Quadratic Functions

[fiery acorn]

Please help me solve this equation with the topic/concepts needed along with listing the formulas later to solving it helping me.

[bold agate]

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[thin needle]

post a picture that can actually be seen w/o having to download anything

[fiery acorn]

its cause its from my phone

is png fine

[keen axle]

screen shot it then

[thin needle]

as long as you post it... and it can be seen w/o having to download it

[fiery acorn]

Should work now

[thin needle]

ok, what have you tried?

[fiery acorn]

Nothing because im so lost. I wrote down the "concepts" that I thought I needed

But idk if i even need them now

[thin needle]

ok... well what are the roots of f in terms of p and/or a?

[fiery acorn]

Does that mean finding the zeroes

[thin needle]

roots and zeros are synonymous, yes

[fiery acorn]

I have not tried that. Would I just set X to 0? I don't remember the formula at all which sucks

[thin needle]

what's the definition of zeroes?

there's no formula, it's just definition and logic

[fiery acorn]

Isn't it related to the y/x intercept

[thin needle]

the zeroes are the x value st f(x)=0.

[keen axle]

x intercept

[thin needle]

as your notes would say

[fiery acorn]

Ok

So would I do

f(x) = a(0-p)(0-3)?

[thin needle]

clearly not

that's plugging in x=0

[thin needle]

.

what does "f(x)=0" look like, given you know explicitly what f(x) is

[fiery acorn]

its setting the equation to be equal to zero isn't it not

[keen axle]

yes

[thin needle]

so do so

[thin needle]

echoes arent needed nor appreciated

[keen axle]

you are simply solving for the roots or solutions

did i ask?

if an explanation does not make sense to the student. You must make it simpler.

Not everyone understands math the way you do.

[thin needle]

Cool, and the courtesy of the server is if someone is already helping, you let them help

[keen axle]

sorry about that @fiery acorn

It is what i was clearly doing

[thin needle]

if they dont understand an explanation, they can simply say that themselves lol

[fiery acorn]

Lmao trust me all good, ill take the help anyways

[keen axle]

he asked a question you made it complicated, i answered his questions.

[fiery acorn]

Ok so

[thin needle]

sure, whatever you say

[fiery acorn]

would it be

[keen axle]

okay bosco

ask your question

[thin needle]

their question was asked ages ago

[keen axle]

the smart guy should be able to help you

he has other questions to ask.

[fiery acorn]

f(x) = a(2.5-p)(2.5-3) = 0

Nah its just i haven't done this in so long

[keen axle]

first lets find the value of p

as it is more easier

[fiery acorn]

Ok

[keen axle]

you are given the axis of symmetry

[fiery acorn]

Correct which is X

[keen axle]

yes

the axis of symmetry’s x coordinate is the vertex to our quadratic

[fiery acorn]

Yeah

[keen axle]

you understand that explanation right

[fiery acorn]

Vertex is the middle point isnt it not

[keen axle]

yes

[fiery acorn]

Yeah

Ok so I would have to plug in 2.5?

[keen axle]

not directly

but we can use that information to find p

[fiery acorn]

Ok

We also have the y-intercept too

[keen axle]

when we have our factored form quadratic

[fiery acorn]

Yeah

[keen axle]

we can find the vertex by adding the roots, p and q and then dividing it by 2

to give us the x coordinate

of the vertex(AOS)

so what do you think we should do

[fiery acorn]

I mean you said the x coordinate of the AOS but AOS x = 2.5

[keen axle]

yes thats the coordinate

[fiery acorn]

Ok but we don't have p and q for the vertex so we'll need to find it

[keen axle]

but we will then have to set it equal to 2.5 and solve for p

we have q

which is 3

(x-3)

[fiery acorn]

Ok

[keen axle]

so we have this formula to find the midpoint of p and q

[fiery acorn]

p + q/2?

[keen axle]

yeah

[fiery acorn]

Oh that

[keen axle]

we have x and q

[fiery acorn]

Whats that formula called

[keen axle]

so now we are solving for p

midpoint formula

[fiery acorn]

Midpoint as in vertex?

[keen axle]

midpoint generally

[fiery acorn]

Is there another term

[keen axle]

its just the the coordinate in between the two values

[fiery acorn]

cause I remember that formula

[thin needle]

the average/mean of p and q

[fiery acorn]

Ok i wrote the formula down

[keen axle]

okay

[fiery acorn]

X would be 2.5

[keen axle]

now just plug in x and q

then solve for p

[fiery acorn]

is it p = 2.5 - 3/2 to solve?

Yeah right?

[keen axle]

no

you would need to multiply the entire equation by two to get rid of the fraction

and then solve for p, then

[fiery acorn]

Yeah I had 2.5 x 2

which is 5

[keen axle]

yeah

and then you get

$p+3=5$

[shell hedge]

ビジヨン

[keen axle]

and you just solve that

[fiery acorn]

Ok so P is 2. But I would have to rearrange that to show. Being 5 - 3 = p which is 2

[keen axle]

yes that’s basically correct

now you have p and q so you have a(x-2)(x-3)

(0,-6) is the y intercept

you have f(x)=a(x-2)(x-3) as your current function right?

[fiery acorn]

Yeah with the p plugged in

[keen axle]

now how would you find a

using the y intercept as information

[fiery acorn]

Don't remember. I had an idea to do f(0) = -6(0-2)0-3) but idk if that would work

[keen axle]

a isnt -6

but we are given (0,-6)

this tells us that when x=0 y is?

[fiery acorn]

-6

[keen axle]

yup

so we have f(0)=?

[fiery acorn]

-6

[keen axle]

yup

now plug in 0 into f(x) and once you solve the right hand side

substitute f(0) with -6

tell me what you get for a

[fiery acorn]

Hm

[keen axle]

what do you have?

$f(0)=a(-2)(-3)$

[shell hedge]

ビジヨン

[fiery acorn]

a should be 6

[keen axle]

you get

$f(0)=6a$

[shell hedge]

ビジヨン

[keen axle]

right?

but remember

$f(0)=-6$

[shell hedge]

ビジヨン

[fiery acorn]

It should be 1 right

[keen axle]

should it?

[fiery acorn]

Wait it was just at 6a

[keen axle]

yeah

[fiery acorn]

how did you isolate it

I thought you did

[keen axle]

but we are told the y intercept is (0,-6)

[fiery acorn]

6/6 = a

[keen axle]

so when x=0 y is -6

and we know f(x) and y are the same thing

[fiery acorn]

Ok

[keen axle]

so replacing f(0) in whichever we end up getting

$6a=-6$

[shell hedge]

ビジヨン

[keen axle]

because we have

$f(0)=6a$ and $f(0)=-6$

[shell hedge]

ビジヨン

[fiery acorn]

a is -1

[keen axle]

yes

you understood right?

the process of how to find p and q?

[fiery acorn]

Yeah

the only thing is

I have to list all the formulas and rules to solve the problem as well

So like I have x = p+q/2

[keen axle]

for the vertex/AOS

[fiery acorn]

Yeah but isn't that not the AOS version

Actually I got it