Summary Of Quadratics

Jananie Chandranathan

Vertex Form

In this unit,we learn the different forms of equations that can represent a parabola. Each of these forms of equations can give us crucial details on determining the shape of the parabola. The equation we will discuss is the vertex form.

Equation for the vertex form: y= a(x-h)^2+k


  • "a" represents the shape of the parabola (tells the direction of opening and indicates whether the graph is compressed or stretched)
  • h represents the transformation of the parabola along the x-axis
  • k represents the transformation along the y-axis
  • h and k represents the vertex of the parabola


First and Second Differences:

  • First differences portray a linear relation since the graph is a line. The first differences are always the same. They are found from the y-values in the table and are calculated by subtracting consecutive y-values.
  • Second differences define a quadratic relation since the graph is a parabola. Second differences are equal and are calculated by subtracting consecutive first differences.


Parts of a Parabola:

  • Vertex
  • Axis of Symmetry
  • y-intercept
  • x-intercept
  • Optimal Value
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Word Problem (Vertex Form)

A soccer ball is kicked into the air. It's height in meters after t seconds is given by

h=-3.5(t-2.3)^2+19


a) What is the maximum height of the ball?

K represents the maximum height, therefore 19m is the maximum height.


b) What was the height of the ball when it was kicked?

h=-3.5 (0-2.3)^2+19

h=-3.5 (-2.3)^2+19

h=-3.5 (5.29)^2+19

h=-18.5+19

h=0.5

The initial height of the ball would be 0.5m.



c) What was the height of the ball after 1 second?

h=-3.5 (1-2.3)^2+19

h=-3.5 (-1.3)^2+19

h=-3.5 (1.69)+19

h=-5.9+19

h=13.1

The height of the ball after 1 second would be 13.1m.

Factored Form

So far, in the unit of quadratics, we have learned about the vertex form and how we use it. Like the vertex form, there is a factored form which is introduced in this unit. The equation in the factored form would be known as y=a(x-r)(x-s). In this equation, the "r" and "s" are the zeroes or x-intercepts. The value of "a" which is indicated in the equation of the factored form, defines the shape and direction of the opening of a parabola. (If "a" is negative, the parabola would be opened downwards and if "a" is positive, the parabola will be opening upwards). The AOS (Axis Of Symmetry is known to find the vertex by using the zeroes (r and s).



Types of Factoring Methods:


Simple Trinomial- Simple trinomials can be divisible, so the leading coefficient will be cancelled out, which causes the leading coefficient to be one. In other words, the standard form: ax^2+bx+c can be factored to get in the factored form. This will result as: x^2+bx+c. This shows us that the "a" value is cancelled. When it is converted to the factored form, r+s has to be =b and (r) (s) should be =c. This is because the value of b defines the sum of two numbers and c represents the product of those exact same two numbers.


You should make notice of the signs of b and c

  • If b and c are positive, r and s will be positive
  • If b is negative and c is positive, then both r and s will be negative
  • If c is negative a is positive, one of r or s will be negative \

After identifying the signs of b and c, we should find the product and sum. We have to discover 2 numbers whose product is c and we need two numbers whose sum is b.


Example: x^2+6x+5

Product: 5(1)

Sum: 5+1

Therefore, the factored form is (x+1) (x+5)


Difference of Squares- Difference of squares is known to be one of the special cases. The form: ax^2+b is known to be the difference of squares. This is factored in the form of: (ax-b) (ax+b). if you have noticed there is a negative sign and a positive sign. This basically defines the name of the specific method of factoring. "Difference" defines different signs which are - and +. Another reason for why there is a positive and negative sign is because when the specific equation is expanded, the middle term will always be cancelled out. The + and - of the number will equal to zero.


Example: x^2-36

sqrt(x^2)=x sqrt(36)=6

(x+6) (x-6)


Greatest Common Factors- When we multiply binomials, we expand. But, in this method, we simplify. When we factor, we have to make the equations be able to multiply. You could do this by looking for the equations GCF.


The "Factor By Group Method" can only be used if the equation consists of 4 terms.


Example: 20x-5+14x+2


Separate the equation into two:

20x-5 14x+2

20x-5 has a common factor of 5

Therefore: 5(4x-1)

14x+2 has a common factor of 2

Therefore: 2(7x+1)

When you put this all together, it would become 5(4x-1)+2(7x+1)


Perfect Square Trinomial- The form of the perfect square trinomial would be known as a^2+2ab+b^2. To explain "a" and "b" should be able to be square rooted. Additionally, when multiplied by 2, "a" and "b" should be equal to the middle term.


Example: 25x^2+200x+4


This is an example of a perfect square because "a" and "b" have a square root. ( 25 has a square root of 5 and 4 has a square root of 2). Also, when multiplying 25(4)(2) it was equal to 200, which was the middle term. The factored form of this would be (5x+2) (5x+2), which can also be written as (5x+2)^2.

Word Problem (Factored Form)

The height, h in metres, of a basketball 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 basketball?

The initial height of the ball is 21m since it represents the y value in the equation.


b) How long does it take the basketball to reach its maximum height?

h=-(t^2-4t-21) AOS: 7-3/2 The basketball reached its maximum height after 2 seconds.

t-7=0 t=7 AOS: 4/2

t+3=0 t=-3 AOS: 2


c) What was the maximum height of the basketball?

h=-(2)^2+4(2)+21 The maximum height of the rocket is 25m.

h=-4+8+21=25


d) What is the height of the basketball after 3 seconds?

h= (-3)^2+4(3)+21 The height of the rocket after 3 seconds was 24m.

h=-9+12+21

h=24

Standard Form

In this unit so far, we have learned 2 different ways to represent a parabola. Vertex form and factored form. Today, I will be showing you the final way to represent a parabola. This way is called the standard form. Unlike the vertex form and the factored form, the standard form can only show what the y-intercept is. The equation for the standard for is y=ax^2+bx+c , in which c is the y-intercept. Again the value of a represents the direction of opening of the parabola, just like all the other a's in vertex form and factored form. If a is positive, the graph opens upward, if a is negative, the graph opens downward. By plugging in zero into the x values of the equation, c will be left over and it will represent the y intercept.



From Standard To Vertex:

How to do it?

The purpose to go from standard form to vertex form is so that we can determine the vertex of the parabola. With just the standard form, all we can do is determine the y-intercept. To find the vertex form with the standard form equation, we must do a little trick called "completing the square"


Example:

Lets say we have an equation of a parabola which is (y=x^2+12x+32)

For us to convert this standard form to vertex form, we are going to complete the square.

Here are the Steps!


1. Move the c value of the equation to the other side: -32+y = x^2 +12x


2. Now if you look at your right side, the b value is a 12. We must now add a number next to bx so that when we factor the right side, we will get a square. So lets add 36 since factoring the right side will give us (x+6)^2. Also if u add 36 to once side u must add it to the other. :

-32+y+36

=x^2 + 12x + 36


3. Now we simplify!

4+y = (x+6)^2.


4. Now we bring 4 to the right side to isolate y :

y = (x+6)^2 - 4


5. Now we have the vertex form!


Quadratic Formula:

The quadratic formula is a very useful formula for determining the zeros of a parabola.The equation is :


How to use it?

Using this formula is very simple. All you have to do is sub in the values for a, b, c and you will get 2 values which will be your 2 zeros.

Word Problem (Standard Form)

The path of a ball is modelled by the equation y = −x+2x+ 3, where x is the horizontal distance, in metres, from a fence and y is the height, in metres, above the ground.


a) What is the maximum height of the ball?

y=-x^2+2x+3

y=-(x^2-2x)+3

y=-(x^2-2x+1-1)+3

y=-(x^2-2x+1)+3+1

y=-(x-1)^2+4

The maximum height of the ball is 1m.


b) At what horizontal distance does the ball reach its maximum height?

It reaches its maximum height at 4m.