Fundamentals of Engineering Exam Review Parabolas (Part 1)

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.

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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 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 1) – Fundamentals of Engineering Exam Review

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