FE Study Guide

Trig Ratios – Fundamentals of Engineering Exam Review

December 20, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV will reinforce your understanding of the Trig Ratios. We will begin by defining SOH CAH TOA and then run through a few example problems using these ratios.

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Right Triangles – Fundamentals of Engineering Exam Review

December 20, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV will reinforce your understanding of the Right Triangle. We will begin by defining the common right triangles and then run through a few example problems.

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Hyperbola – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV will reinforce your understanding of the Hyperbola. We will begin by defining the general equation for a Hyperbola and then run through a few example problems.

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Ellipse – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Ellipses.

We will begin by defining the general equation for an ellipse and then run through a few example problems.

This Fundamentals of Engineering Exam Review of Ellipses is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Ellipses.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Ellipses by reviewing the General Formula.

The general equation of an ellipse that takes the form of X minus H squared

divided by a squared plus y minus K

squared divided by B squared is equal to

1 now please note that in order for the

ellipse to be in the standard form that

the right side must equal 1 the point H

K on the other hand as it was in all the

other general forms such as the circle

on the parabola that represents this

center of the ellipse now a is the the

square root of the term that is below

the X term in the general form equation

and a is the distance that the that the

ellipse moves from the center right and

left so if we have the term say we have

the center is located at H K then the

rightmost point on the graph is going to

be H plus a ok

and the left most point this is the

right point and the leftmost point is

going to be H minus a ok so that’s just

how far the ellipse extends out from the

center to left and right now the the B

term on the other hand is the square

root of the term that is below the Y

term in the general equation and the B

term represents how much the lip

stretches in the up-and-down direction

now once again if we’re given HK is the

center then the top most the top most

point

is going to be h k+ b and the bottom

most point is going to be equal to H K

minus B so if we’re given these terms

the a square terms for given the a

squared term in the B squared term if

we’re given H and K then that’s all we

really need to graph the ellipse and

determine all the points on the ellipse

so let’s quickly look at an example here

let’s say that we were given an equation

we’re given in eclipse with the general

in the general form of X plus 2 squared

divided by 9 plus y minus 4 squared

divided by 25 is equal to 1 so what we

want to do is find the rightmost the

left-most the top and the bottom most

points of this ellipse so we’re lucky

because this is just in the standard

form we easily can see what we’re H and

K is or where the center is and in this

case H is going to be equal to negative

2 and K is going to be equal to 4 so the

center of the Klipsch HK is located at

negative 2 4 and so what we want to do

now is determine what a and B is now we

know this term at the bottom below the X

term is a squared and we know this term

below the Y term is B squared and so all

we need to do is take the square roots

to find out that a is going to equal to

3 and B is going to be equal to 5 so

let’s go ahead and write this down at

the right point we want to find the left

point we want to find the top point and

the bottom point of this ellipse

so all we need to do is know that the

right point is going to be H plus a okay

the left point is going to be H minus a

plus K the top point is going to be H K

plus B and the bottom is going to be H K

minus B so all we need to do is plug in

our information and we find that H plus

a H is negative 2 a is 3 we’re going to

have 1 and we know K is 4 so that the

right point is going to be 1/4 we have h

minus a H again is negative 2 and a is 3

so we got negative 2 minus 3 we’re going

to get negative 5/4 and then now we got

to find the top and now we’re dealing

with the B term and we know that H is

going to be negative 2 and that K is 4

and we want to add 5 to that so it’s

going to be negative 2 9 and now we want

the bottom turn once again H is going to

be negative 2 and we got B is equal to 5

and we – that from K so we get negative

1 we get that the rightmost point is 1

for the leftmost point is negative 5 4

we got the top point at negative 2 9 and

we have the bottom point at negative 2

negative 1 so that’s it guys just a

quick review on how we can determine

some of the information if given an

equation of ellipse if we’re asked to

find Center if we’re asked to find some

certain point on that ellipse we know

the general equation and we know how to

find that information now so if you guys

have any more questions don’t hesitate

to visit the site at engineering

training exam comm where I continue to

put up more information more content to

help you guys out I could definitely use

suggestions though I

I know that you know I’ve been through

this process but the only way I can help

you is if you tell me what you guys need

you know I know I know it’s hard

sometimes to you know just give you guys

the information that you need and that’s

all I want to do so head on over there

and contact me through the contact page

or go ahead and sign up for my free EIT

preparation boot camp and you can

contact me through that mailer

I’m always responding to emails every

day to every single question I get so

don’t hesitate to do that all right guys

so take care tonight and we’ll talk soon

all right

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Circles – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Circles.

He will first start with defining the general equation of a circle and how it relates to the distance Distance formula, then run through a few example problems.

This Fundamentals of Engineering Exam Review of Circles is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Circles.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Circles by revisiting the Distance Formula.

Let’s take at the point say X Y and x1 y1 the

distance formula tells us that the

distance between these two points D D is

equal to x1 minus x squared plus y 1

minus y squared all under square root

now a circle or the equation in the

circle is just a reiteration of the

distance formula all the way around it

is telling us that each point on the

outside of a circle is a fixed distance

R from a central point H K so let’s call

X Y here our central point HK let’s call

our D our set radius so D is equal to R

and we can rewrite this equation to say

that are the fixed radius distance

between H K and any point on a circle is

actually at x1 minus H squared plus y1

minus K squared all under the square

root to see this in the standard form or

in a form that were more used to seeing

all we need to do is square both sides

so by doing that we get R squared is

equal to let’s get rid of the x1 and y1

let’s just call them XY representing any

point XY on the outside of a circle so R

squared is equal

to X minus H squared plus y minus K

squared so this right here is going to

be our lovely equation standard form

equation for circle so this is the

standard form in this form we very

quickly can see what the radius is and

we can also see what the center of the

circle is which is going to be located

at HK whatever those two numbers are so

let’s look at a quick example here and

just illustrate how we can go about

defining the equation of a circle say

we’re given a circle with the radius of

8 and let’s say that we’re told that the

center of that circle is located at

negative 4 7 so let’s just recall that

this term negative 4 7 the point

negative 4 7 is our HK in our standard

equation and this of course is our R so

once again restating the standard form

it’s R squared is equal to X minus H

squared plus y minus K squared and this

is very simple all we need to do is plug

in R all the given information we got 8

squared is equal to X minus negative 4

squared plus y minus 7 squared so this

ultimately plays out to be let’s see X

plus 4 squared plus y minus 7 squared is

equal to 64 so this is an equation

that’s telling us that the circle is

located at negative 4

and has a radius of a square root of 64

which is 8 so let’s look at another

example here let’s say on the other hand

were given the equation of a circle and

we’re asked to find the center and also

the radius of that circle let’s say that

we’re given the equation let’s go with x

squared plus y squared plus 8x plus 7

now when we first glance at this

equation we obviously hope that it’s

going to over already be in the standard

form so we can quickly just pick out the

necessary information that we need which

is the radius and the center of the

circle but in this case obviously it is

not so we do have a little work to do so

the first step or the first 2 steps is

to first get the constant on the right

side of the equation and also group the

terms together group the X and the y

terms together so to do that we get x

squared plus 8x plus y squared is equal

to negative 7 so the next step is for

each variable with two terms we need to

complete the square of those terms and

we look on the left side here and see

that X is the only term that or the only

the only variable that has two terms so

that is the only variable that we need

to complete the square of so let’s

recall to complete the square what we do

is we take half the coefficient of x and

square it so we’ll go 8 divided by 2

squared which is equal to 4 squared

which is equal to 16 so now what we do

is add that we add this term to both

sides of the equation

so we get x squared plus 8x

plus 16 plus y squared is equal to

negative 7 plus 16 so that what that is

if we just lets see now we can go ahead

on this right side and just do this

calculation real quick we got 9 and on

the left side here now we can factor the

X term such that it’s equal to X plus 4

squared plus y squared which is also

equal to Y minus 0 squared is equal to 9

so now we have our equation the same

equation here now we have it in the

standard form of a circle and we can

quickly determine what HK is and we can

also determine what the radius is so

looking at this we know that HK the

center of the circle is going to be

located at negative 4 0 and that the

radius of the circle is the square root

of 9 which is also equal to 3 so the

answer to this this this problem is if

we’re asked given the N equation we’re

asked to determine the center as well as

the radius we can just simply put it

into the standard form pick out the

necessary information and answer the

question so that’s it guys that’s all I

got for you today quick recap of circles

if you guys have any more questions or

suggestions or any other tutorials you

want to see go ahead and visit my site

at engineer and training exam com either

subscribe and get the free AIT

preparation bootcamp that I got or go

ahead and contact me through the contact

form either or to shoot me your

suggestions your feedback just let me

know how everything’s going here if you

guys are getting what you need my

concern is that you guys are at least

getting a little bit of value out of

each one of these videos and you know

that’s what I aim to do is just help you

guys out in your preparation so anyways

that’s it

for today you guys take care and we’ll

be talking soon all right bye

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Distance Formula – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Distance Formula.

He will start with defining the Pythagorean Theorem and how the Distance formula relates, then finish off with a quick example of how to find the distance between any two given points.

This Fundamentals of Engineering Exam Review of Distance Formula is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Distance Formula.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Distance Formula by defining exactly what the Distance Formula is.

So what is the distance formula?

Well essentially
the distance formula is a restatement of
the Pythagorean theorem packaged up into
a simple equation that will allow us to
find the distance between two given
points so let’s consider the two points
negative 2 1 and 1 5 now it’s the
standard process to find the distance to
the of these two points we can simply
graph them and then draw a straight line
between them and then develop a triangle
as such would this be inside a this side
be inside being the hypotenuse which is
the distance between the two points be
inside C so graphically once again it’s
easy to find what C is just using the
Pythagorean theorem which is a squared
plus B squared is equal to C squared so
using this process we can simply find
out that if this is a point 1 5 and this
is point negative 2 1 that we can see
that a is going to be 3 and B is going
to be 4 so if we just plug that into the
Pythagorean theorem here we find that 3
squared plus 4 squared is equal to C
squared and just calculating that out we
find that it is equal to 9 plus 16 which
is equal to 25 which is equal to C
squared and we find that the hypotenuse
or the distance between these two points
is the square root of 25 which is equal
to
five so this this method obviously holds
true in any case however the the
distance formula
like I said repackages this formula into
a simple a simple equation that we can
just quickly plug in any two given
points in a coordinate system to figure
out what the distance is between those
two points
so the distance formula states that D is
equal to the square root of x2 minus x1
squared plus y squared and all that is
under the square root now this is of
course if we’re given x1 y1 and x2 y2 so
all we need to do is plug this given
information into the distance formula
and we will be able to determine the
distance between those two points so
let’s look at a quick example here let’s
say that we’re given the points negative
3 negative 2 and we’re given the point 5
2 now we can go ahead and graph this
make a triangle determine the sides a
and B and solve for the hypotenuse C or
we can go ahead and just take our
distance formula and plug in the
information that we are given which we
will do in this case so D is equal to
the square root of x2 which in this case
is 5 minus x1 which is negative 3 all
that squared plus y2 which is 2 minus y1
which is negative 2 all squared so
working that out we get we get the
square root of 8 squared plus 4 squared
which comes down to the square root of
80 and just thrown that into our
calculator real quick
the distance between these two given
points is eight point nine four so
that’s essentially it guys not much to
it it just once again is a reach the
distance formula is once again a
restatement of the Pythagorean theorem
it just allows us a quick concise way to
plug in two given points and define real
quickly what the distance is between
those two points so if you guys have any
questions go ahead and visit my site at
engineer in training exam comm check out
the other videos I got posted there as
well as here on YouTube leave comments
suggestions you know email contact me
I’m always open to help you guys out in
any way that I’m able once again
everything’s free for you guys to use so
check it out and look forward to talking
to you guys soon all right take care

—

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Parabolas (Part 1) – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Parabolas. This is Part 1 of 2

In this tutorial, we will reinforce your understanding of Parabolas, begining by defining the two general equations of a parabola and then running through a few example problems.

This Fundamentals of Engineering Exam Review of Parabolas is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Parabolas.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Parabolas by defining exactly a Parabola is.

let’s recall that the general form of a quadratic equation f of X is equal to ax squared plus BX plus C when we throw this equation up

onto a graph the output we get is called

a parabola now these are two separate

these would be two separate equations

that this quadratic equation will give

us one of these two so there’s a couple

a couple pieces of information that we

should define right off the bat of a

parabola first of all a parabola has

what we call line of symmetry and what

this means is the graph on either side

of this axis of symmetry is a mirror

image of the other side so there’s a

it’s a didn’t exact exact replica we can

basically just cut it off right here

flip it over and determine what all the

points on the opposite side would be so

that is called the line or axis of

symmetry now when you go and when you go

and graph a parabola you won’t see this

line it’s not a it’s not a it’s not a

point on the graph it’s just the

location or it’s a it’s an invisible

line that just it basically cuts our

parabola in half now the second thing we

need to define is either the top the

peak of the parabola top or bob berry

bottom of the parabola is called the

vertex so there’s two forms two forms of

equations that will give us a parabola

and one of the forms is easier to graph

than the other easier to extract

information from than the other however

the second form that we will be going

over isn’t it’s more common

there is a little bit more work to be

done if we’re asked to find certain

information like the vertex and and and

say the other pieces of information but

not much more work to be done but let’s

first go over the the first form and

that form i want to go over is f of x is

equal to a X minus H squared plus K now

this if we plug in whatever information

we want to plug in here will give us a

graph of a parabola and like I said this

is very simple in determining a couple

keep key characteristics of the parabola

number one depending on the sign of a

here will tell us whether or not the

parabola opens upward or if it opens

downward now if a is negative then just

going open downward if it’s positive

then it’s getting open upward the second

piece of information that we can quickly

extract from this equation is H and K

now the point H K will be the vertex of

our parabola so that’s why just really

quickly looking at this form why it

would be easier because usually we’re

typically asked to graph graph parabolas

or at least define the equation to a

parabola based off a given graph and

with this form we can quickly the graph

it or we can quickly define it so let’s

look at a quick equate example of how we

can quickly define the characteristics

of a parabola let’s say that we’re given

the equation G X is equal to negative X

minus 2 squared minus 1 so Rick real

quick quickly we see that a is equal to

negative 1 and since it’s negative we

know the orientation of this of this

parabola is it’s going to be opening

downward next we could possibly asked

where the vertex is and real quickly

once again all we need to do is define

H&K; is and luckily this the equation is

put in the form that we like and real

quickly we can see that HK or the vertex

is going to be located at 2 negative 1

and next say we’re asked to find the x

and y-intercepts well just by looking at

where the vertex is and knowing that the

orientation is downward we can quickly

just visually I mean probably without

graphing it we can see that the vertex

is going to be at 2 negative 1 and it’s

going to be opening downward so this is

telling us that there’s not going to be

any x-intercepts but there will be a

y-intercept so they turn to determine

that we need to plug into the equation X

is equal to 0 and just solve where that

is going to where the y-intercept is

going to be so that’s fairly easy just

plug in 0 wherever you see X and we got

0 minus 2 squared minus 1 which is equal

to minus negative 2 squared minus 1

which is equal to minus 4 minus 1 minus

5 so now by looking at that we know that

the y-intercept is going to be located

at 0 negative 5 and there’s not going to

be x intercept so like I said it was

fairly simple when we’re given an

equation in this form we can quickly

find the orientation we can quickly find

the vertex and also we can quickly

define where the intercepts are going to

be so now let’s take a look at the

not-so-friendly form not necessarily

harder just just slightly a little bit

more work to be done to determine the

characteristics of the parabola so that

is also called the general form the 4

that were more or used to seeing as the

quadratic equation f of X is equal to ax

squared plus BX plus C now in this form

the sign a is it means the same thing if

a is positive the parabola is getting

open upwards if it’s negative it’s going

to be opening downwards so that that’s

still the same but unlike the previous

form we’re not quickly given the vertex

but there’s a set there’s a set group of

equations that will quickly give us H

and K so if we want to find H and K all

we need to do is the define a and B and

plug them into this equation so H is

going to equal negative B divided by 2a

and then to get K all we do is plug the

value of this B divided by 2 a negative

B divided by 2a into the function and

that will give us our K so to get the

vertex all we need to do is compute the

x-coordinate from a and B and then plug

that into the function to get the

y-coordinate now the y-intercept is easy

all we need to do is plug in 0 to this

equation and the O which would look we

can really quickly show you it’s going

to basically be equal to C so the

y-intercept is going to be 0 C so let’s

take a look at the not-so-friendly form

and an example of the not-so-friendly

form if you will of a parabola of an

equation where we’re asked to also find

the same information as we were asked

before we want to find the orientation

we want to also find the vertex and also

the intercepts of the equation so let’s

say that we’re given the equation f of X

is equal to negative x squared plus 8x

so right off the bat we’re bummed

because it’s not in the form that we

want to see however we know now after

defining this second form or the

quadratic equation that just with a

little bit more work we can find that

same information so right off the bat

let’s define a B and C so a in this case

it’s going to be negative 1 B is going

to be 8 and C is going to be 0 so like I

defined right prior to this we know that

the the y-intercept is going to be 0 C C

is 0 in this case so we already know

that the y-intercept is at 0 0 so we

knocked that out of the way next we can

look at a we can look at the sine of a

and determine if it’s going to be a

parabola open upward or open down where

in this case it’s a negative 1 and we

know that the orientation of the

parabola means it’s going to be downward

opening downward so the next step is to

is to figure out where the vertex is

like I mentioned before to find H K all

we need to do is to find what a and B is

Devine define what the x-coordinate is

and plug that x-coordinate into the

function to determine the y-coordinate

so once again I can restate that as

negative B over 2a is going to be H and

then we want to plug that value into the

function to determine the K so let’s

quickly do this the x-coordinate or H is

going to be equal to 8 negative 8 over 2

times negative 1 which is equal to 4

now we want to plug 4 into the function

and we find that that’s negative 4

squared plus 8 times 4 and we find that

K is now going to be equal to 16 so that

tells us that the vertex now the vertex

is at 4 16 so let’s just quickly graph

this just roughly and look at look at

just see if we’re gonna have any

x-intercepts we know we have a

y-intercept and that’s at 0 0 and we

also know that the vertex is at 4 16 so

let’s just say 4 is right here and let’s

just say 16 is way up here and that we

have our vertex right there it’s going

to open downward and we know our

y-intercept is at 0 so it’s going to be

something like that and also something

like that and since it’s open downward

and the vertex is above the x-axis we

know we’re going to have to two

x-intercepts so let’s go ahead and find

those all we need to do is set the

function equal to zero so we got 0 is

equal to negative x squared plus 8

factored that and we got X minus X plus

8 and to make this statement true we

find that X needs to equal 0 or X needs

to equal 8 so that’s telling us that our

x-intercepts our x-intercepts are going

to be at 0 0 and 8 0

End of Part 1 Fundamentals of Engineering Exam Review of Parabolas

—

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Parabolas (Part 2) – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Parabolas.

This is Part 2 of 2

In this tutorial, we will reinforce your understanding of Parabolas, begining by defining the two general equations of a parabola and then running through a few example problems.

This Fundamentals of Engineering Exam Review of Parabolas is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

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Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Parabolas.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Parabolas by picking up where we left off in part 1.

An equation in our unformed friendly
form and I’ve kind of just adopted that
turn but we give it term but we’re
giving it in the second form and we want
to actually convert it into the the
nicer of the two forms the first form
that we went over well we can actually
do that and let’s go ahead and go
through an example of how you would go
about doing that
let’s say that we’re given a function f
of X is equal to Nu equal to 2x squared
minus 12x plus 3 so this is the
unfriendly form or the quadratic
equation and we want it to be in the
form of f of X is equal to a X minus H
squared plus K and we want it to be in
this form so we can quickly define the
information that we need to define about
the parabola like I said the orientation
the vertex and we also want to quickly
define where the intercepts are so all
we need to do here is essentially just
complete the square however it’s going
to be slightly different than what we
went over in the past tutorials but
essentially it’s going to be it’s the
same concept that the end result is the
same so the first thing we want to do is
make sure that the x squared term has a
coefficient of 1 and so to do that we
need to factor out 2 in this case so we
got 2 is equal to X or multiplied by x
squared minus 6x plus 3 over 2 and now
the next now this is where the process
gets kind of tricky and like I said
we’re going to do we’re going to do the
the process of completing the square
here but we’re going to do a slightly
different and instead of adding the
result to both sides we’re going to add
the result and minus the result on the
same side so let’s do this to complete
the square you take the coefficient of 6
divided by 2
and square it so in this quick case
negative six we get negative three
squared which is going to equal nine so
nine is our is our constant that we now
need to add and subtract from the same
side so let me illustrate this f of X is
equal to two x squared minus 6x plus 9
minus 9 plus three half now you say this
this is net zero nine minus nine is net
zero so it doesn’t affect our equation
or the integrity of our equation it just
allows us to factor this X term right
here so let’s go ahead and do that let’s
factor we got f of X is equal to two and
now we could factor X minus three
squared minus 15 over 2 15 over twos all
I do is convert nine into a fraction
which is going to be 18 over 2 minus 18
over two plus 3 over 2 is minus 15 over
two so next all we need to do is
multiply 2 back through and we got f of
X is equal to two X minus three squared
minus 15 and now check this out we took
our not-so-friendly form and we
converted it into our friendly form and
now we can quickly quickly determine a
lot of things let’s let’s the
orientation is going to be what if you
said a is equal to two so it’s a
positive two so it’s going to be opening
upward and now we also can really
quickly define H K which will be the
last thing we do H K in this instance is
going to be three negative 15
so just just like that we were able to
convert by completing the square and
converting the not-so-friendly form into
our friendly form of the of a parabola
and determine a couple aspects of its of
its orientation so anyways that’s it
that’s a quick recap of parabolas and
the equations that revolve around them
and some of the possible questions that
you will see on the exam so if you guys
have any more questions or want to
review any more of the videos go ahead
and check out engineering training exam
com I have all the videos there as well
as here on my YouTube channel and also
uh don’t don’t hesitate to leave me a
little feedback some some suggestions if
you guys are looking for certain
tutorials or whatnot like I’ve stated in
previous tutorials I’m just here to help
you guys out you know I just want to
give you guys some value some free value
and so you guys can go ahead and prepare
and you know do things a little bit
differently and then hitting the books
you know sometimes these videos help out
sometimes they don’t just let me know if
they are and I’ll be happy to continue
to create them so anyways that’s it for
now guys you guys take care have a good
night and we’ll be talking soon

—

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Parabolas (Part 1) – Fundamentals of Engineering Exam Review

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Parabolas (Part 2) – Fundamentals of Engineering Exam Review

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