Everything you need to know!
Part One: Graphing Vertex form
1. Parts of a Parabola
2. Vertex form
4. Step Patterning
5. Mapping Notation
6. Solving an equation with limited information
7. Working on Word problem
at the end of this section, there will be a video addressing the concept of transformations more in depth
Lets get started!
1. Parts of a Parabola:
AXIS OF SYMMETRY: The vertical line through the vertex, divides the parabola into two mirrored images
ROOTS/ZEROS: The point/points in which a parabola goes through the x-axis of the graph
VERTEX: The point on the graph where the parabola changes direction from positive to negative or negative to positive
MAXIMUM VALUE: The highest point in which a parabola can be graphed before descending
MINIMUM VALUE: The lowest point in which a parabola can be graphed before ascending back to the starting point
OPTIMAL VALUE: Either the highest or lowest point graphed in a parabola, it is not possible for a parabola to have both a maximum and minimum value,
2. Vertex Form:
the vertex form of a quadratic equation is as followed:
when graphing using vertex form, always remember:
the Vertex is of the parabola is (h,k)
the Axis of symmetry is (x=h)
And the Optimal value is (y=k)
based on the vertex form above, I will explain how transformations affect the values a, h, and k
a tells you the direction of opening and compression/stretch
H represents the horizontal translation of the parabola (meaning which direction the parabola will shift depending on the whether the value of h is negative or positive, or how far the graph will of translated based on the value of h)
The value of H in the vertex form equation will be flipped when graphing. for example, lets say that is the value of h in vertex form is -6, when graphed, it will be placed at positive 6
K represents the vertical translation (meaning whether the parabola will be placed higher of lower of the graph, depending on whether the value of K is positive or negative, K represents how many unit up or down the graph is placed, depending on how large the value of is)
4. Graphing using a Step Pattern
One method of graphing a parabola is known the step method. To explain this method, we will be the following example equation: 2(x -1)2 + 5
first, you want to find the vertex of the equation, in this case, knowing that (h,k) equals to the vertex of a parabola, we already know that our vertex is (1,5)
Secondly, whatever number is in front of your x value outside of your brackets (your a value) , you multiple by a step pattern of 1, 3, 5, 7 and so on
So in this case:
Thirdly, once you have multiplied your a value by this pattern, you can start graphing!
When you are graphing, you want to always move one unit on the horizontal axis and shift either up or down based on the product of multiplying your a value and step pattern on your vertical axis (each of the products is a point in the parabola)
Using the example above, after plotting your vertex you would move once to the right on the x axis and 2 spaces up on the vertical axis, you do this to the other side of the parabola and continue this process until you have the desired amount of plot point on your graph
Another popular method of graphing parabolas is known as mapping notation. We will be using the equation: 2(x -3)2 + 1, to demonstrate how to apply and utilise mapping notation:
Before we begin, we must create a table of value for our base parabola: y=x^2
-1 10 0
First, we need to convert the equation in the vertex form above into the mapping notation formula, which is as follows:
Knowing the information we already know based on the equation in vertex form, we can sub in the known values:
Secondly, you must create a second table of values to for your new mapping notation fomula.
In order to fill in the values for x in this table, you must add the a value of your mapping notation and vertex form equation with the corresponding x value of the previous table, the sums will be your x values of the new table:
To calculate the y values of your new table, you must sub in the y values of the previous table into the mapping notation formula 2y+1, the answers to the formula will be your y values:
Here is our new table:
Thirdly, once you have completed both tables, you can now graph. You want to begin by plotting the vertex and then multiplying the value of a by our step pattern: 1, 3, 5 , 7
The products of the equations will be your plotting points. Now all you have to do is plot the graph like you would using a step pattern! ( move once on the horizontal axis, and up or down how many times based on your step pattern)
How to solve an equation with Limited information provided
When limited information is provided, you can still find the y intercept!
to find the y-intercept, simply set x as 0 and solve for y
to solve the equation, set y as 0 and solve for x or can you expand and then simplify to get the standard form, once you have done that, then use the quadratic formula
7. Word problem:
we will be solving the following word problem:
At a baseball game, a fan throws a baseball from the stadium back onto the field. The height in metres of a ball t seconds after being thrown is modelled by the function h=-4.9(t-2)^2+45
a) What is the maximum height of the ball?
The maximum height of the ball is 45 m. I know this because the k value of the vertex form represents how high up the ball is going to reach before it descends because it is the highest point of the graph
b) When did the maximum height occur?
The maximum height occurs at two seconds. I know this because the vertex (h,k) indicates where on the graph the ball reaches its maximum height on the x axis, in this case, it was at two seconds
c) What is the height of the ball after 1 second?
To calculate this, you would have to sub in the number 1 into the value of t in the equation and solve:
The height of the ball after one second is 40.1 meters
d) What is the initial height of the ball?
To calculate this, sub 0 for t and solve
The initial height of the ball is 25.4 meters
Graphing using transformations:
Part two: Factored form
1. Some brief learning goals
2. expanding and simplifying an expression in factored form
3. different types of factoring
4. how to solve for zeros
5. how to find the vertex
6. solving word problems
2. understand how to expand and simplify equations in factored form
3. understand how to solve for the x-intercepts, as well as the vertex
2.How to expand and simplify an expression given in factored form
multiply the terms in this order:
one you have done that, add the two middle terms together, your final answer should be a trinomial
if there are any number outside of the brackets, hold off on multiplying them by anything until the final step, then simply multiply each term by that number.
3. Types of factoring:
Factoring by grouping: similar to common factoring. Instead of factoring everything into one large answer, we group together terms and follow the same steps as common factoring.
Simple factoring: a method of factoring used when the value of a is equal to 1. In order to simple factor, you must find two numbers that when multiplied equal to the value of B, and when added to K. Once you have found your two terms, sub them into the equation (x-r) (x-s). (replace the r and s values)
Factored form: (x+12)(x+8)
complex factoring: Somewhat similar to simple factoring, complex factoring is used when the value of a is equaled to a value larger than 1. In order to use this method, you must first multiply the k and together, the product is what you will be using to find your two terms. now, find two terms that when multiplied equal to your new product, but when added to b.
After this, rewrite your b value as these two terms and factor by grouping to find the factored form
Difference of squares: when factoring a difference of squares, you must square the second and first term and sub it into the expression of (x+r) (x-s)
x² - 9=(x + 3)(x - 3)
Perfect square: when factoring a perfect square, simply square the b them into the expression of (x+r) (x-s)
x2 + 6x + 9
= (x+3) (x-3)
to check if your solution is correct, sub your values into the expression 2ab. if your factored form is correct, the product should equal to your k value
4. How to solve for Zeros
5.How to find the vertex using factored form
6. word problem
a) What are the zeros?
b) Are both zeros valid? Explain what they mean for this problem.
No, both zeros are not valid because in this scenario, the value of x represents time. because time can't be negative, only the positive 4 is valid.
c) When does the disk reach its maximum height?
-4+2 divided by 2= 1
the disk reached the maximum height at one second.
d) What is the maximum height reached by the disk? A-2
the disk reached maximum height at 18 meters.
e) write the equation in standard form
Part Three: Standard form
1. learning goals
2. A quick summary of the unit
3. Completing the square
4. the discriminant
5. Quadratic formula
6. Word problems
7. My Reflection
Learning goals & Quick summary
Use the quadratic equation to find the zeros
Learn how to complete the square
Apply the quadratic equation to various word problems
Summary of the unit:
-the value of a gives you the shape and direction of opening
- the value of c is the y-intercept
- Complete the square to get vertex form
- Use quad form for X intercepts
Completing the square:
Completing the square is a method used to change an equation in standard form to vertex form in order to be able to find the vertex
(to go from standard to vertex form in order the find the vertex)
1. firstly, factor out the coefficient of the x-squared term, leave the c term alone for now (don't factor it along with the b term, leave it outside the brackets)
2. divide the b term by two and then square it; after you have done that, add and subtract your answer to keep equation the same (you cannot change the equation)
3. after that, kick the subtracted term out (to put it with c later on) by multiplying by the term outside of brackets
4. once you have done that, apply the perfect square rules to the values in the brackets, the rewrite it in vertex form.
5. square root of first and last term
Now you have successfully completed the square!
the expression inside the square root is : b^2 -4ac
1. sub in values into the expression
2. Solve the expression
3. determine how many solutions that equation has
An easy way of remembering how many solutions an equation has is knowing that:
if the discriminant equals 0, it has one solution
If the discriminant equals a number greater than zero , it has two solutions
If the discriminant equals a number lower than zero (negative), it has no solutions
The Quadratic formula:
1. sub in the values from the equation in standard into the quad formula
2. focus on the solving for the discriminant first (value inside the square root after the expression has been solved), we will worry about the negative b value and the plus/minus symbol later on
3. once you have solved the discriminant, square root it , make sure to only continue with the rest of the equation if your answer is greater than zero or equaled to zero. ( if your answer is a negative integer that means that there are no x-intercepts)
4. separate the formula into two separate parts now. ( this is where the plus/minus symbol next to the negative b value comes into play)
have one expression be the negative b value + the discriminant square rooted divided by (a value multiplied by two)
have the other expression be the negative b value - the discriminant square rooted divided by (a value multiplied by two)
5. solve the two expressions to find your x-intercepts!
Video with example
The sum of the squares of two consecutive even integers is 452. Find the integers.
let x rep the first even integer
let (x+2) rep the second integer
1. expand the square, then simplyfy and rearrange the equation
2. use your new equation in standard form to sub into the quadractic formula, the two zeros will be your intergers!
For the most part, I enjoyed the unit, even though I struggled with certain concepts from time to time. Math is by far my weakest subject, so this outcome was expected. I discover that my favorite unit of study was the last one regarding standard formula. I also enjoyed working with word problems because found them challenging and once would get an answer right, had a feeling of satisfaction. The area I has issues with the most was the unit on factored form, however, through practice, slowly improved my algebra skills.
- you can solve for the x-intercepts using either factored form or the quadratic equation
- you can complete the square using the skills learned in the factored form unit
- you can convert an equation from standard form to vertex form to help graph