Graphing Vertex Form By Mahimna

Summary of Unit

In this unit we learned

  • 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
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Learning Goals

- I am able to graph if given an equation in vertex form

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

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 modeled by the function h= -4.9(t-2)²+45

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

3.2 Graphing from Vertex Form
This video shows how to graph a quadratic function in vertex form, and also how to find out the stretch and compression of the parabola.


When using vertex form the equation gives you the translations required to graph the equation. The variables of the translations are a,h, and k. A= stretch or compression, H= horizontal translation, and K= vertical translation.


When your graphing an equation in vertex form there is method that consistently shows the right points. It is called the step pattern this pattern is just over 1 up 1 over 2 up 4. If there is an A value then you multiply the A value to 1 then 4 so the pattern will look like over 1 up 1 * A then over 2 up 4 * A.

Factored Form

Summary Of Unit

In this unit we learned how to factor :

  • 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 factor various equations that i have been taught.

- 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

This is video shows how to factor a quadratic equations that trinomials. In the video the person teaching shows how to use the method of decomposition.

Complex Trinomials

another way of factoring complex trinomials is. First multiply the leading coefficient with the last term. The number that is received is the number that you need to find the sum and product of. Next just use grouping to finish the factoring,

Perfect Square

Difference of Square

Finding the vertex and x-intercepts

Another part of factoring is finding the vertex to graph the equation. Finding vertex requires the x-intercepts. After factoring the equation you receive two x intercepts, then you add them and divide by two. Then you have your x value, then you sub in the x value into original equation and solve for the y value.
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This graph was made through using an equation that is in factored form. To graph this equation the x-intercepts would be necessary. The x-intercepts in this graph are -3 and. For the vertex the x value is -0.5 and the y value is -6. The graph also opens up.

Word Problems

1. Rock is thrown down from a cliff that is 180m high. The function h= -5t squared-10t+180 give the approximate height of rock above the water, h= height in meters and t= time in seconds . when the rock reach a ledge that 105 m above the water.



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.


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




The vertex is (1,65)

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

Standard Form

Summary of Unit

In this unit we learn to:

  • 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 achieve vertex form if given an equation in standard form

- 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

Standard form is a way most equations are written in quadratics unit. Standard form looks like ax^2+bx+c. When there is standard form equation different methods from different parts of the quadratics unit can be used solve for a plethora of questions. Standard form is one most frequently used equations in the grade 10 quadratics unit.

Quadratic Formula

The quadratic formula gives x-intercepts if used properly the quadratic formula can be used to achieve vertex form.

Quadratic Formula:

Using the Quadratic Formula

The Quadratic formula will give you two x-intercepts. To get the x value, you need to take the two x intercepts and add them. Then you divide the answer by two. That is your x value, to find the y intercepts you sub the x value into the standard form equation given in the beginning.


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


86.4+6.2+1 y= 93.7

Completing the Square

When your given a standard form equation you have the option of using factoring, quadratic formula and completing the square. This depends on you question. If you need to find the vertex or graph a quadratic equation that is when you use 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





This my video
Completing the Square


The Discriminant is found in the quadratic formula. The Discriminant is d= b squared-4(a)(c)/2(a). The standard form equation will give you the a value, b value, and c value. If d is less than zero then there is no x-intercepts. If d is greater than zero then there are two x-intercepts. If d=0 then there is one x-intercept.

Graphing Standard Form

When you given an equation in standard form and told to graph. You can use the quadratic formula or complete the square. Both ways can achieve the vertex form. The quadratic formula uses x-intercepts. The x-intercepts can easily be turned into the x value. The x value will help you find the y value by subbing the x value into the original equation. For Completing the Square you have to solve and you will receive vertex form. Then you graph the vertex and plot the points.
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This graph shows two parabola that have been graphed by given a quadratic equation in standard form,

Word problems

1). A model rocket is launched from a deck in Jim's backyard. The path followed by the model rocket can be modeled as h=-5t^2+100t+15, where h is height in meters, and t is time in seconds.

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= 306.2+350+15


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




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.


In quadratics I learned a lot of new concepts, but in the beginning of factoring and standard form i didn't understand. throughout the time these concepts were taught i gradually understood how to use factoring, the quadratic formula, completing the square. I did not have difficulty learning it, but it was the application problems that i had some issues with. Overall I have improved, learned, and understood these concepts. Every concept in quadratics is connected with each other. Graphing is connected with factoring because you factor to receive an equation you are able to graph with. Factoring is connected with standard form because you can get the information you need from the equation in standard form and use factoring to get your vertex and x-intercepts. Standard form is connected with graphing because in standard form you learn two methods to achieve vertex form where you can graph the equation. Most questions for standard form is related with graphing. I have learned these techniques to solve a quadratic function but I still have a lot to improve on.
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Mahimna Dave