notation to figure out some more coordinates in order to graph. I also learned that with standard form it requires more solving in order to graph however it is possible with completing the square and the quadratic formula. I also learned that we can convert from standard form to vertex form by completing the square as well and that if we expand vertex form then we get standard form. Those are some of the main points I learned in this quadratics unit.