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Mathematics 8 Online
OpenStudy (sammieell):

I need help with Algebra 2A IM ABOUT TO FAIL THIS CLASSSSS!!!!!!!!

OpenStudy (sammieell):

OpenStudy (sammieell):

i just need someone to walk me through this

OpenStudy (baby456):

when is it due because i am Algebra 2a right now

OpenStudy (sammieell):

it's been due im just behind

OpenStudy (baby456):

let me see because i never did anyprojects in my class just test

OpenStudy (baby456):

uh i get help tyou with the last part

OpenStudy (baby456):

uh which part is confusing you

OpenStudy (sammieell):

all of it

OpenStudy (baby456):

ok so for the last part i will give you examples|dw:1453084137774:dw|

OpenStudy (sammieell):

what is that

OpenStudy (baby456):

roots tutorial.math.lamar.edu/Classes/Alg/ZeroesOfPolynomials.aspx

OpenStudy (baby456):

you should know what vertex is

OpenStudy (coconutjj):

Let's talk about the PDF question.

OpenStudy (coconutjj):

you are given 3 coordinates

OpenStudy (coconutjj):

and what you basically need is a formula

OpenStudy (coconutjj):

the quadratic equation is in the form ax^2 + bx + c

OpenStudy (sammieell):

i dont know how to set it up

OpenStudy (coconutjj):

well every x coordinate we plug into the equation we should get the corresponding y

OpenStudy (baby456):

vertex is usully 0,0 i would help but these are resources i use and if i help toom much i would confuse me because we worked on zeroes not roots just use your scientific caculator and liik online.

OpenStudy (baby456):

or graphing caculator

OpenStudy (sammieell):

yes i have it up

OpenStudy (baby456):

or online graphing caculaator

OpenStudy (coconutjj):

thus for (2.6, 7.9): f(2.6) = a(2.6)^2 + 2.6b + c = 7.9

OpenStudy (coconutjj):

now you tell me the equation for the next 2 coords

OpenStudy (sammieell):

f(4.8)=a(4.8)^2+4.8b+c=12.4 f(9.7)=a(9.7)^2+9.7b+c=15.1

OpenStudy (coconutjj):

yes very well done... now what should we do with these 3 equations to find a,b and c

OpenStudy (sammieell):

i dont know walk me through

OpenStudy (coconutjj):

It looks like a system of equations right. You have all the information you need

OpenStudy (coconutjj):

do you know how to solve 3 equations, 3 unknowns?

OpenStudy (sammieell):

no

OpenStudy (coconutjj):

How about 2 equations, 2 unknowns?

OpenStudy (sammieell):

no

OpenStudy (coconutjj):

oh.. okay.. let's start with something simpler.. y=2x-1 y=4x+3

OpenStudy (coconutjj):

we have 2 strategies: 1. Elimination 2. Substitution

OpenStudy (coconutjj):

we can use both, however one is always easier than the other. depends on the equation

OpenStudy (sammieell):

Substitution

OpenStudy (coconutjj):

yes.. since y is equal to 2x-1 and 4x+3, we can set them equal to each other

OpenStudy (coconutjj):

solve for x and then plug x back in to any of the two equations to find y

OpenStudy (coconutjj):

2x-1 = 4x+3 solving for x, we get -2

OpenStudy (sammieell):

y=−5, x=−2

OpenStudy (coconutjj):

yes very well. in elimination we try and cancel one term by adding or subtraction

OpenStudy (coconutjj):

thus these two terms must have equal coeficcients

OpenStudy (coconutjj):

positive or negative does not matter. same magnitude

OpenStudy (coconutjj):

we 2x-1=y -) 4x+3=y -------------------- -2x-4=0 x=-2

OpenStudy (coconutjj):

ta da.! now in 3 equation 3 unknown, we use the same strategy: EQUATION 1 EQUATION 2 EQUATION 3 1. EQUATION 1 +/- EQUATION 2 = EQUATION A 2. EQUATION 1 +/- EQUATION 3 = EQUATION B 3. EQUATION A +/- EQUATION B = a answer for 1 of the variables

OpenStudy (coconutjj):

do you think you can apply this to the question?

OpenStudy (sammieell):

can you teach how to set it up

OpenStudy (coconutjj):

a(4.8)^2+4.8b+c=12.4 a(9.7)^2+9.7b+c=15.1 a(2.6)^2 + 2.6b + c = 7.9

OpenStudy (coconutjj):

try using elimination method, it will be easier in computation

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